Resources For Teachers For Tutors For Students & Parents Pricing
Algebra 2 Quadratic functions (advanced)

Completing the square (advanced)

20 practice questions 0 video lessons Theory + worked examples

Completing the Square

Texas Algebra II (TEKS) • Standard 2A.4(D) • Quadratic Functions

Completing the Square is the opening topic of Quadratic Functions in the Texas Essential Knowledge and Skills (Algebra II, §111.40). It is aligned to Standard 2A.4(D), which requires students to transform a quadratic function into f(x)=a(x-h)^2+k form by completing the square.

Completing the square turns a quadratic into vertex form \(a(x-h)^2+k\) by adding \(\left(\dfrac{b}{2}\right)^2\).

Texas Algebra II (TEKS) › Quadratic Functions › Completing the Square  —  Standard 2A.4(D)

Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

Completing the square rewrites \(x^2+bx\) as a perfect square by adding \(\left(\dfrac{b}{2}\right)^2\):
\[x^2+bx+\left(\dfrac{b}{2}\right)^2=\left(x+\dfrac{b}{2}\right)^2.\]

This produces vertex form \(a(x-h)^2+k\), where \((h,k)\) is the vertex.

If \(a\neq 1\), factor \(a\) out of the \(x\)-terms first.
Vertex form from completing the square Completing the square rewrites a quadratic in vertex form, exposing the vertex. x y vertex (-3,-4) y=(x+3)²-4
Vertex form exposes the vertex \((-3,-4)\).
Completing the square Completing the square Completing the square x² + bx → add (b/2)² x² + bx + (b/2)² = (x + b/2)² gives vertex form a(x-h)² + k
The completing-the-square step.

The key step:

\[x^2+bx+\left(\dfrac{b}{2}\right)^2=\left(x+\dfrac{b}{2}\right)^2\]
add half the coefficient of x squared to complete the square
Add and subtract \((b/2)^2\) so the value doesn't change.

How to complete the square

  1. If \(a\neq1\), factor \(a\) from the \(x^2\) and \(x\) terms.
  2. Take half of \(b\) and square it.
  3. Add and subtract that value to form a perfect square.
  4. Write as \(a(x-h)^2+k\).
Example 1 — To vertex form
Write \(x^2+6x+5\) in vertex form.
Solution

Half of \(6\) is \(3\); add and subtract \(3^2=9\).

\((x^2+6x+9)-9+5\)
\(=\)\((x+3)^2-4\)
vertex form is x plus 3 squared minus 4
Example 2 — Solve by completing the square
Solve \(x^2-4x-1=0\).
Solution

Move the constant, then complete the square.

\(x^2-4x\)\(=\)\(1\)
\(x^2-4x+4\)\(=\)\(5\)
\((x-2)^2\)\(=\)\(5\)
\(x\)\(=\)\(2\pm\sqrt5\)
x equals 2 plus or minus root 5
Example 3 — Find the vertex
Find the vertex of \(y=x^2+8x+10\).
Solution

Complete the square: half of \(8\) is \(4\).

\((x^2+8x+16)-16+10\)
\(y\)\(=\)\((x+4)^2-6\)

The vertex is \((-4,-6)\).

the vertex is negative 4 comma negative 6
Example 4 — Leading coefficient
Write \(2x^2+8x+3\) in vertex form.
Solution

Factor \(2\) from the \(x\)-terms first.

\(2(x^2+4x)+3\)
\(=\)\(2(x^2+4x+4)-8+3\)
\(=\)\(2(x+2)^2-5\)
vertex form is 2 times x plus 2 squared minus 5

Common pitfalls

Add and subtract \((b/2)^2\) — don't change the expression's value.
Factor \(a\) first when the leading coefficient isn't \(1\).
Watch the sign of \(h\): \((x+3)^2\) has vertex at \(x=-3\).

Frequently asked questions

What is completing the square?

Adding \((b/2)^2\) to make \(x^2+bx\) a perfect square trinomial.

What is vertex form?

\(a(x-h)^2+k\), where \((h,k)\) is the vertex.

What if the leading coefficient isn't \(1\)?

Factor it out of the \(x\)-terms before completing the square.

Why add and subtract the same value?

So the expression stays equal while forming the perfect square.