All courses Algebra I · A-SSE.3.b 58 of 76
Complete the square to reveal maximum or minimum values of quadratic functions.

Complete the square for a monic quadratic

Problem
Rewrite \(x^{2}+8x+3\) in vertex form by completing the square. Show the half-coefficient and square calculation, compensate for the added value, and identify the vertex.
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Hint

Compute b/2 and(b/2)² exactly.

x²+bx=(x+b/2)²−(b/2)². Preserve equivalence by adding and subtracting the same square.

Solution walkthrough

01

Take half the x-coefficient

\[\begin{aligned} x^2+8x+3 \\ 8/2=4 \end{aligned}\]

For a monic quadratic, the binomial constant used to complete the square is half the signed linear coefficient.

02

Square and compensate

\[\begin{aligned} 4^2=16 \\ x^2+8x+16-16+3 \end{aligned}\]

Adding 16 creates a perfect-square trinomial. Subtracting the same 16 keeps the expression equivalent, and the original constant 3 remains.

03

Write vertex form

\[\begin{aligned} x^2+8x+16=(x+4)^2 \\ (x+4)^2-16+3=(x+4)^2-13 \end{aligned}\]

The completed trinomial is x plus 4 squared, and the outside constants combine to -13.

04

Identify and verify the vertex

\[\begin{aligned} \text{vertex}=(-4,-13) \\ (x+4)^2-13=x^2+8x+16-13=x^2+8x+3 \end{aligned}\]

Vertex form is (x-h)²+k. Since x+4=x-(-4), h=-4 and k=-13. Expansion returns the original expression.

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Another way

  1. Use h=-b/(2a)=-4 and evaluate the original at -4 to get k=-13.

!

Common mistake

Do not add and subtract only the half-coefficient 4. The completing value is its square, 16.

Solution walkthrough video