Complete The Square

#1 Make sure your equation is in standard form:

Y = aX2+bX+c or 0 = ax2+bx+c

Y = 2X2 + 4X + 8

#2 Set Y = 0 0 = 2X2 + 4X + 8

#3 Make the coefficient of X2 equal to 1 by factoring, NOT division. (Don’t forget about this factor later)

-4 = X2 + 2X

-4 -4

#4 Move the “C” value to make way for a value that makes a perfect square.

(2)

0 = X2 + 2X + 4

-4 = X2 + 2X #4 Move the “C” value to make way for a value that makes a perfect square.

#5 Find the new value of “C” by the following method:

2

2

b

12

22

+1 +1

-3 = X2 + 2X +1

#6 We now have a perfect square on the right, so change it to the factored form.

Hint: It’s just

-3 = (X+1)2

2

2

bx

#6 We now have a perfect square on the left, so change it to the factored form.

-3 = (X+1)2

#7 Now set the equation back to zero.

0 = (X+1)2 +3

+3 +3

#8 Did you factor in the beginning? Better distribute it back.

(2)

0 = 2(X+1)2 +6

#9 We set Y = 0, so 0 = Y Y = 2(X+1)2 +6

We did it, we completed the square and converted the equation into Vertex Form!

So …. What’s so great about that?

Y = 2(X + 1)2 + 6

Y = a(X - h)2 +k

Our Vertex Form Equation

The Format of Vertex Form

Note: “a” is not the same “a” value of the standard form.

Since our “a” is positive, the graph is going up!

(h,k) is the vertex! So our vertex is: (-1,6)

Set Y = 0 and solve for X to find the Roots! (Where the graph intercepts the X-Axis).

6)1(2 2 xy

6)1(20 2 x2)1(26 x2)1(3 x

13 x

x 31

Subtract 6 from Both Sides

Given

Set Y =0

Finding the Roots

Divide Both Sides by 2

Take the Square Root of Both Sides

Subtract 1 from Both Sides

The Roots are )0,31( )0,31( &

Since you can not take an even root of a negative number, the roots are imaginary. This means our graph never goes through the x-axis.