Algebra (Period 4)

The Quadratic Formula: 13-4

Make sure to read the first post on the Quadratic Formula BEFORE or AFTER you watch these You Tube Videos....

How about creating your own video!! (For extra Credit!!)

Now if Emily can recite the Quadratic Formula after singing Twinkle, Twinkle Little Star... shouldn't you?

Algebra (Period 4)

The Quadratic Formula: 13-4

Method 5:
THE QUADRATIC FORMULA

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

x =
-b ± √ (b² - 4ac)
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!


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!


EXAMPLE: x² + 8x = 48 


You can move the 48 over or move the x² + 8x over and you'll get the same answers!
Let's look at:


"a" coefficient positive *****vs***** "a" coefficient negative



x² + 8x - 48 = 0 ****VS**** -x² - 8x + 48 = 0


First x² + 8x - 48 = 0
-8 ± √[64 - 4(1)(-48)]

2(1)



-8 ± √(64 + 192)

2



-8 ± √(256)
2



-8 ± 16
2



(-8 + 16)/2 and (-8 - 16)/2



8/2 and -24/2
4 and -12

Now let's compare -x² - 8x + 48 = 0


8 ± √[64 - 4(-1)(48)]

-2(1)

8 ± √(64 + 192)

-2

8 ± √(256)
-2


8 ± 16
-2

(8 + 16)/-2 and (8 - 16)/-2
24/-2 and -8/-2
-12 and 4

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

Algebra (Period 4)

Completing the Square: 13-3
METHOD 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

That is, take half of the b coefficient ,
or in this case (- 10/2 = -5)


Step 2) Square b/2

(b/2)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

√(x - 5)² = √ 25

x - 5 = ± 5

Step 6) ADD 5 TO BOTH SIDES

x - 5 = ± 5
+ 5 = +5
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)
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
√ (x - 5)² = √ 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
√ (x - 5)² = √ 43
x - 5 = ± √ 43
x = 5 ± √ 43
x = 5 + √ 43 and x = 5 - √ 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)(-3/2)]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
√[(x - 3/4)² ] = ±√ [17/16]

x - 3/4 = ±[√17] /4

x = 3/4 ±[√ 17] /4

x = (3 ± [√17]) /4
or written better x =
3 ± √ 17
4