Completing The Square

Completing the square rewrites a quadratic into a form like $(x - h)^2 + k$. That makes the graph easier to read, and it gives you a reliable way to solve quadratic equations when factoring is not convenient.

If the quadratic part starts with $x^2 + bx$, the key identity 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.

What Completing The Square Means

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:

$$\left(\frac{b}{2}\right)^2$$

Why Half Then Square Works

Start with

$$x^2 + bx$$

Add $\left(\frac{b}{2}\right)^2$:

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

Now the trinomial factors as

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

So the original expression can be rewritten as

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

You are not changing the quantity. You are only changing the form.

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

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

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

Move $4$ to the other side:

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

Take square roots:

$$x + 3 = \pm 2$$

Then solve for $x$:

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

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

When The Coefficient Of $x^2$ Is Not $1$

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

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 $2$.

Common Mistakes

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

When Students Use Completing The Square

You will usually see this method 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 Quick Check

After you complete the square, expand your answer and make sure you recover the original expression exactly.

For example, if you claim

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

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

Try A Similar Problem

Try $x^2 - 8x + 1$. Half of $-8$ is $-4$, so the square part should involve $(x - 4)^2$.

If you want a useful next comparison, solve the same quadratic with the quadratic formula and check that both methods lead to the same roots.