Wize High School Algebra I Textbook (Common Core) > Completing the Square

Completing the Square
Given a quadratic equation in standard form y=x2+bx+cy=x^2+bx+cy=x2+bx+c, we often want to rewrite it in vertex form y=(x−h)2+ky=(x-h)^2+ky=(x−h)2+k. This process is called completing the square.

Steps for Completing the Square for y=x2+bx+c\bco{y=x^2+bx+c}y=x2+bx+c
Focus on the first 2 terms: y=[x2+bx]+cy=[\colorbox{yellow}{$x^2+bx$}]+cy=[x2+bx​]+c
Add and subtract (b2)2\left(\dfrac{b}{2}\right)^2(2b​)2: y=[x2+bx+(b2)2−(b2)2]+cy=\left[x^2+bx\colorbox{yellow}{$+\left(\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^2$}\right]+cy=​x2+bx+(2b​)2−(2b​)2​​+c
Rewrite (factor) the first 3 terms as a perfect square: y=[(x+b2)2−(b2)2]+cy=\left[\colorbox{yellow}{$\left(x+\dfrac{b}{2}\right)^2$}-\left(\dfrac{b}{2}\right)^2\right]+cy=​(x+2b​)2​−(2b​)2​+c
Simplify the constant terms: y=(x+b2)2−(b2)2+cy=\left(x+\dfrac{b}{2}\right)^2\colorbox{yellow}{$-\left(\dfrac{b}{2}\right)^2+c$}y=(x+2b​)2−(2b​)2+c​

Write it Down
Short-cut Formula for Completing the Square for y=x2+bx+cy=x^2+bx+cy=x2+bx+c:
x2+bx+c=(x+b2)2−(b2)2+c\boxed{x^2+bx+c=\left(x+\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^2+c}x2+bx+c=(x+2b​)2−(2b​)2+c​

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Example: Completing the Square
Write y=x2−12x+7y=x^2-12x+7y=x2−12x+7 in vertex form by completing the square.
1. Focus on the first 2 terms: y=[x2+bx]+c\bco{y=[\colorbox{yellow}{$x^2+bx$}]+c}y=[x2+bx​]+c

y=[x2−12x]+7y=[x^2-12x]+7y=[x2−12x]+7

2. Add and subtract (b2)2\bco{\left(\dfrac{b}{2}\right)^2}(2b​)2: y=[x2+bx+(b2)2−(b2)2]+c\bco{y=\left[x^2+bx\colorbox{yellow}{$+\left(\dfrac{b}{2}\right)^2-\left(\dfrac{b}{2}\right)^2$}\right]+c}y=​x2+bx+(2b​)2−(2b​)2​​+c

In our equation b=−12b=-12b=−12 
b2=−122=−6\dfrac{b}{2}=\dfrac{-12}{2}=-62b​=2−12​=−6
(b2)2=(−6)2=36\left(\dfrac{b}{2}\right)^2=(-6)^2=36(2b​)2=(−6)2=36
y=[x2−12x+36−36]+7y=\left[x^2-12x+36-36\right]+7y=[x2−12x+36−36]+7

3. Rewrite (factor) the first 3 terms as a perfect square: y=[(x+b2)2−(b2)2]+c\bco{y=\left[\colorbox{yellow}{$\left(x+\dfrac{b}{2}\right)^2$}-\left(\dfrac{b}{2}\right)^2\right]+c}y=​(x+2b​)2​−(2b​)2​+c

y=[(x−6)2−36]+7
y=\left[\left(x-6\right)^2-36\right]+7
y=[(x−6)2−36]+7

4. Simplify the constant terms: y=(x+b2)2−(b2)2+c\bco{y=\left(x+\dfrac{b}{2}\right)^2
\colorbox{yellow}{$-\left(\dfrac{b}{2}\right)^2+c$}}y=(x+2b​)2−(2b​)2+c​

y=(x−6)2−36+7y=(x−6)2−29\begin{aligned}
y&=\left(x-6\right)^2-36+7\\
y&=(x-6)^2-29
\end{aligned}
yy​=(x−6)2−36+7=(x−6)2−29​

So, by completing the square, we get the vertex form equation y=(x−6)2−29y=(x-6)^2-29y=(x−6)2−29

Practice: Completing the Square
Write y=x2+10x−14y=x^2+10x-14y=x2+10x−14 in vertex form by completing the square. Then determine the vertex of this quadratic relation.

Enter the vertex form of the equation in the form

y=(x-h)^2+k

Enter the vertex in the form

(x,y)

I don't know

Practice: Completing the Square
Write y=x2−18x−11y=x^2-18x-11y=x2−18x−11 in vertex form by completing the square. Then, graph this quadratic relation.

Enter the vertex form equation in the form

y=(x-h)^2+k

Check the solution to see if you graphed it properly.

I don't know

Practice: Completing the Square
Write y=x2−11x−6y=x^2-11x-6y=x2−11x−6 in vertex form by completing the square. Then, determine the vertex and y-intercept of this quadratic relation.

Enter the vertex form equation in the form

y=(x-h)^2+k

Enter the vertex in the form

(x,y)

Enter the y-intercept as a point in the form

(x,y)

I don't know

Practice: Completing the Square
Write the following in vertex form by completing the square.

a) x2+8xx^2+8xx2+8x

b) x2+16x−1x^2+16x-1x2+16x−1

c) x2−6x+17x^2-6x+17x2−6x+17

d) x2+9x+7x^2+9x+7x2+9x+7

e) x2−21x−3x^2-21x-3x2−21x−3

a) Enter the vertex form in the form

(x-h)^2+k

b) Enter the vertex form in the form

(x-h)^2+k

c) Enter the vertex form in the form

(x-h)^2+k

d) Enter the vertex form in the form

(x-h)^2+k

e) Enter the vertex form in the form

(x-h)^2+k

I don't know

Wize High School Algebra I Textbook (Common Core) > Completing the Square

Completing the Square ( $x^2+bx+c$ )

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