Completing The Square

When factoring a quadratic is awkward, completing the square is the method that always works. It rewrites a quadratic into a form like $(x - h)^2 + k$, which makes the graph easier to read and gives a reliable way to solve quadratic equations by taking square roots.

The core identity, when the quadratic part starts with $x^2 + bx$, is:

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

You add the exact term needed to make a square, then subtract the same term so the value stays unchanged.

When To Reach For This Method

You will usually use completing the square when you need to:

1. Solve a quadratic that does not factor nicely
2. Rewrite a quadratic into vertex form
3. Find the maximum or minimum value of a quadratic function
4. Understand where the quadratic formula comes from

A perfect square trinomial comes from squaring a binomial:

$$\left(x + p\right)^2 = x^2 + 2px + p^2$$

or

$$\left(x - p\right)^2 = x^2 - 2px + p^2$$

Completing the square means rewriting part of a quadratic so it matches one of those patterns exactly. The fast rule is: in $x^2 + bx$, take half of $b$, then square it. That gives the needed constant $\left(\frac{b}{2}\right)^2$.

The Steps

1. Isolate the quadratic part

Keep the $x^2$ and $x$ terms together. If the quadratic starts as $ax^2 + bx + c$ with $a \ne 1$, factor $a$ out of the $x^2$ and $x$ terms first, because the half-then-square shortcut applies directly only after the quadratic part has leading coefficient $1$.

2. Half the linear coefficient, then square

In $x^2 + bx$, take $b/2$ and square it. Adding $\left(\frac{b}{2}\right)^2$ to $x^2 + bx$ gives

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

3. Add and balance

Add that value and subtract the same amount so the expression stays equal:

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

You are not changing the quantity, only the form.

4. Rewrite the square and simplify

Turn the trinomial into a square such as $(x + 3)^2$ and combine the constants.

A Full Worked Example: Rewrite And Solve $x^2 + 6x + 5 = 0$

Start with

$$x^2 + 6x + 5$$

Focus on $x^2 + 6x$. Half of $6$ is $3$, and $3^2 = 9$, so $9$ is the term that completes the square. Add and subtract $9$:

$$x^2 + 6x + 5 = x^2 + 6x + 9 - 9 + 5$$

Group the square and simplify:

$$= \left(x + 3\right)^2 - 4$$

Now the structure is clear. The vertex is $(-3, -4)$, so the graph reaches its minimum when $x = -3$.

To solve $x^2 + 6x + 5 = 0$, set the rewritten form equal to zero:

$$\left(x + 3\right)^2 - 4 = 0$$

Move $4$ across:

$$\left(x + 3\right)^2 = 4$$

Take square roots:

$$x + 3 = \pm 2$$

Then solve:

$$x = -1 \text{ or } x = -5$$

One rewrite gave both the vertex and the solutions. That is the main practical reason this method is useful.

Where Students Get Stuck, And How To Check

The hardest single step is handling a leading coefficient. If the quadratic starts as $ax^2 + bx + c$ with $a \ne 1$, factor $a$ out of the quadratic part first. For example,

$$2x^2 + 8x + 1$$

becomes

$$2\left(x^2 + 4x\right) + 1$$

Inside the parentheses, half of $4$ is $2$, so you add $4$ there:

$$2\left(x^2 + 4x + 4\right) + 1 - 8$$

That simplifies to

$$2\left(x + 2\right)^2 - 7$$

The balancing term is $-8$, not $-4$, because the added $4$ was inside parentheses multiplied by $2$.

To self-check at the end, expand your answer and make sure you recover the original. If you claim

$$x^2 + 6x + 5 = \left(x + 3\right)^2 - 4$$

then expanding gives $x^2 + 6x + 9 - 4 = x^2 + 6x + 5$, which confirms the rewrite.

Watch for these recurring slips:

1. Squaring before halving. For $x^2 + 10x$, the needed term is $25$, not $100$.
2. Forgetting to balance. If you add a value to make a square, subtract the same total value.
3. Skipping the leading coefficient step when the quadratic starts with $2x^2$ or $3x^2$.
4. Losing the sign. $(x - 4)^2$ expands to $x^2 - 8x + 16$, not $x^2 + 8x + 16$.

To rehearse the leading-coefficient-free case, work through $x^2 - 8x + 1$: half of $-8$ is $-4$, so the square part involves $(x - 4)^2$. Then solve the same quadratic with the quadratic formula and confirm both methods give the same roots.