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Algebra 2 Quadratic functions (advanced)

Completing the square (advanced)

20 practice questions 0 video lessons Theory + worked examples

Completing the Square

California Algebra 2 • Standard A-REI.4a • Quadratic Functions

Completing the Square is the opening topic of Quadratic Functions in the California Common Core State Standards. It is aligned to Standard A-REI.4a, which requires students to solve quadratic equations by inspection, completing the square, and the quadratic formula.

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

California Algebra 2 › Quadratic Functions › Completing the Square  —  Standard A-REI.4a

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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.