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All courses Algebra I · A-REI.4.a 51 of 76
Complete the square to transform quadratics and derive the quadratic formula.

Complete the square for a quadratic of the form x^2 + bx + c

Problem
Rewrite \(x^{2}+8x+5\) in completed-square form. Use half the linear coefficient to build a perfect-square trinomial and compensate for the value introduced.
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Hint

Take half of the \(x\) coefficient and square it. For \(x^2+8x+5\), half of \(8\) is \(4\), and \(4^2=16\).

Completing the square rewrites \(x^2+bx\) as \((x+b/2)^2\) after adding and subtracting the same square value.

Solution walkthrough

01

Find the completing value

\[\begin{aligned} b=8 \\ b/2=4 \\ (b/2)^2=16 \end{aligned}\]

For x²+bx, the perfect-square term is built from half the linear coefficient. Half of 8 is 4, and its square is 16.

02

Add and compensate

\[x^2+8x+5=x^2+8x+16-16+5\]

Add 16 to create a perfect-square trinomial and subtract the same 16 so the original expression's value does not change.

03

Factor the square and combine constants

\[\begin{aligned} x^2+8x+16=(x+4)^2 \\ -16+5=-11 \\ (x+4)^2-11 \end{aligned}\]

The first three terms factor because (x+4)² expands to x²+8x+16. The remaining constants total -11.

04

Expand to check equivalence

\[(x+4)^2-11=x^2+8x+16-11=x^2+8x+5\]

Expanding returns the original quadratic, confirming the completed-square form (x+4)²-11.

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Another way

  1. Match x²+8x+5=(x+p)²+q: 2p=8 gives p=4, and p²+q=5 gives q=-11.

!

Common mistake

Do not use 8 itself inside the square or add 16 without subtracting it. Complete the square with half of 8, then compensate for the introduced 16.

Solution walkthrough video