Completing the square: steps, vertex form, circles

Q: Where does the quadratic formula come from?

It is completing the square done once on ax^2 + bx + c = 0 with letters instead of numbers. Divide by a, move \frac{c}{a}, add \frac{b^2}{4a^2} to both sides, take square roots, and you get x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

Q: What if the right side becomes negative?

If you reach (x - h)^2 = a negative number, there are no real solutions, because no real square is negative. In Algebra 2 you can still solve it with i: (x - 1)^2 = -9 gives x = 1 \pm 3i.

The key idea

A perfect square trinomial always has the same pattern: the last number is half the middle coefficient, squared. Completing the square means adding exactly that number so the pattern appears.

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

For example, half of 10 is 5 and $5^2 = 25$ , so $x^2 + 10x + 25 = (x + 5)^2$ .

The steps for solving

If the $x^2$ coefficient is not 1, divide every term by it.
Move the constant to the right side.
Add $\left(\frac{b}{2}\right)^2$ to both sides.
Write the left side as a squared binomial.
Take the square root of both sides, with $\pm$ , and solve.

Worked examples

Example 1: solve by completing the square

Problem Solve $x^2 + 6x - 7 = 0$ .

Move the constant to the right.
$x^2 + 6x = 7$
Half of 6 is 3, and $3^2 = 9$ . Add 9 to both sides.
$x^2 + 6x + 9 = 16$
The left side is now a perfect square.
$(x + 3)^2 = 16$
Square root of both sides. Remember both signs: $x + 3 = 4$ or $x + 3 = -4$ .
Subtract 3 in each case: $x = 1$ or $x = -7$ . Check: $1 + 6 - 7 = 0$ and $49 - 42 - 7 = 0$ .

Answer $x = 1$ or $x = -7$

Example 2: when the leading coefficient is not 1

Problem Solve $2x^2 - 8x - 5 = 0$ . Give exact answers.

Divide every term by 2.
$x^2 - 4x - \frac{5}{2} = 0$
Move the constant.
$x^2 - 4x = \frac{5}{2}$
Half of $-4$ is $-2$ , and $(-2)^2 = 4$ . Add 4 to both sides.
$x^2 - 4x + 4 = \frac{13}{2}$
Factor the left side.
$(x - 2)^2 = \frac{13}{2}$
Take square roots: $x - 2 = \pm\sqrt{\frac{13}{2}}$ . Since $\sqrt{\frac{13}{2}} = \frac{\sqrt{26}}{2}$ , add 2 to get the answer.

Answer $x = 2 \pm \frac{\sqrt{26}}{2}$ , which is about 4.55 or $-0.55$

Example 3: vertex form

Problem Write $y = 3x^2 + 12x + 5$ in vertex form and give the vertex.

Factor the 3 out of the x terms only.
$3x^2 + 12x + 5 = 3(x^2 + 4x) + 5$
Inside the parentheses, half of 4 is 2, and $2^2 = 4$ . Add 4 and subtract 4 inside so nothing changes.
$3(x^2 + 4x) + 5 = 3(x^2 + 4x + 4 - 4) + 5$
Move the $-4$ out. It gets multiplied by the 3 on its way out.
$3(x^2 + 4x + 4 - 4) + 5 = 3(x + 2)^2 - 12 + 5$
Combine the constants.
$3(x + 2)^2 - 12 + 5 = 3(x + 2)^2 - 7$

Answer $y = 3(x + 2)^2 - 7$ , so the vertex is $(-2, -7)$

Example 4 (test-hard): center and radius of a circle

Problem Find the center and radius of the circle $x^2 + y^2 + 10x - 4y - 7 = 0$ .

Group the x terms and the y terms, and move the constant.
$x^2 + 10x + y^2 - 4y = 7$
Complete the square twice: add $5^2 = 25$ for x and $(-2)^2 = 4$ for y, on both sides.
$(x^2 + 10x + 25) + (y^2 - 4y + 4) = 7 + 25 + 4$
Factor each group.
$(x + 5)^2 + (y - 2)^2 = 36$
Compare with $(x - h)^2 + (y - k)^2 = r^2$ . Here $h = -5$ , $k = 2$ , and $r^2 = 36$ .

Answer Center $(-5, 2)$ , radius 6

Common mistakes

Adding the number to one side only. If you add 9 on the left, add 9 on the right too. Fix: write "+9" on both sides in the same step.
Completing the square before dividing by a. With $2x^2 - 8x$ , half of $-8$ is not the right number. Fix: make the $x^2$ coefficient 1 first, or factor it out of the x terms.
Forgetting the factor outside the parentheses. In $3(x^2 + 4x + 4 - 4)$ , the $-4$ leaves as $-12$ , not $-4$ . Fix: multiply by the outside factor when you move a number out.
Taking only the positive square root. $(x + 3)^2 = 16$ has two answers. Fix: write $\pm$ as soon as you take the root.
Reading the circle radius as $r^2$ . In $(x + 5)^2 + (y - 2)^2 = 36$ , the radius is 6, not 36. Also, the center is $(-5, 2)$ : the signs flip.

Quick methods

Practice

5 practice questions

What number should be added to $x^2 - 14x$ to make a perfect square trinomial?
1. $7$
2. $14$
3. $49$
4. $196$
Show answer
Answer: $49$
Half of $-14$ is $-7$ , and $(-7)^2 = 49$ . So $x^2 - 14x + 49 = (x - 7)^2$ . 7 is half of b but not squared, and 196 squares b without halving.
Which is $x^2 + 8x + 10$ written in vertex form?
1. $(x + 4)^2 - 6$
2. $(x + 4)^2 + 10$
3. $(x - 4)^2 - 6$
4. $(x + 8)^2 - 54$
Show answer
Answer: $(x + 4)^2 - 6$
$x^2 + 8x + 16 - 16 + 10 = (x + 4)^2 - 6$ . $(x + 4)^2 + 10$ forgets to subtract the 16 you added. $(x - 4)^2$ has the wrong sign inside.
Solve $x^2 - 2x - 4 = 0$ by completing the square.
1. $1 \pm \sqrt{5}$
2. $-1 \pm \sqrt{5}$
3. $1 \pm \sqrt{3}$
4. $2 \pm \sqrt{5}$
Show answer
Answer: $1 \pm \sqrt{5}$
$x^2 - 2x = 4$ . Add 1: $(x - 1)^2 = 5$ , so $x = 1 \pm \sqrt{5}$ . $1 \pm \sqrt{3}$ subtracts the 1 instead of adding it.
What is the vertex of $y = -2x^2 + 12x - 11$ ?
1. $(3, 7)$
2. $(-3, 7)$
3. $(3, -11)$
4. $(6, -11)$
Show answer
Answer: $(3, 7)$
$-2(x^2 - 6x) - 11 = -2(x^2 - 6x + 9) + 18 - 11 = -2(x - 3)^2 + 7$ . The $-9$ leaves the parentheses as $+18$ because it is multiplied by $-2$ .
What is the radius of the circle $x^2 + y^2 - 6x + 8y = 0$ ?
Show answer
Answer: 5
Add 9 and 16 to both sides: $(x - 3)^2 + (y + 4)^2 = 25$ . So $r^2 = 25$ and the radius is 5. The center is $(3, -4)$ .

Frequently asked questions

When should I complete the square instead of factoring?

Factor first if the numbers work out quickly. Complete the square when the quadratic does not factor nicely, when a problem asks for vertex form, or when you need a circle's center and radius. When the x coefficient is even, it is often faster than the quadratic formula.

Where does the quadratic formula come from?

It is completing the square done once on $ax^2 + bx + c = 0$ with letters instead of numbers. Divide by a, move $\frac{c}{a}$ , add $\frac{b^2}{4a^2}$ to both sides, take square roots, and you get $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

Why do I add the number to both sides?

An equation stays true only if both sides change the same way. When solving, you add to both sides. When rewriting an expression like $y = \ldots$ , there is no other side, so you add and subtract the same number instead, which adds zero.

What if the right side becomes negative?

If you reach $(x - h)^2 =$ a negative number, there are no real solutions, because no real square is negative. In Algebra 2 you can still solve it with $i$ : $(x - 1)^2 = -9$ gives $x = 1 \pm 3i$ .