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 0 = 2X2 + 4X + 8 #2 Set Y = 0 (2) #3 Make the coefficient of X2 equal to 1 by factoring, NOT division. (Don’t forget about this factor later) 0 = X2 + 2X + 4 -4 -4 #4 Move the “C” value to make way for a value that makes a perfect square. -4 = X2 + 2X

#4 Move the “C” value to make way for a value that makes a perfect square. -4 = X2 + 2X +1 +1 #5 Find the new value of “C” by the following method: -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

#6 We now have a perfect square on the left, so change it to the factored form. -3 = (X+1)2 +3 +3 #7 Now set the equation back to zero. 0 = (X+1)2 +3 (2) #8 Did you factor in the beginning? Better distribute it back. 0 = 2(X+1)2 +6 Y = 2(X+1)2 +6 #9 We set Y = 0, so 0 = Y We did it, we completed the square and converted the equation into Vertex Form!

So …. What’s so great about that?

Our Vertex Form Equation Y = 2(X + 1)2 + 6 The Format of Vertex Form Y = a(X - h)2 +k 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). Note: “a” is not the same “a” value of the standard form.

Finding the Roots Given Set Y =0 Subtract 6 from Both Sides Divide Both Sides by 2 Take the Square Root of Both Sides Subtract 1 from Both Sides The Roots are &

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.