Wize High School Grade 10 Math Textbook > Completing the Square

Completing the Square
Given a quadratic equation in standard form y=ax2+bx+cy=ax^2+bx+cy=ax2+bx+c, we often want to rewrite it in vertex form y=a(x−h)2+ky=a(x-h)^2+ky=a(x−h)2+k. This process is called completing the square.

Steps for Completing the Square for y=ax2+bx+c\bco{y=ax^2+bx+c}y=ax2+bx+c
Factor aaa out of the first 2 terms: y=a[x2+bax]+cy=\colorbox{yellow}{$a$}\left[x^2+\dfrac{b}{a}x\right]+cy=a​[x2+ab​x]+c
Add and subtract (b2a)2\left(\dfrac{b}{2a}\right)^2(2ab​)2: y=a[x2+bax+(b2a)2−(b2a)2]+cy=a\left[x^2+\dfrac{b}{a}x\colorbox{yellow}{$+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2$}\right]+cy=a​x2+ab​x+(2ab​)2−(2ab​)2​​+c
Rewrite (factor) the first 3 terms as a perfect square: y=a[(x+b2a)2−(b2a)2]+cy=a\left[\colorbox{yellow}{$\left(x+\dfrac{b}{2a}\right)^2$}-\left(\dfrac{b}{2a}\right)^2\right]+cy=a​(x+2ab​)2​−(2ab​)2​+c
Multiply aaa into the brackets: y=a(x+b2a)2−a(b2a)2+cy=\colorbox{yellow}{$a$}\left(x+\dfrac{b}{2a}\right)^2-\colorbox{yellow}{$a$}\left(\dfrac{b}{2a}\right)^2+cy=a​(x+2ab​)2−a​(2ab​)2+c
Simplify the constant terms: y=a(x+b2a)2−a(b2a)2+cy=a\left(x+\dfrac{b}{2a}\right)^2\colorbox{yellow}{$-a\left(\dfrac{b}{2a}\right)^2+c$}y=a(x+2ab​)2−a(2ab​)2+c​

Write it Down
Short-cut Formula for Completing the Square for ax2+bx+cax^2+bx+cax2+bx+c:
ax2+bx+c=a(x+b2a)2−a(b2a)2+c\boxed{ax^2+bx+c=a\left(x+\dfrac{b}{2a}\right)^2-a\left(\dfrac{b}{2a}\right)^2+c}ax2+bx+c=a(x+2ab​)2−a(2ab​)2+c​

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Example: Completing the Square
Write y=2x2−12x+11y=2x^2-12x+11y=2x2−12x+11 in vertex form by completing the square.

1. Factor a\bco{a}a out of the first 2 terms: y=a[x2+bax]+c\bco{y=\colorbox{yellow}{$a$}\left[x^2+\dfrac{b}{a}x\right]+c}y=a​[x2+ab​x]+c

y=2[x2−6x]+11y=2[x^2-6x]+11y=2[x2−6x]+11

2. Add and subtract (b2a)2\bco{\left(\dfrac{b}{2a}\right)^2}(2ab​)2: y=a[x2+bax+(b2a)2−(b2a)2]+c\bco{y=a\left[x^2+\dfrac{b}{a}x\colorbox{yellow}{$+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2$}\right]+c}y=a​x2+ab​x+(2ab​)2−(2ab​)2​​+c

ba=−6\dfrac{b}{a}=-6ab​=−6
b2a=−62=−3\dfrac{b}{2a}=\dfrac{-6}{2}=-32ab​=2−6​=−3
(b2a)2=(−3)2=9\left(\dfrac{b}{2a}\right)^2=(-3)^2=9(2ab​)2=(−3)2=9
y=2[x2−6x+9−9]+11y=2\left[x^2-6x+9-9\right]+11y=2[x2−6x+9−9]+11

3. Rewrite (factor) the first 3 terms as a perfect square: y=a[(x+b2a)2−(b2a)2]+c\bco{y=a\left[\colorbox{yellow}{$\left(x+\dfrac{b}{2a}\right)^2$}-\left(\dfrac{b}{2a}\right)^2\right]+c}y=a​(x+2ab​)2​−(2ab​)2​+c

y=2[(x−3)2−9]+11y=2[(x-3)^2-9]+11y=2[(x−3)2−9]+11

4. Multiply a\bco{a}a into the brackets: y=a(x+b2a)2−a(b2a)2+c\bco{y=\colorbox{yellow}{$a$}\left(x+\dfrac{b}{2a}\right)^2-\colorbox{yellow}{$a$}\left(\dfrac{b}{2a}\right)^2+c}y=a​(x+2ab​)2−a​(2ab​)2+c

y=2(x−3)2+2(−9)+11y=2(x-3)^2+2(-9)+11y=2(x−3)2+2(−9)+11

5. Simplify the constant terms: y=a(x+b2a)2−a(b2a)2+c\bco{y=a\left(x+\dfrac{b}{2a}\right)^2\colorbox{yellow}{$-a\left(\dfrac{b}{2a}\right)^2+c$}}y=a(x+2ab​)2−a(2ab​)2+c​

y=2(x−3)2−18+11y=2(x−3)2−7\begin{array}{rcl}
y&=&2(x-3)^2-18+11\\
y&=&2(x-3)^2-7
\end{array}yy​==​2(x−3)2−18+112(x−3)2−7​

Practice: Completing the Square
Write y=3x2+18x−1y=3x^2+18x-1y=3x2+18x−1 in vertex form by completing the square.

Enter the vertex form equation in the form

y=(x-h)^2+k

I don't know

Practice: Completing the Square
Write y=−4x2+8x−5y=-4x^2+8x-5y=−4x2+8x−5 in vertex form by completing the square. Then, determine the vertex 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)

I don't know

Practice: Completing the Square
Write y=2x2+5x+11y=2x^2+5x+11y=2x2+5x+11 in vertex form by completing the square. Then state the vertex and graph this quadratic relation.

Enter the vertex form equation in the form

y=(x-h)^2+k

Enter the vertex in the form

(x,y)

Check the solution to see if you graphed it properly

I don't know

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

a) 2x2+8x2x^2+8x2x2+8x

b) 4x2+16x−14x^2+16x-14x2+16x−1

c) 3x2−6x+173x^2-6x+173x2−6x+17

d) 3x2+9x+73x^2+9x+73x2+9x+7

e) 5x2−21x−35x^2-21x-35x2−21x−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 Grade 10 Math Textbook > Completing the Square

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

Popular Courses

Completing the Square

Steps for Completing the Square for $\bco{y=ax^2+bx+c}$

Example: Completing the Square

1. Factor $\bco{a}$ out of the first 2 terms: $\bco{y=\colorbox{yellow}{$a$}\left[x^2+\dfrac{b}{a}x\right]+c}$

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

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

4. Multiply $\bco{a}$ into the brackets: $\bco{y=\colorbox{yellow}{$a$}\left(x+\dfrac{b}{2a}\right)^2-\colorbox{yellow}{$a$}\left(\dfrac{b}{2a}\right)^2+c}$

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

Practice: Completing the Square

Practice: Completing the Square

Practice: Completing the Square

Practice: Completing the Square

Wize High School Grade 10 Math Textbook > Completing the Square

Completing the Square (ax2+bx+cax^2+bx+cax2+bx+c)

Popular Courses

Completing the Square

Steps for Completing the Square for y=ax2+bx+c\bco{y=ax^2+bx+c}y=ax2+bx+c

Example: Completing the Square

1. Factor a\bco{a}a out of the first 2 terms: y=a[x2+bax]+c\bco{y=\colorbox{yellow}{$a$}\left[x^2+\dfrac{b}{a}x\right]+c}y=a​[x2+ab​x]+c

2. Add and subtract (b2a)2\bco{\left(\dfrac{b}{2a}\right)^2}(2ab​)2: y=a[x2+bax+(b2a)2−(b2a)2]+c\bco{y=a\left[x^2+\dfrac{b}{a}x\colorbox{yellow}{$+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2$}\right]+c}y=a​x2+ab​x+(2ab​)2−(2ab​)2​​+c

3. Rewrite (factor) the first 3 terms as a perfect square: y=a[(x+b2a)2−(b2a)2]+c\bco{y=a\left[\colorbox{yellow}{$\left(x+\dfrac{b}{2a}\right)^2$}-\left(\dfrac{b}{2a}\right)^2\right]+c}y=a​(x+2ab​)2​−(2ab​)2​+c

4. Multiply a\bco{a}a into the brackets: y=a(x+b2a)2−a(b2a)2+c\bco{y=\colorbox{yellow}{$a$}\left(x+\dfrac{b}{2a}\right)^2-\colorbox{yellow}{$a$}\left(\dfrac{b}{2a}\right)^2+c}y=a​(x+2ab​)2−a​(2ab​)2+c

5. Simplify the constant terms: y=a(x+b2a)2−a(b2a)2+c\bco{y=a\left(x+\dfrac{b}{2a}\right)^2\colorbox{yellow}{$-a\left(\dfrac{b}{2a}\right)^2+c$}}y=a(x+2ab​)2−a(2ab​)2+c​

Practice: Completing the Square

Practice: Completing the Square

Practice: Completing the Square

Practice: Completing the Square

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

Steps for Completing the Square for $\bco{y=ax^2+bx+c}$

1. Factor $\bco{a}$ out of the first 2 terms: $\bco{y=\colorbox{yellow}{$a$}\left[x^2+\dfrac{b}{a}x\right]+c}$

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

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

4. Multiply $\bco{a}$ into the brackets: $\bco{y=\colorbox{yellow}{$a$}\left(x+\dfrac{b}{2a}\right)^2-\colorbox{yellow}{$a$}\left(\dfrac{b}{2a}\right)^2+c}$

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