Algebra Honors (Periods 5 & 6)

Completing the Square: 12-2
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! We saw that in the previous section.
perfect square = k ( when k ≥ 0)

So what if that side is not a perfect BINOMIAL SQUARED?

It may be possible to trasform it into one... by COMPLETING THE SQUARE
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 - 3x -18 = 0

(Head's up-- I wouldn't use this method here because I can see that it factors easily... into (x-6)(x+3) = 0 so the solution set is {-3,6}
However, knowing what I need to get for my solutions might be a good way to practice Completing the square..
so

x² - 3x = 18

Step 1) b/2

That is, take half of the b coefficient ,
or in this case (- 3/2)



Step 2) Square b/2

(b/2)²
(-3/2 ⋅ -3/2 = 9/4)


Step 3) Add (b/2)² to both sides of the equation

x² - 3x + 9/4 = 18 +9/4

Step 4) Factor to a binomial square

(x - 3/2)² = 18 + 9/4
 = 81/4


Step 5) Square root each side and solve

√(x - 3/2)² = √ 81/4

x - 3/2 = ± 9/2
x = 3/2 ± 9/2
x = 3/2 + 9/2 AND x = 3/2 - 9/2
x = 6 and x = -3
{-3, 6}
We got the same solutions !! Yay!!


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
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