Solve quadratic equations by completing the square, and use completing the square to rewrite a quadratic in vertex form (A.EI.6).

What this topic is asking

This part of A.EI.6 asks you to solve a quadratic by completing the square and to use the same process to rewrite a quadratic in vertex form. On the Virginia Algebra I SOL these are Equations and Inequalities items: complete the square, solve, or identify the perfect-square binomial. They appear as fill-in-the-blank and multiple choice, and the method underlies both the quadratic formula and the vertex of a parabola.

The core move: half of b, squared

A perfect-square trinomial $x^2 + bx + c$ comes from squaring $(x + \frac{b}{2})$, which means the constant must be $c = \left(\frac{b}{2}\right)^2$. So to complete the square on $x^2 + bx$, add $\left(\frac{b}{2}\right)^2$:

because half of $8$ is $4$ and $4^2 = 16$. The number inside the binomial is always half of $b$.

Solving by completing the square

The critical balance point: whatever you add to complete the square on the left, you must add to the right side too, or the equation is no longer equivalent.

When the leading coefficient is not 1

If $a \ne 1$, factor $a$ out of the $x^2$ and $x$ terms before completing the square. For $2x^2 + 8x + 3 = 0$: factor, $2(x^2 + 4x) + 3 = 0$, complete the square inside (add and subtract $4$ inside, tracking the factor of $2$), and proceed. This step keeps the coefficient of $x^2$ at $1$ inside the parentheses, which is what the half-of-$b$ rule requires.

Producing vertex form

Completing the square is also how you rewrite a quadratic function in vertex form $a(x - h)^2 + k$, which displays the vertex $(h, k)$ directly. For $y = x^2 + 6x + 5$: complete the square, $y = (x^2 + 6x + 9) + 5 - 9 = (x + 3)^2 - 4$, so the vertex is $(-3, -4)$. Here you add and subtract the same constant on one side to keep $y$ unchanged, rather than adding to both sides as when solving.

Why the constant is (b/2) squared

The half-of-$b$, square-it rule is not a trick to memorize, it is forced by the algebra of squaring a binomial. Expanding $(x + p)^2$ gives $x^2 + 2px + p^2$, so the coefficient of $x$ is $2p$ and the constant is $p^2$. If you are handed $x^2 + bx$ and want it to be the start of $(x + p)^2$, you match $2p = b$, which means $p = \frac{b}{2}$, and then the constant you need is $p^2 = \left(\frac{b}{2}\right)^2$. So "half of $b$" comes from undoing the $2p$, and "square it" comes from the $p^2$. This same derivation, applied to the general $ax^2 + bx + c = 0$, produces the quadratic formula, which is why completing the square is the backbone of every quadratic method: factoring, the formula, and the vertex all trace back to it.

How the SOL examines this topic

• Fill-in-the-blank. State the constant that completes the square, or solve by completing the square and type the solutions.
• Multiple choice. Identify the perfect-square binomial or the vertex form.
• Drag-and-drop. Order the steps of the method.

Try this

Q1. What constant completes the square for $x^2 + 10x$? [1 point]

• Cue. Half of $10$ is $5$; $5^2 = 25$.

Q2. Solve $x^2 - 2x - 3 = 0$ by completing the square. [2 points]

• Cue. $x^2 - 2x + 1 = 4 \Rightarrow (x - 1)^2 = 4 \Rightarrow x = 3$ or $x = -1$.