Example of Completing the Square

Completing the Square

Here is how to Solve by Completing the Square.
The Quadratic Equation in Standard Form is
\(\boxed{ y=x^2+bx+c }\)

To Solve by Completing the Square we add and subtract
\( \textcolor{blue}{({b \over 2})^2} \) which yields:
\(\boxed{ y=x^2+bx+ \textcolor{blue}{({b \over 2})^2} + c – \textcolor{blue}{({b \over 2})^2} }\)

Completing the Square yields:
\(\boxed{ y=(x+b/2)^2 + c – ({b \over 2})^2 }\)

which is the Complete the Square Formula.

Another Example of Completing the Square

We are given the Quadratic Equation below in Standard Form
\( y=x^2- 6x-7 \) .

First, add and subtract
\( ({ – b \over 2a})^2= ({ -(-6)\over (2*1)})^2 = 3^2=9 \) .

Thus, we have:
\( y=(x^2- 6x+ \textcolor{blue}{9})-7- \textcolor{blue}{9} \)

This allows us to Solve via Completing the Square:
\( y=(x-3)^2-16 \) .

To solve \( (x-3)^2-16 =0 \) we first add 16:

\( (x-3)^2=16 \) . Next, we take the Square Root:

\( x-3=\pm 4 \) . Adding 3 to \( \pm 4 \) yields the 2 solutions:

\( x=7 , x=-1 \) .

Easy, wasn’t it?

Tip: When using the Complete the Square Solver to solve
\(x^2-6x-7=0\) we must enter the 3 coefficients as
\(a=1, b=-6\) and \(c=-7\).

Then, the Solver will first Complete the Square to find the Vertex \( (h,k)=(3,-16) \) .

Thus, the Vertex Form of the Parabola is \( y=(x-3)^2-16 \) .

Solving \( (x-3)^2-16=0 \) yields the two zeros: \( x=7 , x=-1 \) .

Get it now? Try the above Complete the Square Solver again.