Algebra (Period 5)

Solving Quadratic Equations By Using the Quadratic Formula 9-5

Let's all sing it to "Pop Goes the Weasel!"

x = -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!

EXAMPLE: x² + 8x = 48 

FIRST DETERMINE BY USING THE DISCRIMINANT THE NUMBER OF ROOTS WE'LL FIND:

b2 - 4ac
= 64 - 4(1)(-48)
= 64 + 192
= +256

so 2 REAL ROOTS

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 (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!

if it's positive, 2 roots

if it's zero - 1 root

if it's negative - no real roots

DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA!

You should know that for the quadratic formula, you don't need the "a" coefficient to be positive!

That's important if you use factoring, SQRTing each side, and completing the square.

But for the quadratic formula, either way, you'll get the same roots!

You can move the 48 over or move the x² + 8x over and you'll get the same answers!

"a" coefficient positive vs "a" coefficient negative

x² + 8x - 48 = 0
VS
-x² - 8x + 48 = 0

So all the signs are simply the opposite of each other and therefore the answers are the same

Algebra Honors ( Period 4)

Solving Quadratic Equations By Using the Quadratic Formula 9-5

Let's all sing it to "Pop Goes the Weasel!"

x = -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!

EXAMPLE: x² + 8x = 48 

FIRST DETERMINE BY USING THE DISCRIMINANT THE NUMBER OF ROOTS WE'LL FIND:

b2 - 4ac
= 64 - 4(1)(-48)
= 64 + 192
= +256

so 2 REAL ROOTS

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 (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!

if it's positive, 2 roots

if it's zero - 1 root

if it's negative - no real roots

DON'T FORGET TO PUT THE QUADRATIC IN STANDARD FORM BEFORE PLUGGING THE VALUES INTO THE QUADRATIC FORMULA!

You should know that for the quadratic formula, you don't need the "a" coefficient to be positive!

That's important if you use factoring, SQRTing each side, and completing the square.

But for the quadratic formula, either way, you'll get the same roots!

You can move the 48 over or move the x² + 8x over and you'll get the same answers!

"a" coefficient positive vs "a" coefficient negative

x² + 8x - 48 = 0
VS
-x² - 8x + 48 = 0

So all the signs are simply the opposite of each other and therefore the answers are the same

Algebra ( Period 5)

COMPLETING THE SQUARE 9-4

Now this is completely new to you!!!

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!

Why is this good? 

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.

But here's how you can make it one!

Step 1) b/2

Take half of the b coefficient (- 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 = +/- 5

Step 6) ADD 5 TO BOTH SIDES to get x by itself

x = 5 +/- 5

Step 7) Simplify if possible

x = 5 + 5 and x = 5 - 5

x = 10 and x = 0

So the 2 roots (solutions/zeros/x intercepts) are 0 and 10.

YOU DON'T NEED TO GRAPH THE PARABOLA, BUT IF YOU DID, IT WOULD CROSS THE X AXIS AT 0 AND 10. I don't know where the vertex is, but I don't need to because it's not the solution to the quadratic (although I certainly could find the vertex by using x = -b/2a AND I do know it's halfway between the roots, so the AOS = 5)

Notice that I could also factor x² - 10x = 0 to get the solution more easily. So don't complete the square if the quadratic factors easily!

IF THERE IS A "c", first move the c constant to the other side of the equation before completing the square:

x² - 10x - 11 = 0

 x² - 10x - 11 + 11 = 0 + 11

 x² - 10x = 11

NOW COMPLETE THE SQUARE AS ABOVE:

x² - 10x + 25 = +25 + 11

 (x - 5)² = 36

 SQRT (x - 5)² = SQRT 36

 x - 5 = +/- 6

 x = 5 +/- 6

x = 11 and x = -1

Again, this one factored easily so I wouldn't have even used completing the square. ALWAYS CHECK IF IT FACTORS FIRST!

 Now an example that DOES NOT FACTOR: x² - 10x - 18 = 0

x² - 10x - 18 = 0

x² - 10x - 11 + 18 = 0 + 18

x² - 10x = 18

NOW COMPLETE THE SQUARE AS ABOVE:

 x² - 10x + 25 = +25 + 18

 (x - 5)² = 43

 SQRT (x - 5)² = SQRT 43

 x - 5 = +/- SQRT 43

 x = 5 +/- SQRT 43

 x = 5 + SQRT 43 and x = 5 - SQRT 43

When there is an IRRATIONAL square root, always SIMPLIFY if possible!

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 find the completing the square term and add it to both sides:

[(-3/2)(1/2)]² = 9/16 (instead of dividing by 2, when you have a fraction, multiply by 1/2)

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

BECAUSE IT'S DIFFICULT TO COMPLETE THE SQUARE WITH AN "a" COEFFICIENT, THERE'S ONE MORE WAY TO FIND THE ROOTS (solutions, x intercepts, zeros) THAT ALWAYS WORKS!

Algebra Honors ( Period 4)

COMPLETING THE SQUARE 9-4

Now this is completely new to you!!!

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!

Why is this good? 

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.

But here's how you can make it one!

Step 1) b/2

Take half of the b coefficient (- 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 = +/- 5

Step 6) ADD 5 TO BOTH SIDES to get x by itself

x = 5 +/- 5

Step 7) Simplify if possible

x = 5 + 5 and x = 5 - 5

x = 10 and x = 0

So the 2 roots (solutions/zeros/x intercepts) are 0 and 10.

YOU DON'T NEED TO GRAPH THE PARABOLA, BUT IF YOU DID, IT WOULD CROSS THE X AXIS AT 0 AND 10. I don't know where the vertex is, but I don't need to because it's not the solution to the quadratic (although I certainly could find the vertex by using x = -b/2a AND I do know it's halfway between the roots, so the AOS = 5)

Notice that I could also factor x² - 10x = 0 to get the solution more easily. So don't complete the square if the quadratic factors easily!

IF THERE IS A "c", first move the c constant to the other side of the equation before completing the square:

x² - 10x - 11 = 0

 x² - 10x - 11 + 11 = 0 + 11

 x² - 10x = 11

NOW COMPLETE THE SQUARE AS ABOVE:

x² - 10x + 25 = +25 + 11

 (x - 5)² = 36

 SQRT (x - 5)² = SQRT 36

 x - 5 = +/- 6

 x = 5 +/- 6

x = 11 and x = -1

Again, this one factored easily so I wouldn't have even used completing the square. ALWAYS CHECK IF IT FACTORS FIRST!

 Now an example that DOES NOT FACTOR: x² - 10x - 18 = 0

x² - 10x - 18 = 0

x² - 10x - 11 + 18 = 0 + 18

x² - 10x = 18

NOW COMPLETE THE SQUARE AS ABOVE:

 x² - 10x + 25 = +25 + 18

 (x - 5)² = 43

 SQRT (x - 5)² = SQRT 43

 x - 5 = +/- SQRT 43

 x = 5 +/- SQRT 43

 x = 5 + SQRT 43 and x = 5 - SQRT 43

When there is an IRRATIONAL square root, always SIMPLIFY if possible!

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 find the completing the square term and add it to both sides:

[(-3/2)(1/2)]² = 9/16 (instead of dividing by 2, when you have a fraction, multiply by 1/2)

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

BECAUSE IT'S DIFFICULT TO COMPLETE THE SQUARE WITH AN "a" COEFFICIENT, THERE'S ONE MORE WAY TO FIND THE ROOTS (solutions, x intercepts, zeros) THAT ALWAYS WORKS!

Algebra (Period 5)

Perfect Squares 8-9 
SOLVING QUADRATICS USING SQUARE ROOTING

We’ve been solving all quadratics with GRAPHING or the ZERO PRODUCTS PROPERTY after we get our happy face on and factor.

There are several other ways to solve 2nd degree equations:
Square rooting both sides – Chapter 8-9 Example 4 on p. 525 with Chapter 10-2
Completing the square and then square rooting both sides - Chapter 9-4
The Quadratic Formula – Chapter 9-5
(FYI: we’ll be skipping Chapter 9-6 and 9-7)

Think about a simple equation like x² = 36.
We could get our happy face on and factor:
x² – 36 = 0   (x + 6)(x-6) = 0
x = -6 and x = 6

However, there’s a much easier way to solve this quadratic: simply take the square root of each side!
x² = 36
√ x² = ±√36

WHY ±?
You need both of the roots (solutions, zeros, x-intercepts).
Why does this work so well?
There are only TWO TERMS: the quadratic term and the constant term…the linear x term is missing.

Consider a different equation such as x² + 5x = 36
Notice that I can no longer just take the square root of each side.
I would have to use the Zero Products Property:
x² + 5x -36 = 0
(x + 9)(x – 4) = 0
x = -9 and x = 4
If you have a BINOMIAL SQUARED, you can also use this approach:

(5x – 7)² = 16
√(5x – 7)² = ±√16
5x – 7 = ±4

5x – 7 = 4
and
5x – 7 = -4
solve
x = 11/5 and x = 3/5

If you have a TRINOMIAL SQUARE, factor it to a BINOMIAL SQUARED, and then use this approach:

25x² + 40x +16 = 49
(5x + 4)² = 49
√(5x + 4)² = ±√49
5x + 4 = ±7
5x + 4 = -7 and 5x + 4 = 7
solve both
x = -11/5 and x = 3/5

This works because you still have only 2 terms, one on each side, that you can take the square root of.

WHAT HAPPENS IF THE CONSTANT IS NOT A PERFECT SQUARE?
You’ll get an IRRATIONAL ANSWER.

There are 2 ways to give that answer.
1. Use the square root button on a calculator and ROUND (usually to the tenths or hundredths)
2. SIMPLIFY the square root (SEE ABOVE CHAPTER 10-2)

Algebra Honors ( Period 4)

Perfect Squares 8-9 
SOLVING QUADRATICS USING SQUARE ROOTING

We’ve been solving all quadratics with GRAPHING or the ZERO PRODUCTS PROPERTY after we get our happy face on and factor.

There are several other ways to solve 2nd degree equations:
Square rooting both sides – Chapter 8-9 Example 4 on p. 525 with Chapter 10-2
Completing the square and then square rooting both sides - Chapter 9-4
The Quadratic Formula – Chapter 9-5
(FYI: we’ll be skipping Chapter 9-6 and 9-7)

Think about a simple equation like x² = 36.
We could get our happy face on and factor:
x² – 36 = 0   (x + 6)(x-6) = 0
x = -6 and x = 6

However, there’s a much easier way to solve this quadratic: simply take the square root of each side!
x² = 36
√ x² = ±√36

WHY ±?
You need both of the roots (solutions, zeros, x-intercepts).
Why does this work so well?
There are only TWO TERMS: the quadratic term and the constant term…the linear x term is missing.

Consider a different equation such as x² + 5x = 36
Notice that I can no longer just take the square root of each side.
I would have to use the Zero Products Property:
x² + 5x -36 = 0
(x + 9)(x – 4) = 0
x = -9 and x = 4
If you have a BINOMIAL SQUARED, you can also use this approach:

(5x – 7)² = 16
√(5x – 7)² = ±√16
5x – 7 = ±4

5x – 7 = 4
and
5x – 7 = -4
solve
x = 11/5 and x = 3/5

If you have a TRINOMIAL SQUARE, factor it to a BINOMIAL SQUARED, and then use this approach:

25x² + 40x +16 = 49
(5x + 4)² = 49
√(5x + 4)² = ±√49
5x + 4 = ±7
5x + 4 = -7 and 5x + 4 = 7
solve both
x = -11/5 and x = 3/5

This works because you still have only 2 terms, one on each side, that you can take the square root of.

WHAT HAPPENS IF THE CONSTANT IS NOT A PERFECT SQUARE?
You’ll get an IRRATIONAL ANSWER.

There are 2 ways to give that answer.
1. Use the square root button on a calculator and ROUND (usually to the tenths or hundredths)
2. SIMPLIFY the square root (SEE ABOVE CHAPTER 10-2)

Algebra (Period 5)

Simplifying Radical Expressions 10-2

SQUARE ROOTS (RADICALS)

MAIN CONCEPT:

Square rooting "undoes" squaring!   It's the inverse operation!!!... Just as subtraction undoes addition     Just as division undoes multiplication

If you square a square root:
(√243)² = 243 (what you started with)

If you square root something squared:√243² = 243 (what you started with)
If you multiply a square root by the same square root:
(√243)(√243) = 243 (what you started with)

IN SUMMARY:

(√243)² = √243² = (√243)(√243) = 243

1) RADICAL sign: The root sign, which looks like a check mark.

If there is no little number on the radical, you assume it's the square root
But many times there will be a number there and then you are finding the root that the number says.

For example, if there is a 3 in the "check mark," you are finding the cubed root.
One more example: The square root of 64 is 8. The cubed root of 64 is 4. The 6th root of 64 is 2.

2)RADICAND : Whatever is under the RADICAL sign
In the example above, 64 was the radicand in every case.

3) ROOT (the answer): the number/variable that was squared (cubed, raised to a power  to get the RADICAND (whatever is under the radical sign) In the example above, the roots were 8, 4, and 2.

4) SQUARE ROOTS: (What we primarily cover in Algebra I) The number that is squared to get to the radicand.

Every POSITIVE number has 2 square roots - one positive and one negative.
Example: The square root of 25 means what number squared = 25
Answer: Either positive 5 squared OR negative 5 squared = 25

5) PRINCIPAL SQUARE ROOT: The positive square root.

Generally, we assume the answer is the principal square root unless there is a negative sign in front of the radical sign or you’re solving to find both roots for a quadratic (We’ll deal with that in Chapter 13-2).

6) ± sign in front of the root denotes both the positive and negative roots at one time!

Example: √ 25 = ±5

7) ORDER OF OPERATIONS with RADICALS: Radicals function like parentheses when there is an operation under the radical. In other words, if there is addition under the radical, you must do that first (like you would do parentheses first) before finding the root.

EXAMPLE: √ (36 + 64) = 10 not 14!!!!

First add 36 + 64 = 100

Then find √100 = 10

Radicals by themselves function as exponents in order of operations
(that makes sense because they undo exponents).
Actually, roots are FRACTIONAL EXPONENTS!
Square roots = 1/2 power,
Cubed roots = 1/3 power,
Fourth roots = 1/4 power, etc.

So √25 = 25^½ = 5

EXAMPLE: 3 + 4√25
you would do powers first...in this case square root of 25 first!|
3 + 4(5)
Now do the multiplication
3 + 20
Now do the addition
23

8) THE SQUARE ROOT OF ANYTHING SQUARED IS ITSELF!!!

EXAMPLE: √ 5² = 5

√ (a -7)² = a - 7

RATIONAL SQUARE ROOTS: Square roots of perfect squares are RATIONAL

REVIEW OF NUMBER SYSTEMS:

Rational numbers are decimals that either terminate or repeat

which means they can be restated into a RATIO a/b of two integers a and b where b is not zero.

Natural numbers: 1, 2, 3, ... are RATIOnal because you can put them over 1

Whole numbers: 0, 1, 2, 3,....are RATIOnal because you can put them over 1

Integers: ....-3, -2, -1, 0, 1, 2, 3,....are RATIOnal because you can put them over 1

Rational numbers = natural, whole, integers PLUS all the bits and pieces in between that can be expressed as repeating or terminating decimals: 2/3, .6, -3.2, -10.7 bar, etc.

Real numbers: all of these!

In Algebra II you will find out that there are Imaginary Numbers!
Square roots of NEGATIVE numbers are IMAGINARY

IRRATIONAL SQUARE ROOTS:  Square roots of a nonperfect squares are IRRATIONAL -

They cannot be stated as the ratio of two integers -
As decimals, they never terminate and never repeat -
you round them and use approximately sign.

MOST FAMOUS OF ALL IRRATIONAL NUMBERS IS PI!

  

note-- Square roots are MOSTLY IRRATIONAL!

There are fewer perfect squares than nonperfect!

Here are some perfect squares: 0, 4, 9, 16, 25, 36, etc.

PERFECT SQUARES CAN ALSO BE TERMINATING DECIMALS!

EXAMPLE: √( .04) is rational because it is ± .2

But all the square roots in between these perfect squares are IRRATIONAL

For example, the square root of 2, the square root of 3, the square root of 5, etc.
You can estimate irrational square roots.

For example, the square root of 50 is close to 7 because the square root of 49 is 7.
You can estimate that the square root of 50 is 7.1 and then square 7.1 to see what you get. 

If that's too much, try 7.05 and square that.
This works much better with a calculator! 

And obviously, a calculator will give you irrational square roots to whatever place your calculator goes to.
Remember: These will never end or repeat

(even though your calculator only shows a certain number of places physically!)

SIMPLIFYING RADICALS

SIMPLIFYING NONPERFECT NUMBERS UNDER THE RADICAL:

A simplified radical expression is one where there is no perfect square left under the radical sign
You can factor the expression under the radical to find any perfect squares in the number:

EXAMPLE: √50 = √(25 * 2)

Next, simplify the SQRT of the perfect square and leave the nonperfect factor under the radical:

√(25 * 2) = √25 * √2 = 5√2

We usually read the answer as "5 rad 2"

HELPFUL HINTS:
When you are factoring the radicand,
you're looking for the LARGEST PERFECT SQUARE
that is a FACTOR of the radicand.
So start with:
Does 4 go into it?
Does 9 go into it?
Does 16 go into it?
Does 25 go into it?
etc.

Another method: Inverted Division or Factor Trees
Factor the radicand completely into its prime factors (remember this from Pre-Algebra?)
Find the prime factorization either way in order from least to greatest.
Circle factors in PAIRS
Every time you have a pair, you have a factor that is squared!
Then, you can take that factor out of the radical sign.
Remember that you are just taking one of those factors out!

Example: √ 250
Prime factorization = 2 x 5 x 5 x 5
Circle the first two 5's
5 x 5 is 25 and so you can take the square root of 25 = 5 out of the radicand
Everything else is not in a pair (squared) so it must remain under the radical

Final answer: √ 250 = 5√(2 x 5) = 5 √10

If there is a number in front of the radicand, simply multiply it by whatever you take out.

Algebra Honors ( Period 4)

Simplifying Radical Expressions 10-2

SQUARE ROOTS (RADICALS)

MAIN CONCEPT:

Square rooting "undoes" squaring!   It's the inverse operation!!!... Just as subtraction undoes addition     Just as division undoes multiplication

If you square a square root:
(√243)² = 243 (what you started with)

If you square root something squared:√243² = 243 (what you started with)
If you multiply a square root by the same square root:
(√243)(√243) = 243 (what you started with)

IN SUMMARY:

(√243)² = √243² = (√243)(√243) = 243

1) RADICAL sign: The root sign, which looks like a check mark.

If there is no little number on the radical, you assume it's the square root
But many times there will be a number there and then you are finding the root that the number says.

For example, if there is a 3 in the "check mark," you are finding the cubed root.
One more example: The square root of 64 is 8. The cubed root of 64 is 4. The 6th root of 64 is 2.

2)RADICAND : Whatever is under the RADICAL sign
In the example above, 64 was the radicand in every case.

3) ROOT (the answer): the number/variable that was squared (cubed, raised to a power  to get the RADICAND (whatever is under the radical sign) In the example above, the roots were 8, 4, and 2.

4) SQUARE ROOTS: (What we primarily cover in Algebra I) The number that is squared to get to the radicand.

Every POSITIVE number has 2 square roots - one positive and one negative.
Example: The square root of 25 means what number squared = 25
Answer: Either positive 5 squared OR negative 5 squared = 25

5) PRINCIPAL SQUARE ROOT: The positive square root.

Generally, we assume the answer is the principal square root unless there is a negative sign in front of the radical sign or you’re solving to find both roots for a quadratic (We’ll deal with that in Chapter 13-2).

6) ± sign in front of the root denotes both the positive and negative roots at one time!

Example: √ 25 = ±5

7) ORDER OF OPERATIONS with RADICALS: Radicals function like parentheses when there is an operation under the radical. In other words, if there is addition under the radical, you must do that first (like you would do parentheses first) before finding the root.

EXAMPLE: √ (36 + 64) = 10 not 14!!!!

First add 36 + 64 = 100

Then find √100 = 10

Radicals by themselves function as exponents in order of operations
(that makes sense because they undo exponents).
Actually, roots are FRACTIONAL EXPONENTS!
Square roots = 1/2 power,
Cubed roots = 1/3 power,
Fourth roots = 1/4 power, etc.

So √25 = 25^½ = 5

EXAMPLE: 3 + 4√25
you would do powers first...in this case square root of 25 first!|
3 + 4(5)
Now do the multiplication
3 + 20
Now do the addition
23

8) THE SQUARE ROOT OF ANYTHING SQUARED IS ITSELF!!!

EXAMPLE: √ 5² = 5

√ (a -7)² = a - 7

RATIONAL SQUARE ROOTS: Square roots of perfect squares are RATIONAL

REVIEW OF NUMBER SYSTEMS:

Rational numbers are decimals that either terminate or repeat

which means they can be restated into a RATIO a/b of two integers a and b where b is not zero.

Natural numbers: 1, 2, 3, ... are RATIOnal because you can put them over 1

Whole numbers: 0, 1, 2, 3,....are RATIOnal because you can put them over 1

Integers: ....-3, -2, -1, 0, 1, 2, 3,....are RATIOnal because you can put them over 1

Rational numbers = natural, whole, integers PLUS all the bits and pieces in between that can be expressed as repeating or terminating decimals: 2/3, .6, -3.2, -10.7 bar, etc.

Real numbers: all of these!

In Algebra II you will find out that there are Imaginary Numbers!
Square roots of NEGATIVE numbers are IMAGINARY

IRRATIONAL SQUARE ROOTS:  Square roots of a nonperfect squares are IRRATIONAL -

They cannot be stated as the ratio of two integers -
As decimals, they never terminate and never repeat -
you round them and use approximately sign.

MOST FAMOUS OF ALL IRRATIONAL NUMBERS IS PI!

  

note-- Square roots are MOSTLY IRRATIONAL!

There are fewer perfect squares than nonperfect!

Here are some perfect squares: 0, 4, 9, 16, 25, 36, etc.

PERFECT SQUARES CAN ALSO BE TERMINATING DECIMALS!

EXAMPLE: √( .04) is rational because it is ± .2

But all the square roots in between these perfect squares are IRRATIONAL

For example, the square root of 2, the square root of 3, the square root of 5, etc.
You can estimate irrational square roots.

For example, the square root of 50 is close to 7 because the square root of 49 is 7.
You can estimate that the square root of 50 is 7.1 and then square 7.1 to see what you get. 

If that's too much, try 7.05 and square that.
This works much better with a calculator! 

And obviously, a calculator will give you irrational square roots to whatever place your calculator goes to.
Remember: These will never end or repeat

(even though your calculator only shows a certain number of places physically!)

SIMPLIFYING RADICALS

SIMPLIFYING NONPERFECT NUMBERS UNDER THE RADICAL:

A simplified radical expression is one where there is no perfect square left under the radical sign
You can factor the expression under the radical to find any perfect squares in the number:

EXAMPLE: √50 = √(25 * 2)

Next, simplify the SQRT of the perfect square and leave the nonperfect factor under the radical:

√(25 * 2) = √25 * √2 = 5√2

We usually read the answer as "5 rad 2"

HELPFUL HINTS:
When you are factoring the radicand,
you're looking for the LARGEST PERFECT SQUARE
that is a FACTOR of the radicand.
So start with:
Does 4 go into it?
Does 9 go into it?
Does 16 go into it?
Does 25 go into it?
etc.

Another method: Inverted Division or Factor Trees
Factor the radicand completely into its prime factors (remember this from Pre-Algebra?)
Find the prime factorization either way in order from least to greatest.
Circle factors in PAIRS
Every time you have a pair, you have a factor that is squared!
Then, you can take that factor out of the radical sign.
Remember that you are just taking one of those factors out!

Example: √ 250
Prime factorization = 2 x 5 x 5 x 5
Circle the first two 5's
5 x 5 is 25 and so you can take the square root of 25 = 5 out of the radicand
Everything else is not in a pair (squared) so it must remain under the radical

Final answer: √ 250 = 5√(2 x 5) = 5 √10

If there is a number in front of the radicand, simply multiply it by whatever you take out.