Completing the Square, SAT math lesson

Some quadratics refuse to factor nicely, yet one small move makes them easy. If you can rewrite an equation so that one side is a perfect square, like $(x + 3)^2 = 25$ , you can undo the square directly and read off both solutions. Completing the square is that move. It also turns $y = x^2 - 8x + 13$ into a form that shows at a glance where the parabola turns, which the SAT asks about often.

See also Completing the Square in Foundations for Calculus and Quadratic Functions and Vertex Form in Precalculus.

Key terms

Perfect-square trinomial

Definition A perfect-square trinomial is a polynomial that equals the square of a binomial. For every real number $k$ and every real $x$ ,

\begin{aligned} (x + k)^2 &= x^2 + 2kx + k^2, \\[0.4em] (x - k)^2 &= x^2 - 2kx + k^2. \end{aligned}

So a monic quadratic $x^2 + bx + c$ is a perfect square exactly when its constant term is the square of half the $x$ coefficient, that is when

c = \left(\frac{b}{2}\right)^2,

and then it equals $\left(x + \frac{b}{2}\right)^2$ .

In plain words Halve the middle coefficient and square it. If you get the constant term, the expression is a square in disguise.

Example In $x^2 - 10x + 25$ half of $-10$ is $-5$ , and $(-5)^2 = 25$ matches the constant term. So $x^2 - 10x + 25 = (x - 5)^2$ . By contrast $x^2 - 10x + 20$ is not a perfect square, because its constant term is $20$ and not $25$ .

Square-root method

Definition For a real number $c$ the equation $z^2 = c$ has

two real solutions $z = \sqrt{c}$ and $z = -\sqrt{c}$ when $c > 0$ ,
exactly one real solution $z = 0$ when $c = 0$ ,
no real solutions when $c < 0$ , since no real number has a negative square.

The symbol $\sqrt{c}$ means only the nonnegative root, so the minus branch has to be written separately. The same applies when $z$ is an expression, for example $z = x - h$ .

In plain words If something squared is $16$ , that something is $4$ or $-4$ . Forgetting the negative branch is the most common way to lose half the answer.

Example Solve $(x - 3)^2 = 16$ . The quantity $x - 3$ is either $4$ or $-4$ .

Branch	Equation	Solution
plus	$x - 3 = 4$	$x = 7$
minus	$x - 3 = -4$	$x = -1$

Both check, since $(7 - 3)^2 = 16$ and $(-1 - 3)^2 = 16$ .

Completing the square

Definition To complete the square in $x^2 + bx$ , add $\left(\frac{b}{2}\right)^2$ , which produces the perfect square

x^2 + bx + \left(\frac{b}{2}\right)^2 = \left(x + \frac{b}{2}\right)^2.

In an equation you add this number to both sides, so the solution set does not change. In an expression you add it and subtract it at once, so the value does not change.

In plain words You build the missing piece of a square, and you keep the balance by adding the same amount on the other side.

Example Solve $x^2 + 6x = 16$ . Half of $6$ is $3$ , and $3^2 = 9$ , so add $9$ to both sides.

\begin{aligned} x^2 + 6x + 9 &= 16 + 9 \\[0.4em] (x + 3)^2 &= 25. \end{aligned}

By the square-root method $x + 3 = 5$ or $x + 3 = -5$ , so $x = 2$ or $x = -8$ .

There is a helpful picture when $x$ is a positive length. Take an $x$ by $x$ square and attach two $3$ by $x$ strips to two of its sides, one along the right and one along the bottom. The total area is $x^2 + 6x$ , and the shape is a large $(x + 3)$ by $(x + 3)$ square with a $3$ by $3$ corner missing. That missing corner is the $9$ you add. The picture only works for positive lengths, but the algebra does not need it. The identity holds for every real $x$ , which is why the negative solution $x = -8$ is just as valid as $x = 2$ . Check it, $(-8)^2 + 6(-8) = 64 - 48 = 16$ .

Vertex form

Definition A quadratic function is in vertex form when it is written as

f(x) = a(x - h)^2 + k, \qquad a \ne 0.

Its graph is a parabola with vertex $(h, k)$ . If $a > 0$ , the smallest value of $f$ is $k$ , reached at $x = h$ . If $a < 0$ , the largest value of $f$ is $k$ , reached at $x = h$ . Completing the square turns $ax^2 + bx + c$ into this form.

In plain words A square is never negative, so with $a > 0$ the expression is smallest when the square is zero. That happens at $x = h$ , and the output there is $k$ .

Example Rewrite $y = x^2 - 8x + 13$ . Half of $-8$ is $-4$ , and $(-4)^2 = 16$ . Add and subtract $16$ .

\begin{aligned} y &= x^2 - 8x + 16 - 16 + 13 \\[0.4em] &= (x - 4)^2 - 3. \end{aligned}

The vertex is $(4, -3)$ , and the minimum value of $y$ is $-3$ . Watch the sign. The form $(x - 4)^2$ puts the vertex at $x = 4$ , not at $x = -4$ .

Nonunit leading coefficient

Definition When the coefficient $a$ of $x^2$ is not $1$ , deal with it before completing the square.

In an equation $ax^2 + bx + c = 0$ with $a \ne 0$ , divide both sides by $a$ . The new equation has the same solutions and a leading coefficient of $1$ .
In a function $f(x) = ax^2 + bx + c$ , factor $a$ out of the $x$ terms only. Do not divide, because that would change the function. The number you add inside the parentheses is multiplied by $a$ , so you must subtract $a$ times it outside.

\begin{aligned} &ax^2 + bx + c \\[0.4em] &= a\left(x^2 + \frac{b}{a}x\right) + c \\[0.4em] &= a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a}. \end{aligned}

In plain words The square trick works on $x^2$ with coefficient $1$ . Get the $a$ out of the way first, and remember it still multiplies whatever you add inside.

Example Solve $2x^2 - 8x - 10 = 0$ . Divide by $2$ to get $x^2 - 4x - 5 = 0$ , so $x^2 - 4x = 5$ . Add $4$ to both sides.

(x - 2)^2 = 9.

So $x - 2 = 3$ or $x - 2 = -3$ , which gives $x = 5$ or $x = -1$ .

Now a function. For $f(x) = 2x^2 + 12x + 7$ factor $2$ from the first two terms. Inside, half of $6$ is $3$ and $3^2 = 9$ . Adding $9$ inside really adds $2 \cdot 9 = 18$ , so subtract $18$ outside.

\begin{aligned} f(x) &= 2(x^2 + 6x + 9) - 18 + 7 \\[0.4em] &= 2(x + 3)^2 - 11. \end{aligned}

The vertex is $(-3, -11)$ . Expanding $2(x + 3)^2 - 11$ gives back $2x^2 + 12x + 7$ , which confirms the work.

Check yourself

Which expression is equivalent to $x^2 + 10x + 18$ ? (A) $(x + 5)^2 + 18$ , (B) $(x + 5)^2 - 7$ , (C) $(x + 10)^2 - 82$ , (D) $(x + 5)^2 + 7$

Show the answer

Choice (B) is correct. Half of $10$ is $5$ , and $(x + 5)^2 = x^2 + 10x + 25$ . To get $18$ instead of $25$ you subtract $7$ . Choice (A) adds the $25$ without taking it back. Choice (C) uses $10$ instead of half of it, and its expansion has middle term $20x$ . Choice (D) has the sign of the correction wrong.

If $(x - 4)^2 = 49$ and $x < 0$ , what is the value of $x$ ?

Show the answer

The answer is $-3$ . The square-root method gives $x - 4 = 7$ or $x - 4 = -7$ , so $x = 11$ or $x = -3$ . Only $-3$ is negative. Taking just the positive root would have missed the answer completely.

The function $f(x) = 3x^2 - 12x + 5$ can be written as $f(x) = 3(x - h)^2 + k$ , where $h$ and $k$ are constants. What is the value of $k$ ? (A) $1$ , (B) $9$ , (C) $17$ , (D) $-7$

Show the answer

Choice (D) is correct. Factor $3$ from the $x$ terms to get $3(x^2 - 4x) + 5$ . Adding $4$ inside adds $3 \cdot 4 = 12$ , so subtract $12$ outside.

\begin{aligned} f(x) &= 3(x - 2)^2 - 12 + 5 \\[0.4em] &= 3(x - 2)^2 - 7. \end{aligned}

Choice (A) subtracts only $4$ , forgetting that the $3$ multiplies it. Choice (B) adds $4$ instead of subtracting it, and choice (C) adds $12$ instead of subtracting it.

Completing the Square

Theory, part 49

Practice, part 50

Key terms

Perfect-square trinomial

Square-root method

Completing the square

Vertex form

Nonunit leading coefficient

Check yourself

How this lesson connects

Builds on

Used later in