Geometry PreView: Continued
Polygons: 4-5
Polygon - closed figure with at least 3 sides- no curves - no overlapping lines
Named by the number of sides (called laterals) or their number of angles
Polygon literally means MANY (poly) ANGLES (gon)
Quadrilaterals - literally means 4 laterals (sides) and angles - 360 degrees (2 triangles!)
Trapezoids - Only 1 set of parallel lines
Kite - no parallel lines - 2 adjacent sides are congruent
Parallelogram - Opposite sides parallel and congruent
Types of parallelograms:
Rhombus - all sides congruent
Rectangle - 4 right angles
Square - Rectangle with all sides congruent (so it's also a rhombus)
Regular polygons - all sides and angles congruent
2 famous ones - equilateral triangle and squares
To name a polygon-- we name its consecutive vertices IN ORDER.
A diagonal of a polygon is a segment joining two NONCONSECUTIVE vertices.
The PERIMETER of a figure is the distance around it. Thus, the perimeter of a polygon is the sum of the lengths of its sides. So to find the PERIMETER of a regular figure, you just need to know one side and multiply by the total number of sides.
Circles: 4-6
A circle is the set of all points in a plane at a given distance from a given point O called the center
A segement joining the center to a point on the circle is called a radius (plural: radii) of the circle. All radii of a given circle have the same length
A segment joining two points on a circle is called a chord, and a chord passing through the center is a diameter of the circle. The ends of a diameter divide the circle into two semicircles. The length of a diameter is called the diameter of the circle.
The perimeter of a circle is called the circumference. “Sir Cumference” – it is the ‘fence’ around it!!
The quotient Circumference ÷ diameter can be showed to be the same for all circles-- Regardless of their size. This quotient is denoted by a Greek letter π
No decimal gives π exactly!! It is non-terminating (it never ends) and non repeating!! I like to use 3.14159. In our book 3.14 is concerned a fairly good approximation.
C ÷ d = π
Formulas you need to know:
Let C = circumference d = diameter, and r = radius
Then
C = πd d = C ÷ π
C = 2πr r = C ÷ (2π)
A polygon is inscribed in a circle if all of its vertices are on the circle.
It can be shown that three points not on a line determine a circle.
There is one and only one circle that passes through the three given points.
Wednesday, April 29, 2009
Tuesday, April 28, 2009
Algebra Period 3 (Tuesday)
SOLVING SYSTEMS OF EQUATIONS: 8-1 TO 8-3
(2 equations with 2 variables)
You cannot solve an equation with 2 variables - you can find multiple coordinates that work
TO SOLVE MEANS THE
ONE COORDINATE THAT WORKS FOR BOTH EQUATIONS
There are 3 ways to find that point:
1. Graph both equations: Where the 2 lines intersect is the solution
2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation
3. Addition method: Eliminate one of the variables by multiplying the equations by that magical number that will make one of the variables the ADDITIVE INVERSE of the other
Example solved all 3 ways:
Find the solution to the following system:
2x + 3y = 8 and 5x + 2y = -2
1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point (you should get (-2, 4))
2. SUBSTITUTION:
Isolate whatever variable seems easiest
I will isolate y in the second equation: 2y = -5x - 2
y = -5/2 x - 1
Plug this -5/2 x - 1 where y is in the other equation
2x + 3(-5/2 x - 1) = 8
2x - 15/2 x - 3 = 8
4/2 x - 15/2 x - 3 = 8
-11/2 x - 3 = 8
-11/2 x = 11
-2/11(-11/2 x) = 11(-2/11)
x = -2
Plug into whichever equation is easiest to find y
3. ADDITION:
Multiply each equation so that one variable will "drop out" (additive inverse)
I will eliminate the x, but could eliminate the y if I wanted to
5(2x + 3y) = (8)5
-2(5x + 2y) = (-2)-2
10x + 15y = 40
-10x - 4y = 4
ADD TO ELIMINATE THE x term:
11y = 44
y = 4
Plug into whichever equation is easiest to find x
NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!
(2 equations with 2 variables)
You cannot solve an equation with 2 variables - you can find multiple coordinates that work
TO SOLVE MEANS THE
ONE COORDINATE THAT WORKS FOR BOTH EQUATIONS
There are 3 ways to find that point:
1. Graph both equations: Where the 2 lines intersect is the solution
2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation
3. Addition method: Eliminate one of the variables by multiplying the equations by that magical number that will make one of the variables the ADDITIVE INVERSE of the other
Example solved all 3 ways:
Find the solution to the following system:
2x + 3y = 8 and 5x + 2y = -2
1. GRAPH BOTH LINES: Put both in y = mx + b form and graph
Read the intersection point (you should get (-2, 4))
2. SUBSTITUTION:
Isolate whatever variable seems easiest
I will isolate y in the second equation: 2y = -5x - 2
y = -5/2 x - 1
Plug this -5/2 x - 1 where y is in the other equation
2x + 3(-5/2 x - 1) = 8
2x - 15/2 x - 3 = 8
4/2 x - 15/2 x - 3 = 8
-11/2 x - 3 = 8
-11/2 x = 11
-2/11(-11/2 x) = 11(-2/11)
x = -2
Plug into whichever equation is easiest to find y
3. ADDITION:
Multiply each equation so that one variable will "drop out" (additive inverse)
I will eliminate the x, but could eliminate the y if I wanted to
5(2x + 3y) = (8)5
-2(5x + 2y) = (-2)-2
10x + 15y = 40
-10x - 4y = 4
ADD TO ELIMINATE THE x term:
11y = 44
y = 4
Plug into whichever equation is easiest to find x
NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!!
Math 6 H Periods 1, 6 & 7 (Monday)
Geometry PreView
(Some of the symbols and notations cannot be displayed here. You will need to find notes from another student or even check your textbook-- what a thought!!)
Points, Lines Planes: 4-1
All the figures that we study in geometry are made up of points. We usually picture a single point by making a dot and labeling it with a capital letter
The words point, line, and plane are undefined words in geometry- we can only describe them.
Among the most important geometric figures that we study are straight lines- or simply –lines.
You probably know this important fact about lines:
Two points determine exactly one line
This means that through two points P and Q we can draw one line and only one line,
Notice the use of arrowheads to show that a line extends without end in either direction
Three points my or may not line on the same line. Three or more points that do lie on the same line are called collinear. Points not on the same line are called noncollinear.
If we take a point P on a line and all the points on the line that lie on one side of P we have a ray with endpoint P. We name a ray by naming first its endpoint and then any other point on it.
It is important to remember that the endpoint is always named first.
Ray PQ is not the same ray as Ray QP ( check your textbook)
If we take two points P and Q on a line and all the points that lie between P and Q we have a segment denoted by the points P and Q are called the endpoints of
Just as two points determine a line, three noncollinear points in space determine a flat surface called a plane. We can name a plane by naming any three noncollinear points in it. Because a plane extends without limit in all directions of the surface, we can show only part of it
Lines in the same plane that do not intersect are called parallel lines. Two segments or rays are parallel if they are part of parallel lines.
is parallel to may be written as
Planes that do not intersect are called parallel planes
Two nonparallel lines that do not intersect are called skew lines
Angles and Angle Measure: 4-3
An angle is a figure formed by two rays with the same endpoints. The common endpoint is called the vertex. The rays are called the sides.
We may name an angle by giving its vertex letter if this is the only angle with that vertex, or my listing letters for points on the two sides with the vertex letter in the middle. We use the symbol that looks like a 'less than ' symbol for angle.
To measure segments we use a rule to mark off unit lengths. To measure angles, we use a protractor that is marked off in units of angle measure called degrees.
To use a protractor, place its center point at the vertex of the angle to be measured and one of its zero points on the side.
We often label angels with their measures. When angles have equal measures we can write
We say that Angle A and Angle B are congruent angles and we write Angle A equal sign with a ~ over it.
If two lines intersect so that the angles they form are all congruent, the lines are perpendicular. We use the symbol (Upside down T to mean “is perpendicular to.”
Angles formed by perpendicular lines each have measure of 90 degrees . A 90 degree angle is called a right angle. A small square is often used to indicate a right angle in a diagram
An acute angle is an angle with measure less than 90 degrees . An obtuse angle has measure between 90 degrees and 180 degrees
Two angles are complementary if the sum of the measures is 90 degrees
Two angles are supplementary if the sum of their measures is 180 degrees
Triangles: 4-4
A triangle is the figure formed when three points, not on a line are jointed by segments.
Triangle ABC
Having segments as its sides
Each of the Points A, B, C is called a vertex
(plural: Vertices)
Each of the angles Angle A ,Angle B , AngleC is called an angle of the triangle ABC
In any triangle- the sum of the lengths of any two sides is greater than the length of the third side
The sum of the measures of the angles is 180 degrees
There are several ways to name triangles. One way is by angles
Acute Triangle
3 acute angles
Right Triangle
1 right angle
Obtuse Triangle
1 obtuse angle
Triangles can be classified by their sides
Scalene Triangle
no 2 sides congruent
Isosceles Triangle
at least 2 side
congruent
Equilateral Triangle
all 3 sides
congruent
The longest side of a triangle is opposite the largest angle and the shortest side is opposite the smallest angle. Two angles are congruent if and only if the sides opposite them are congruent.
(Some of the symbols and notations cannot be displayed here. You will need to find notes from another student or even check your textbook-- what a thought!!)
Points, Lines Planes: 4-1
All the figures that we study in geometry are made up of points. We usually picture a single point by making a dot and labeling it with a capital letter
The words point, line, and plane are undefined words in geometry- we can only describe them.
Among the most important geometric figures that we study are straight lines- or simply –lines.
You probably know this important fact about lines:
Two points determine exactly one line
This means that through two points P and Q we can draw one line and only one line,
Notice the use of arrowheads to show that a line extends without end in either direction
Three points my or may not line on the same line. Three or more points that do lie on the same line are called collinear. Points not on the same line are called noncollinear.
If we take a point P on a line and all the points on the line that lie on one side of P we have a ray with endpoint P. We name a ray by naming first its endpoint and then any other point on it.
It is important to remember that the endpoint is always named first.
Ray PQ is not the same ray as Ray QP ( check your textbook)
If we take two points P and Q on a line and all the points that lie between P and Q we have a segment denoted by the points P and Q are called the endpoints of
Just as two points determine a line, three noncollinear points in space determine a flat surface called a plane. We can name a plane by naming any three noncollinear points in it. Because a plane extends without limit in all directions of the surface, we can show only part of it
Lines in the same plane that do not intersect are called parallel lines. Two segments or rays are parallel if they are part of parallel lines.
is parallel to may be written as
Planes that do not intersect are called parallel planes
Two nonparallel lines that do not intersect are called skew lines
Angles and Angle Measure: 4-3
An angle is a figure formed by two rays with the same endpoints. The common endpoint is called the vertex. The rays are called the sides.
We may name an angle by giving its vertex letter if this is the only angle with that vertex, or my listing letters for points on the two sides with the vertex letter in the middle. We use the symbol that looks like a 'less than ' symbol for angle.
To measure segments we use a rule to mark off unit lengths. To measure angles, we use a protractor that is marked off in units of angle measure called degrees.
To use a protractor, place its center point at the vertex of the angle to be measured and one of its zero points on the side.
We often label angels with their measures. When angles have equal measures we can write
We say that Angle A and Angle B are congruent angles and we write Angle A equal sign with a ~ over it.
If two lines intersect so that the angles they form are all congruent, the lines are perpendicular. We use the symbol (Upside down T to mean “is perpendicular to.”
Angles formed by perpendicular lines each have measure of 90 degrees . A 90 degree angle is called a right angle. A small square is often used to indicate a right angle in a diagram
An acute angle is an angle with measure less than 90 degrees . An obtuse angle has measure between 90 degrees and 180 degrees
Two angles are complementary if the sum of the measures is 90 degrees
Two angles are supplementary if the sum of their measures is 180 degrees
Triangles: 4-4
A triangle is the figure formed when three points, not on a line are jointed by segments.
Triangle ABC
Having segments as its sides
Each of the Points A, B, C is called a vertex
(plural: Vertices)
Each of the angles Angle A ,Angle B , AngleC is called an angle of the triangle ABC
In any triangle- the sum of the lengths of any two sides is greater than the length of the third side
The sum of the measures of the angles is 180 degrees
There are several ways to name triangles. One way is by angles
Acute Triangle
3 acute angles
Right Triangle
1 right angle
Obtuse Triangle
1 obtuse angle
Triangles can be classified by their sides
Scalene Triangle
no 2 sides congruent
Isosceles Triangle
at least 2 side
congruent
Equilateral Triangle
all 3 sides
congruent
The longest side of a triangle is opposite the largest angle and the shortest side is opposite the smallest angle. Two angles are congruent if and only if the sides opposite them are congruent.
Algebra Period 3 (Monday)
Simplify, Multiply, and Divide RATIONAL EXPRESSIONS 10-1, 10-2, and 10-3
Rational Expressions = Expressions in fraction format (division) with a variable in the denominator
You have already been simplifying, multiplying and dividing these throughout this year!
SIMPLIFY: 10-1
You will need to FACTOR (Chapter 6) both the numerator and denominator and "cross out" common factors in both (their quotient is 1!)
EXAMPLE: Simplify
y2 + 3y + 2 =
y2 - 1
(y + 2)(y + 1) =
(y - 1)(y + 1)
y + 2
y - 1
MULTIPLY: 10-2
FACTOR if possible, cross cancel if possible, multiply numerators, then denominators, simplify
EXAMPLE:
(y + 4)3[y2 + 4y + 4] =
[(y + 2) 3(y2 + 8y + 16)]
(y + 4) 3][ (y + 2) 2 =
(y + 2) 3(y + 4) 2
y + 4
y + 2
DIVIDE: 10-3
Same as the previous example, only this time you will need to
“FLIP the SECOND” fraction,
FACTOR then
MULTIPLY!!!!!
EXAMPLE:
x + 1 ÷ x + 1 =
x2 - 1 x2 - 2x + 1
( x + 1) ( x2 - 2x + 1) =
(x + 1)( x - 1) (x + 1)
( x + 1)( x - 1)(x - 1) =
(x + 1)( x - 1)(x + 1)
x - 1
x + 1
Add and subtract rational expressions with LIKE DENOMINATORS: 10-4
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify
When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)
When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x2 - 16 x – 4
First, factor the denominators if possible
2x + x
(x + 4)(x - 4) x – 4
Find the LCM = (x + 4)(x - 4);
2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)
Restate both fractions with the LCM
2x + x2 + 4x
(x + 4)(x - 4)
Now add the numerators
x2 + 6x =
(x + 4)(x - 4)
Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)
Rational Expressions = Expressions in fraction format (division) with a variable in the denominator
You have already been simplifying, multiplying and dividing these throughout this year!
SIMPLIFY: 10-1
You will need to FACTOR (Chapter 6) both the numerator and denominator and "cross out" common factors in both (their quotient is 1!)
EXAMPLE: Simplify
y2 + 3y + 2 =
y2 - 1
(y + 2)(y + 1) =
(y - 1)(y + 1)
y + 2
y - 1
MULTIPLY: 10-2
FACTOR if possible, cross cancel if possible, multiply numerators, then denominators, simplify
EXAMPLE:
(y + 4)3[y2 + 4y + 4] =
[(y + 2) 3(y2 + 8y + 16)]
(y + 4) 3][ (y + 2) 2 =
(y + 2) 3(y + 4) 2
y + 4
y + 2
DIVIDE: 10-3
Same as the previous example, only this time you will need to
“FLIP the SECOND” fraction,
FACTOR then
MULTIPLY!!!!!
EXAMPLE:
x + 1 ÷ x + 1 =
x2 - 1 x2 - 2x + 1
( x + 1) ( x2 - 2x + 1) =
(x + 1)( x - 1) (x + 1)
( x + 1)( x - 1)(x - 1) =
(x + 1)( x - 1)(x + 1)
x - 1
x + 1
Add and subtract rational expressions with LIKE DENOMINATORS: 10-4
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify
When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)
When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x2 - 16 x – 4
First, factor the denominators if possible
2x + x
(x + 4)(x - 4) x – 4
Find the LCM = (x + 4)(x - 4);
2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)
Restate both fractions with the LCM
2x + x2 + 4x
(x + 4)(x - 4)
Now add the numerators
x2 + 6x =
(x + 4)(x - 4)
Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)
Saturday, April 25, 2009
Algebra Period 3 (Friday)
How about one more video..
let me know if you find one you think everyone in our class would like
Solving Rational Expressions: 13-5
YOU NEED TO GET RID OF ALL DENOMINATORS!
After you do that, you may end up with a Quadratic that you can solve using any of your methods.
Easy one first
(a+1)/2 = 1/a
You could find the LCM of the denominators which would be 2a but what about using cross products—as you did with proportions—last year.
then the above becomes
a(a + 1) = 2(1)
0r
a2 + a = 2
a2 + a - 2= 0 using ZERO products property
Factor
(a + 2)(a-1) = 0
so a = -2 and a= 1
When you MUST Find the LCM of the denominators:
MULTIPLY EACH & EVERY TERM ON BOTH SIDES BY THE LCM!
You should end up with NO DENOMINATORS!
(Otherwise, you don't have the right LCM!)
EXAMPLE:
4/x – 4/(x+2) = 1
In this case the LCM is discovered just by multiplying both of the existing denominators. It is as if they are relatively prime!!
The LCM is x(x + 2)
Multiply each term on BOTH sides by x(x + 2)
[x(x +2)4]/x – [4 x(x +2)]/(x+2) = 1 [x(x +2)]
Simplify the denominators with the numerators:
(x + 2) 4 - x(4) = x(x + 2)
SIMPLIFY more:
4x + 8 - 4x = x2 + 2x
USE THE ZERO PRODUCTS PROPERTY:
x2 + 2x - 8 = 0
FACTOR OR QUADRATIC FORMULA:
(x + 4)(x - 2) = 0
x = -4 and x = 2
Solving Radical Equations 13-6
THIS IS A REVIEW OF CHAPTER 11
You square both sides to get rid of the radical sign
For these problems, you'll get a quadratic on one side after you square
BE SURE TO CHECK BOTH ANSWERS TO MAKE SURE THEY BOTH CHECK!
x – 5 = SQRT (x+ 7)
squaring both sides gives us
(x -5)2 = [SQRT(x + 7)]2
x2 - 10x + 25 = x + 7
x2- 11x + 18 = 0
(x – 9)(x -2) = 0
or x = 9 and x = 2
BUT when you check you discover
x -5 = SQRT ( x + 7)
9 – 5 =?= SQRT (9 + 7)
4 = 4
BUT
x -5 = SQRT ( x + 7)
2 – 5 =?= SQRT (2 + 7)
-3 DOES NOT EQUAL 3
so only solution is x = 9
Solve
[SQRT ( 27 - 3x)] + 3 = x
move the 3 to the other side using transformations
[SQRT ( 27 - 3x)] = x -3
NOW squaring both sides
[SQRT ( 27 - 3x)]2 = (x -3)2
27 – 3x = x2 - 6x + 9
move everything to the right side – to use the ZERO PRODUCTS PROPERTY
0 = x2 - 3x – 18
0 = ( x – 6) (x +3)
so x = 6 or x = -3
WE MUST CHECK again:
Start with the ORIGINAL equation
[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = 6
[SQRT ( 27 – 3(6))] + 3 =? = 6
[SQRT ( 27 - 18)] + 3 =? = 6
[SQRT ( 9)] + 3 =?= 6
3 + 3 = 6 YES
But we discover
[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = -3
[SQRT ( 27 – 3(-3))] + 3 =?= -3
[SQRT ( 27 +9)] + 3 =?= -3
[SQRT ( 36)] + 3 =?= -3
6 + 3 DOES NOT EQUAL -3
so only x = 6 is the solution
let me know if you find one you think everyone in our class would like
Solving Rational Expressions: 13-5
YOU NEED TO GET RID OF ALL DENOMINATORS!
After you do that, you may end up with a Quadratic that you can solve using any of your methods.
Easy one first
(a+1)/2 = 1/a
You could find the LCM of the denominators which would be 2a but what about using cross products—as you did with proportions—last year.
then the above becomes
a(a + 1) = 2(1)
0r
a2 + a = 2
a2 + a - 2= 0 using ZERO products property
Factor
(a + 2)(a-1) = 0
so a = -2 and a= 1
When you MUST Find the LCM of the denominators:
MULTIPLY EACH & EVERY TERM ON BOTH SIDES BY THE LCM!
You should end up with NO DENOMINATORS!
(Otherwise, you don't have the right LCM!)
EXAMPLE:
4/x – 4/(x+2) = 1
In this case the LCM is discovered just by multiplying both of the existing denominators. It is as if they are relatively prime!!
The LCM is x(x + 2)
Multiply each term on BOTH sides by x(x + 2)
[x(x +2)4]/x – [4 x(x +2)]/(x+2) = 1 [x(x +2)]
Simplify the denominators with the numerators:
(x + 2) 4 - x(4) = x(x + 2)
SIMPLIFY more:
4x + 8 - 4x = x2 + 2x
USE THE ZERO PRODUCTS PROPERTY:
x2 + 2x - 8 = 0
FACTOR OR QUADRATIC FORMULA:
(x + 4)(x - 2) = 0
x = -4 and x = 2
Solving Radical Equations 13-6
THIS IS A REVIEW OF CHAPTER 11
You square both sides to get rid of the radical sign
For these problems, you'll get a quadratic on one side after you square
BE SURE TO CHECK BOTH ANSWERS TO MAKE SURE THEY BOTH CHECK!
x – 5 = SQRT (x+ 7)
squaring both sides gives us
(x -5)2 = [SQRT(x + 7)]2
x2 - 10x + 25 = x + 7
x2- 11x + 18 = 0
(x – 9)(x -2) = 0
or x = 9 and x = 2
BUT when you check you discover
x -5 = SQRT ( x + 7)
9 – 5 =?= SQRT (9 + 7)
4 = 4
BUT
x -5 = SQRT ( x + 7)
2 – 5 =?= SQRT (2 + 7)
-3 DOES NOT EQUAL 3
so only solution is x = 9
Solve
[SQRT ( 27 - 3x)] + 3 = x
move the 3 to the other side using transformations
[SQRT ( 27 - 3x)] = x -3
NOW squaring both sides
[SQRT ( 27 - 3x)]2 = (x -3)2
27 – 3x = x2 - 6x + 9
move everything to the right side – to use the ZERO PRODUCTS PROPERTY
0 = x2 - 3x – 18
0 = ( x – 6) (x +3)
so x = 6 or x = -3
WE MUST CHECK again:
Start with the ORIGINAL equation
[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = 6
[SQRT ( 27 – 3(6))] + 3 =? = 6
[SQRT ( 27 - 18)] + 3 =? = 6
[SQRT ( 9)] + 3 =?= 6
3 + 3 = 6 YES
But we discover
[SQRT ( 27 - 3x)] + 3 = x
Plug in for x = -3
[SQRT ( 27 – 3(-3))] + 3 =?= -3
[SQRT ( 27 +9)] + 3 =?= -3
[SQRT ( 36)] + 3 =?= -3
6 + 3 DOES NOT EQUAL -3
so only x = 6 is the solution
Thursday, April 23, 2009
Algebra Period 3 (Thursday)
The Quadratic Formula
Wait until you see and hear these videos--
This is the one from class
this is to the Flintstones
pretty funny stuff
So what do you think? Create one of your own...
THE QUADRATIC FORMULA:
-b + or - SQRT b2 - 4ac
2a
-b plus or minus the square root of b squared minus 4ac all over 2a
Notice how the first part is the x value of the vertex -b/2a
The plus or minus square root of b squared minus 4ac represents
how far away the two x intercepts (or roots) are from the vertex!!!!
Very few real world quadratics can be solved by factoring or square rooting each side.
And completing the square always works, but it long and cumbersome!
All quadratics can be solved by using the QUADRATIC FORMULA.
(you will find out that some quadratics have NO REAL solutions, which means that there are no x intercepts - the parabola does not cross the x axis! Think about what kinds of parabolas would do this....ones that are smiles that have a vertex above the x or ones that are frowns that have a vertex below the x axis. You will find out in Algebra II that these parabolas have IMAGINARY roots)
So now you know 5 ways that you know to find the roots:
1. graph
2. factor if possible
3. square root each side
4. complete the square - that's what the quadratic formula is based on!
5. plug and chug in the Quadratic Formula -
This method always works if there's a REAL solution!
DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA! ax2 + bx + c = 0
DISCRIMINANTS - a part of the Quadratic Formula that helps you to understand the graph of the parabola even before you graph it!
the discriminant is b2 - 4ac
(the radicand in the Quadratic Formula, but without the SQRT)
Depending on the value of the radicand, you will know
HOW MANY REAL ROOTS IT HAS
1) Some quadratics have 2 real roots (x intercepts or solutions) - Graph crosses x axis twice
2) Some have 1 real root (x intercept or solution) - Vertex is sitting on the x axis
3) Some have NO real roots (no x intercepts or solutions) - vertex either is above the x axis and is a smiley face (a coefficient is positive) or
the vertex is below the x axis and is a frown face (a coefficient is negative)
In both of these cases, the parabola will NEVER CROSS (intercept) the x axis!
b2 -4ac > 0 if it's positive, 2 roots
b2 -4ac = 0 if it's zero - 1 root
b2 -4ac < 0 if it's negative - no real roots
Wait until you see and hear these videos--
This is the one from class
this is to the Flintstones
pretty funny stuff
So what do you think? Create one of your own...
THE QUADRATIC FORMULA:
-b + or - SQRT b2 - 4ac
2a
-b plus or minus the square root of b squared minus 4ac all over 2a
Notice how the first part is the x value of the vertex -b/2a
The plus or minus square root of b squared minus 4ac represents
how far away the two x intercepts (or roots) are from the vertex!!!!
Very few real world quadratics can be solved by factoring or square rooting each side.
And completing the square always works, but it long and cumbersome!
All quadratics can be solved by using the QUADRATIC FORMULA.
(you will find out that some quadratics have NO REAL solutions, which means that there are no x intercepts - the parabola does not cross the x axis! Think about what kinds of parabolas would do this....ones that are smiles that have a vertex above the x or ones that are frowns that have a vertex below the x axis. You will find out in Algebra II that these parabolas have IMAGINARY roots)
So now you know 5 ways that you know to find the roots:
1. graph
2. factor if possible
3. square root each side
4. complete the square - that's what the quadratic formula is based on!
5. plug and chug in the Quadratic Formula -
This method always works if there's a REAL solution!
DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA! ax2 + bx + c = 0
DISCRIMINANTS - a part of the Quadratic Formula that helps you to understand the graph of the parabola even before you graph it!
the discriminant is b2 - 4ac
(the radicand in the Quadratic Formula, but without the SQRT)
Depending on the value of the radicand, you will know
HOW MANY REAL ROOTS IT HAS
1) Some quadratics have 2 real roots (x intercepts or solutions) - Graph crosses x axis twice
2) Some have 1 real root (x intercept or solution) - Vertex is sitting on the x axis
3) Some have NO real roots (no x intercepts or solutions) - vertex either is above the x axis and is a smiley face (a coefficient is positive) or
the vertex is below the x axis and is a frown face (a coefficient is negative)
In both of these cases, the parabola will NEVER CROSS (intercept) the x axis!
b2 -4ac > 0 if it's positive, 2 roots
b2 -4ac = 0 if it's zero - 1 root
b2 -4ac < 0 if it's negative - no real roots
Math 6 H Periods 1, 6 & 7
Review of Solving Combined Operations 8-5
Follow these general procedures to solving equations:
Step 1: Simplify each side of the equation
Step 2: If there are still indicated additions or subtrations, use the inverse operations to undo them
Step 3: If there are indicated multiplication or divisions involving the variable, use the the inverse operations to undo them
YOU MUST ALWAYS PERFORM THE SAME OPERATION ON BOTH SIDES OF THE EQUATION.
DO ONTO ONE SIDE WHAT YOU WOULD DO TO THE OTHER... hmmm... where have you heard that before??
Balance, balance , balance... it's all a balancing act!!
Solve the equation 3/2(n) + 7 = 22
subtract 7 FROM BOTH SIDES
(3/2)n = 15
Multiply BOTH SIDES by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
simplify before you multiply out
n = 10
40 - (5/3)n = 15
There are two ways to solve this one
add (5/3)n to both sides
40 = (5/3)n + 15
then subtract 15 from BOTH SIDES
25 = 5/3(n)
Multiply both sides by 3/5, the reciprocal of 5/3
(3/5)25 = 5/3(n)(3/5)
Simplify before multiplying
15 = n
OR
40 - (5/3)n = 15
subtract 30 from both sides
- (5/3)n = 15 -40
need to ask yourself who wins? the negative... and by how much 25
- (5/3)n = -25
multiply by -3/5, which is the reciprocal of -5/3
(-3/5)-(5/3)n = -25(-3/5)
n = 15
we arrived at the same solution
Follow these general procedures to solving equations:
Step 1: Simplify each side of the equation
Step 2: If there are still indicated additions or subtrations, use the inverse operations to undo them
Step 3: If there are indicated multiplication or divisions involving the variable, use the the inverse operations to undo them
YOU MUST ALWAYS PERFORM THE SAME OPERATION ON BOTH SIDES OF THE EQUATION.
DO ONTO ONE SIDE WHAT YOU WOULD DO TO THE OTHER... hmmm... where have you heard that before??
Balance, balance , balance... it's all a balancing act!!
Solve the equation 3/2(n) + 7 = 22
subtract 7 FROM BOTH SIDES
(3/2)n = 15
Multiply BOTH SIDES by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
simplify before you multiply out
n = 10
40 - (5/3)n = 15
There are two ways to solve this one
add (5/3)n to both sides
40 = (5/3)n + 15
then subtract 15 from BOTH SIDES
25 = 5/3(n)
Multiply both sides by 3/5, the reciprocal of 5/3
(3/5)25 = 5/3(n)(3/5)
Simplify before multiplying
15 = n
OR
40 - (5/3)n = 15
subtract 30 from both sides
- (5/3)n = 15 -40
need to ask yourself who wins? the negative... and by how much 25
- (5/3)n = -25
multiply by -3/5, which is the reciprocal of -5/3
(-3/5)-(5/3)n = -25(-3/5)
n = 15
we arrived at the same solution
Wednesday, April 22, 2009
Algebra Period 3 (Tues/Wed)
Solving Quadratics by Completing the Square 13-3
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!
When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!
This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!
EXAMPLE: x2 - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square
Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)
Step 2: Square b/2
(-5 x -5 = 25)
Step 3: Add (b/2)2 to both sides of the equation
(x2 - 10x + 25 = +25)
Step 4: Factor to a binomial square
(x - 5)2 = 25
Step 5: Square root each side and solve
SQRT (x - 5)2 = SQRT 25
x - 5 = + and - 5
ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x2 - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10
x2-4x -7 = 0
add 7 to both sides
x2-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x2-4x + 4 = 7 +4
x2- 4x + 4 = 11
(x -2)2 = 11
take the square root of both sides
SQRT(x -2)2 = + or – SQRT 11
x - 2 = + or - SQRT 11
IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:
Example: 2x2 - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x2 - 3x = 1
Divide each term by the "a" coefficient:
x2 - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]2 = (-3/4)2
= 9/16
x2 - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)2 = 8/16 + 9/16
(x - 3/4)2 = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4
When our textbook ask you to complete the square
as in x2 - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)2 = 9
so the answer is
x2 - 6x + 9
Okay, up until now the methods of solving quadratics should have been fairly familiar to you—but this method is a NEW strategy!!
When does the "square root = +/- square root" method work well?
When the side with the variable is a PERFECT SQUARE!
So what if that side is not a perfect BINOMIAL SQUARED?
You can follow steps to make it into one!
This method is great because then you can just square root each side to find the roots!
THIS METHOD ALWAYS WORKS!
EXAMPLE: x2 - 10x = 0
Not a TRINOMIAL SQUARE so it would not factor to a BINOMIAL SQUARED. (However, we know that we could just factor out an x to solve—I want you to see how completing the square works—even with easy ones)
Here's how you can make this into a trinomial square
Step 1: b/2
Take half of the b coefficient in this case (- 10/2 = -5)
Step 2: Square b/2
(-5 x -5 = 25)
Step 3: Add (b/2)2 to both sides of the equation
(x2 - 10x + 25 = +25)
Step 4: Factor to a binomial square
(x - 5)2 = 25
Step 5: Square root each side and solve
SQRT (x - 5)2 = SQRT 25
x - 5 = + and - 5
ADD 5 TO BOTH SIDES
x = 5 + and - 5
x = 5 + 5 and x = 5 - 5
x = 10 and x = 0
Now the check for this one—just because we are learning this strategy--is to factor and make sure you get the same results.
that is
x2 - 10x = 0 we know becomes
x(x-10) = 0
x = 0 and x = 10
x2-4x -7 = 0
add 7 to both sides
x2-4x =7
take –b/2 or -4/2 = -2 now square that (-2),sup>2 = 4
add 4 to both sides
x2-4x + 4 = 7 +4
x2- 4x + 4 = 11
(x -2)2 = 11
take the square root of both sides
SQRT(x -2)2 = + or – SQRT 11
x - 2 = + or - SQRT 11
IF THERE IS AN "a" COEFFICIENT, YOU MUST DIVIDE EACH TERM BY IT BEFORE YOU CAN COMPLETE THE SQUARE:
Example: 2x2 - 3x - 1 = 0
Move the 1 to the other side of the equation:
2x2 - 3x = 1
Divide each term by the "a" coefficient:
x2 - 3/2 x = 1/2
Now follow the step to find the completing the square term and add it to both sides:
-b/2 = (-3/2)(1/2) remember (instead of dividing by 2, when you have a fraction, multiply by 1/2)= – 3/4
NOTE: REMEMBER this term you will use it again!!
square that [(-3/2)(1/2)]2 = (-3/4)2
= 9/16
x2 - 3/2 x + 9/16 = 1/2 + 9/16
(x - 3/4)2 = 8/16 + 9/16
(x - 3/4)2 = 17/16
SQRT[(x - 3/4)2 ] = + or - SQRT [17/16]
x - 3/4 = + or -[SQRT 17] /4
x = 3/4 + or -[SQRT 17] /4
x = 3 + or - [SQRT 17] /4
When our textbook ask you to complete the square
as in x2 - 6x
all it is asking is that you follow the steps
-b/2 is -6/2 = -3
then square that (-3)2 = 9
so the answer is
x2 - 6x + 9
Friday, April 17, 2009
Algebra Period 3 (Monday)
Introduction to Quadratic Equations 13-1
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
We learned from Chapter 12 that a quadratic function is a function that can be defined by an equation of the form
ax2 + bx + c = y where a is not equal to 0. This is a parabola when the domain is the set of REAL numbers.
When y = 0 in the quadratic function ax2 + bx + c = y we have an equation of the form ax2 + bx + c = 0. An equation that can be written in this form is called a quadratic equation.
STANDARD FORM is ax2 + bx + c = 0
4x2 + 7x = 5 write in standard form and determine a, b, and c
4x2 + 7x – 5 = 0
a= 4
b = 7
c = -5
CHAPTER 13 gives you several different ways to SOLVE QUADRATICS
Solving a quadratic means to find the x intercepts of a parabola.
There are different ways of asking the exact same question:
Find the.....
x intercepts = the roots = the solutions = the zeros of a quadratic
We'll answer this question one of the following ways:
1) Read them from the graph (read the x intercepts) That’s were y = 0 or where the parabola crosses the x-axis!!
but...graphing takes time and sometimes the intercepts are not integers
2) Set y or f(x) = 0 and then factor (we did this in Chapter 6)
but...some quadratics are not factorable
3) Square root each side (+ or - square root on the answer side)
but...sometimes the variable side is not a perfect square (it's irrational)
4) If not a perfect square on the variable side, complete the square, then solve using #3 method
Now this method ALWAYS works, but...it takes a lot of time and can get complicated
5) Quadratic Formula (works for EVERY quadratic)
Really easy if you just memorize the formula and how to use it! :)
Section 13-1
Reading the x intercepts from a graph or factoring and solving using the zero products property.
METHOD 1:
Where the graph crosses the x axis is/are the x intercepts. (Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
METHOD 2:
When we factored in Chapter 6 and set each piece equal to zero, we were finding the x value when y was zero.
That means we were finding these two roots!
y2 – 5y = 6 = 6y – 18
first put this in standard form
y2 – 11y + 24 = 0
(y -8) (y-3) = 0
y = 8 or y = 3
Substitute to verify that 8 and 3 are solutions!!
More Solving Quadratic Equations 13-2
You did this in Chapter 11 for Pythagorean Theorem!
If there is no x term, it's easiest to just square root both sides to solve!
DIFFERENT FROM PYTHAGOREAN: NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANYMORE. NEED THE + OR - SYMBOL!!
This is also different from what we did when we had radical equations. Before we squared both sides to solve. It looks like these, but only after we squared both sides. Before we had to carefully check each answer—we had changed the equations by squaring. However, always check your solutions for any mistakes!!
3x2 = 18
divide both sides by 3 and get: x2 = 6
square root each side and get x = + SQRT 6 or - SQRT 6
(x - 5)2 = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2
(x + 2)2 = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7
FORMULAS THAT ARE QUADRATICS:
Many formulas have a variable that is squared: compounded interest, height of a projectile (ball)
The formula for a projectile is h = -5t2 + v0t
We can find when a projectile is a ground level ( h= 0) by solving for
0 =-5t2 + v0t \If the projectile begins its flight at height c, its approximate height at time t is h = -5t2+v0t + c We can find when it hits the ground by solving 0 = -5t2 + v0t + c
For example: a slow-pitch softball player hits a pitch when the ball is 2 m above the ground. The ball pops up with an initial velocity of 9m/s If the ball is allowed to drop to the ground, how long will it be in the air?
When the ball hits the ground h = 0 so
0 = -5t2 + 9t + 2 or
-5t2 + 9t + 2 = 0
( 5t + 1)(t -2) = 0
5t = 1 and t = 2
t = -1/5 can’t be a solution since the answer should be positive t must be 2 seconds
Check out Purple Math for great help on quadratics
Algebra Period 3 ( Review)
Quadratic Equations Review
Maybe another teacher reviewing how to graph quadratics might be just what you need:
So did that help?
Maybe another teacher reviewing how to graph quadratics might be just what you need:
So did that help?
Thursday, April 9, 2009
Math 6 Honors Periods 1, 6 & 7 (Thursday)
Combined Operations 8-5
In order to solve an equation of the form
ax + b = c or ax –b = c or b – ax = c
where a, b, c are given numbers and x is the variable, we must use more than one transformation
Solve the equation 3n - 5 = 10 + 6
Simplify the numerical expression
3n - 5 = 10 + 6
3n – 5 = 16
add 5 to both sides
3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5
3n = 21
divide both sides by 3 (or multiply each side by the reciprocal of 3)
3n/3 = 21/3
n = 7
General procedures for solving equations
Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them
The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”
Solve the equation
(3/2)n + 7 = 22
subtract 7 from both sides
(3/2)n + 7 - 7 = 22 - 7
(3/2)n =15
multiply both sides by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
n = 10
Solve the equation
40 – (5/3)n = 15
add (5/3)n to both sides
40 – (5/3)n + (5/3)n = 15 + (5/3)n
40 = 15 + (5/3)n
subtract 15 from both sides
40 – 15 = 15-15 + (5/3)n
25 = (5/3)n multiply both sides by 3/5
(3/5)(25) = (5/3)n (3/5)
15 = n
In order to solve an equation of the form
ax + b = c or ax –b = c or b – ax = c
where a, b, c are given numbers and x is the variable, we must use more than one transformation
Solve the equation 3n - 5 = 10 + 6
Simplify the numerical expression
3n - 5 = 10 + 6
3n – 5 = 16
add 5 to both sides
3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5
3n = 21
divide both sides by 3 (or multiply each side by the reciprocal of 3)
3n/3 = 21/3
n = 7
General procedures for solving equations
Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them
The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”
Solve the equation
(3/2)n + 7 = 22
subtract 7 from both sides
(3/2)n + 7 - 7 = 22 - 7
(3/2)n =15
multiply both sides by 2/3, the reciprocal of 3/2
(2/3)(3/2)n = 15(2/3)
n = 10
Solve the equation
40 – (5/3)n = 15
add (5/3)n to both sides
40 – (5/3)n + (5/3)n = 15 + (5/3)n
40 = 15 + (5/3)n
subtract 15 from both sides
40 – 15 = 15-15 + (5/3)n
25 = (5/3)n multiply both sides by 3/5
(3/5)(25) = (5/3)n (3/5)
15 = n
Wednesday, April 8, 2009
Math 6 H Periods 1, 6 & 7 (Wednesday)
Equations: Decimals and Fractions 8-4
You can use transformations to solve equations which involve decimals or fractions
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5
Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.80
How would we solve the following: (2/3)x = 6?
Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1
Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.) Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1. [Can you picture the poster in the front of the room?]
So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.
If an equation has the form
(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18
(3/1)(1/3)y = 18(3/1)
y =18(3)
y = 54
Solve the equation: (6/7)n = 8
(7/6)(6/7)n = 8(7/6)
n = 8(7/6)
simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check
(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]
Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8
Try:
1. (1/7)a = 13
2. b/8 = 16
3. 3.6d = 0.9
4. (3/8)f = 129
You can use transformations to solve equations which involve decimals or fractions
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5
Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.80
How would we solve the following: (2/3)x = 6?
Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1
Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.) Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1. [Can you picture the poster in the front of the room?]
So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.
If an equation has the form
(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18
(3/1)(1/3)y = 18(3/1)
y =18(3)
y = 54
Solve the equation: (6/7)n = 8
(7/6)(6/7)n = 8(7/6)
n = 8(7/6)
simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check
(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]
Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8
Try:
1. (1/7)a = 13
2. b/8 = 16
3. 3.6d = 0.9
4. (3/8)f = 129
Tuesday, April 7, 2009
Math 6 H Periods 1, 6 & 7 (Tuesday)
Equations: Multiplication and Division 8-3
If an equation involves multiplication or division, the following transformations are used to solve the equation:
Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.
Transformations by division: Divide both sides of the equation by the same nonzero number.
Remember: Do undo on one side what you would do undo the other!!
Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”
Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3
n = 8
Or you could have use the reciprocal of 3-- which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution
Solve 5x/5 = 53/5
x = 10 3/5
Solve n/4 = 7
(4)(n/4) = 7(4)
n = 28
How could you know for sure your answer is correct?
Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57
2. 714 = 7t
3. Solve A = bh for h
4. Solve P = 4s for s
5. Solve C = 2πr for r
If an equation involves multiplication or division, the following transformations are used to solve the equation:
Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.
Transformations by division: Divide both sides of the equation by the same nonzero number.
Remember: Do undo on one side what you would do undo the other!!
Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”
Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3
n = 8
Or you could have use the reciprocal of 3-- which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution
Solve 5x/5 = 53/5
x = 10 3/5
Solve n/4 = 7
(4)(n/4) = 7(4)
n = 28
How could you know for sure your answer is correct?
Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57
2. 714 = 7t
3. Solve A = bh for h
4. Solve P = 4s for s
5. Solve C = 2πr for r
Algebra Period 3 (Monday)
Direct Variation, Indirect Variation and Joint Variation: 12-5 TO 12-7
DIRECT VARIATION
As x increases, y also increases (graphs as a LINE)
or as x decreases, y also decreases
x and y go in the SAME DIRECTION
For example: y = 2x
the "2" is called the constant of variation and is "k" in the formula:
y = kx
To find k, you need one (x, y) coordinate to solve.
It's like solving for "m" in y = mx + b, but there is no b!
Example: y varies directly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR DIRECT VARIATION: y = kx
PLUG IN THE (x, y):
8 = k(2)
SOLVE FOR k:
k = 4
WRITE THE EQUATION:
y = 4x
PLUG IN VARIABLE GIVEN:
y = 4(3)
SOLVE FOR MISSING VARIABLE:
y = 12
INDIRECT VARIATION
As x increases, y decreases (graphs as a RATIONAL function...I'll show you it in class)
x and y go in OPPOSITE DIRECTIONS
For example: y = 2/x
the "2" is still called the constant of variation and is "k" in the formula:
y = k/x
To find k, you need one (x, y) coordinate to solve.
Example: y varies indirectly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR INDIRECT VARIATION:
y = k/x
PLUG IN THE (x, y):
8 = k/2
SOLVE FOR k:
k = 16
WRITE THE EQUATION:
y = 16/x
PLUG IN VARIABLE GIVEN:
y = 16/3
SOLVE FOR MISSING VARIABLE:
y = 5 1/3
JOINT VARIATION
TWO VARIABLES VARY WITH y AT THE SAME TIME!
As x AND z increase, y also increases (graphs as a LINEAR function)
The PRODUCT xz and y go in the SAME DIRECTION
For example: y = 2xz
the "2" is still called the constant of variation and is "k" in the formula:
y = kxz
To find k, you need one (x, y, z) coordinate to solve.
Example: y varies jointly with x and z. When y = 12, x = 2, and z = 3. Find y when x = 3 and z = 5
FORMULA FOR JOINT VARIATION:
y = kxz
PLUG IN THE (x, y, z):
12 = k(2)(3)
SOLVE FOR k:
k = 2
WRITE THE EQUATION:
y = 2xz
PLUG IN VARIABLE GIVEN:
y = 2(3)(5)
SOLVE FOR MISSING VARIABLE:
y = 30
DIRECT VARIATION
As x increases, y also increases (graphs as a LINE)
or as x decreases, y also decreases
x and y go in the SAME DIRECTION
For example: y = 2x
the "2" is called the constant of variation and is "k" in the formula:
y = kx
To find k, you need one (x, y) coordinate to solve.
It's like solving for "m" in y = mx + b, but there is no b!
Example: y varies directly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR DIRECT VARIATION: y = kx
PLUG IN THE (x, y):
8 = k(2)
SOLVE FOR k:
k = 4
WRITE THE EQUATION:
y = 4x
PLUG IN VARIABLE GIVEN:
y = 4(3)
SOLVE FOR MISSING VARIABLE:
y = 12
INDIRECT VARIATION
As x increases, y decreases (graphs as a RATIONAL function...I'll show you it in class)
x and y go in OPPOSITE DIRECTIONS
For example: y = 2/x
the "2" is still called the constant of variation and is "k" in the formula:
y = k/x
To find k, you need one (x, y) coordinate to solve.
Example: y varies indirectly with x. When y = 8, x = 2. Find y when x = 3.
FORMULA FOR INDIRECT VARIATION:
y = k/x
PLUG IN THE (x, y):
8 = k/2
SOLVE FOR k:
k = 16
WRITE THE EQUATION:
y = 16/x
PLUG IN VARIABLE GIVEN:
y = 16/3
SOLVE FOR MISSING VARIABLE:
y = 5 1/3
JOINT VARIATION
TWO VARIABLES VARY WITH y AT THE SAME TIME!
As x AND z increase, y also increases (graphs as a LINEAR function)
The PRODUCT xz and y go in the SAME DIRECTION
For example: y = 2xz
the "2" is still called the constant of variation and is "k" in the formula:
y = kxz
To find k, you need one (x, y, z) coordinate to solve.
Example: y varies jointly with x and z. When y = 12, x = 2, and z = 3. Find y when x = 3 and z = 5
FORMULA FOR JOINT VARIATION:
y = kxz
PLUG IN THE (x, y, z):
12 = k(2)(3)
SOLVE FOR k:
k = 2
WRITE THE EQUATION:
y = 2xz
PLUG IN VARIABLE GIVEN:
y = 2(3)(5)
SOLVE FOR MISSING VARIABLE:
y = 30
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