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Math101
Printable cheat sheet
AlgebraGrades 9–12

Completing the Square

Completing the square rewrites a quadratic as a perfect square plus a constant, revealing its vertex and solutions.

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Completing the square turns a quadratic expression into a form whose geometry and algebra are easier to read.

The perfect-square pattern

Expanding a squared binomial gives

$$ (x+p)^2=x^2+2px+p^2. $$

Therefore an expression $x^2+bx$ becomes part of a perfect square when we add

$$ \left(\frac b2\right)^2. $$

Half the coefficient of $x$, then square it. This single pattern drives the method.

Worked example: convert to vertex form

Solving a quadratic by completing the square

The method can isolate a square and then use square roots. Solve

$$ x^2+4x-1=0. $$

Move the constant and complete the square:

$$ x^2+4x=1, $$
$$ x^2+4x+4=5, $$
$$ (x+2)^2=5. $$

Take both square-root branches:

$$ x=-2\pm\sqrt5. $$

The $\pm$ is necessary because both a positive and negative number can have the same square.

Connection to the quadratic formula

The quadratic formula can be derived by completing the square on

$$ ax^2+bx+c=0. $$

After dividing by $a$, moving the constant, and completing the square, solving for $x$ produces

$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. $$

This explains why the discriminant appears: it is the expression left under the square root.

Choosing among quadratic methods

Factoring is fastest when factors are obvious. The quadratic formula works uniformly. Completing the square is especially useful when vertex form or geometric features are required, or when a quadratic lacks easy integer factors.

All three methods agree because they transform the same equation without changing its solutions.

Common mistakes

Adding a number without subtracting it. Rewriting an expression must preserve its value.

Using half of $b$ instead of its square. Add $(b/2)^2$.

Factoring $a$ from the constant too. Initially factor it only from the $x^2$ and $x$ terms unless you deliberately factor the entire expression.

Forgetting the outside multiplier. A square added inside brackets is also multiplied by $a$.

Taking only one square root. Solving $(x-h)^2=q$ requires $x-h=\pm\sqrt q$ when $q>0$.

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