Quadratic Equations

Lessons

Senior 3 · Mathematics

Quadratic Equations

5 lessons

Lesson 2 of 5

Perfect squares and completing the square

25 min

When factorising fails

Try to factorise $x^{2} + 6 x - 3$ . You need two whole numbers with product $- 3$ and sum $6$ , but the only candidates, $3$ and $- 1$ or $- 3$ and $1$ , have sums $\pm 2$ . The expression does not factorise with whole numbers, yet the equation $x^{2} + 6 x - 3 = 0$ has two real roots. This lesson gives a method that works for every quadratic: rewrite it so that the unknown appears inside a single square.

Definition

Square root property: if $X^{2} = k$ and $k \ge 0$ , then $X = \sqrt{k}$ or $X = -\sqrt{k}$ .

Notation: the two answers are written together as $X = \pm\sqrt{k}$ , read "plus or minus root $k$ ".

Example: $x^{2} = 49$ gives $x = \pm 7$ . If $k \lt 0$ , as in $x^{2} = - 4$ , there is no real solution, because no real number has a negative square.

Important

Why both signs: for $k \ge 0$ ,

\begin{aligned} X^2 = k &\iff X^2 - k = 0 \\ &\iff (X - \sqrt{k})(X + \sqrt{k}) = 0 \\ &\iff X = \sqrt{k} \ \text{ or } \ X = -\sqrt{k}. \end{aligned}

This is the zero-product rule again, with a difference of two squares. Forgetting the minus sign loses a root.

Worked example 1: equations that are already squares

Solve (a) $(x - 3)^{2} = 16$ (b) $2 (x + 1)^{2} = 10$ (c) $(x + 4)^{2} = - 9$ .

Solution

(a) $x - 3 = \pm 4$ , so $x = 3 + 4 = 7$ or $x = 3 - 4 = - 1$ .

(b) Divide by 2 first: $(x + 1)^{2} = 5$ , so $x + 1 = \pm\sqrt{5}$ and $x = -1 \pm \sqrt{5}$ . As decimals (2 d.p.), $x \approx 1.24$ or $x \approx -3.24$ .

Definition

Perfect square trinomial: a three-term expression that is the square of a binomial.

Notation: $x^{2} + 2 p x + p^{2} = (x + p)^{2}$ and $x^{2} - 2 p x + p^{2} = (x - p)^{2}$ .

Example: $x^{2} + 6 x + 9 = (x + 3)^{2}$ . The constant term 9 is the square of half the coefficient of $x$ : $\left(\frac{6}{2}\right)^2 = 9$ .

Completing the square: the picture

Draw $x^{2} + 6 x$ as areas: a square of side $x$ , and $6 x$ split into two strips of $3 \times x$ , one on the right and one below. The shape is almost a square of side $x + 3$ ; only the corner $3 \times 3 = 9$ is missing. Adding that 9 completes the square:

x^{2} + 6 x + 9 = (x + 3)^{2} .

The dashed corner, $3^{2} = 9$ , is what $x^{2} + 6 x$ needs to become $(x + 3)^{2}$ .

In general, half the coefficient of $x$ gives the side of the corner, so

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

The method (when $a = 1$ )

Move the constant term to the right-hand side.
Add $\left(\frac{b}{2}\right)^2$ to both sides.
Write the left-hand side as $\left(x + \frac{b}{2}\right)^2$ and simplify the right-hand side.
Use the square root property, then solve for $x$ .

If $a \ne 1$ , first divide every term by $a$ .

Worked example 2: completing the square

Solve $x^{2} + 6 x - 3 = 0$ . Give exact answers, then answers to 2 decimal places.

Solution

\begin{aligned} x^2 + 6x &= 3 \\ x^2 + 6x + 9 &= 3 + 9 \\ (x + 3)^2 &= 12 \\ x + 3 &= \pm\sqrt{12} = \pm 2\sqrt{3} \\ x &= -3 \pm 2\sqrt{3}. \end{aligned}

Since $\sqrt{3} \approx 1.732$ , $x \approx 0.46$ or $x \approx -6.46$ . Check $x \approx 0.464$ : $0.464^2 + 6(0.464) - 3 \approx 0.215 + 2.784 - 3 \approx 0$ . ✓

Worked example 3: when $a$ is not 1

Solve $2 x^{2} - 8 x + 3 = 0$ , giving the roots to 2 decimal places.

Solution Divide every term by 2, then complete the square with $\left(\frac{-4}{2}\right)^2 = 4$ :

\begin{aligned} x^2 - 4x + \tfrac{3}{2} &= 0 \\ x^2 - 4x &= -\tfrac{3}{2} \\ x^2 - 4x + 4 &= -\tfrac{3}{2} + 4 \\ (x - 2)^2 &= \tfrac{5}{2} \\ x &= 2 \pm \sqrt{\tfrac{5}{2}}. \end{aligned}

$\sqrt{2.5} \approx 1.581$ , so $x \approx 3.58$ or $x \approx 0.42$ .

Worked example 4: an odd coefficient of $x$

Solve $x^{2} - 5 x + 2 = 0$ exactly.

Solution Half of $- 5$ is $-\frac{5}{2}$ , and its square is $\frac{25}{4}$ .

\begin{aligned} x^2 - 5x &= -2 \\ x^2 - 5x + \tfrac{25}{4} &= -2 + \tfrac{25}{4} = \tfrac{17}{4} \\ \left(x - \tfrac{5}{2}\right)^2 &= \tfrac{17}{4} \\ x - \tfrac{5}{2} &= \pm\tfrac{\sqrt{17}}{2} \\ x &= \frac{5 \pm \sqrt{17}}{2}. \end{aligned}

To 2 d.p., $x \approx 4.56$ or $x \approx 0.44$ .

What completing the square shows about the graph

Completing the square also rewrites the expression itself: $x^{2} + 6 x - 3 = (x + 3)^{2} - 12$ . A square is never negative, so the smallest value of the expression is $- 12$ , when $x = - 3$ . That is the lowest point (the vertex) of the parabola $y = x^{2} + 6 x - 3$ . You will study the vertex in the unit on quadratic functions; here it explains why the two roots sit symmetrically on either side of $x = - 3$ .

The roots $-3 \pm 2\sqrt{3}$ are the same distance, $2\sqrt{3} \approx 3.46$ , from the line $x = - 3$ through the vertex.

Warning

Three slips to avoid. (1) Add $\left(\frac{b}{2}\right)^2$ to both sides, not just the left. (2) Write $\pm$ when you take the square root, or you lose a root. (3) When $a \ne 1$ , divide by $a$ before completing the square: $2 x^{2} - 8 x$ is not $(\sqrt{2}x - 2)^2 - 4$ .

Tip

Halve it, square it, add it. Halve the coefficient of $x$ , square the result, and add it to both sides. The half you found is the number that goes inside the bracket.

Khan Academy: Solving quadratic equations by completing the square (14 min). Watch how the constant is moved across, half the coefficient of x is squared and added to both sides, and the plus-or-minus square root finishes the solution.

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Practice

Give exact answers, and decimals to 2 d.p. where the roots are not whole numbers. The last three are harder.

$x^{2} = 81$
$(x + 2)^{2} = 25$
What must be added to $x^{2} - 10 x$ to make a perfect square? Write the square.
Solve $x^{2} + 4 x - 5 = 0$ by completing the square.
Solve $x^{2} - 6 x + 2 = 0$ .
Solve $2 x^{2} + 4 x - 7 = 0$ .
Show that $x^{2} + 3 x + 5 = 0$ has no real roots.
Write $x^{2} - 8 x + 10$ in the form $(x - p)^{2} - q$ and state the least value of the expression.

Show answersHide answers

$x = \pm 9$
$x + 2 = \pm 5$ : $x = 3$ or $x = - 7$ .
$\left(\frac{-10}{2}\right)^2 = 25$ ; $x^{2} - 10 x + 25 = (x - 5)^{2}$ .
$x^{2} + 4 x + 4 = 9$ , $(x + 2)^{2} = 9$ , $x + 2 = \pm 3$ : $x = 1$ or $x = - 5$ .
$(x - 3)^{2} = 7$ : $x = 3 \pm \sqrt{7} \approx 5.65$ or $0.35$ .
$x^2 + 2x = \frac{7}{2}$ , $(x + 1)^2 = \frac{9}{2}$ : $x = -1 \pm \frac{3}{\sqrt{2}} = -1 \pm \frac{3\sqrt{2}}{2} \approx 1.12$ or $- 3.12$ .
$\left(x + \frac{3}{2}\right)^2 = \frac{9}{4} - 5 = -\frac{11}{4}$ . A square cannot be negative, so there are no real roots.
$(x - 4)^{2} - 16 + 10 = (x - 4)^{2} - 6$ . The least value is $- 6$ , when $x = 4$ .

Sources and credits

10.1: Solve Quadratic Equations Using the Square Root Property, from Elementary Algebra by Lynn Marecek, MaryAnne Anthony-Smith and Andrea Honeycutt Mathis, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
10.2: Solve Quadratic Equations by Completing the Square, from Elementary Algebra by Lynn Marecek, MaryAnne Anthony-Smith and Andrea Honeycutt Mathis, OpenStax, provided by LibreTexts, licensed under CC BY 4.0 (licence read on the page, 30 September 2026). Adapted by Ganzaa.
The area diagram and the section on the vertex. Written by Ganzaa.
Video: Solving quadratic equations by completing the square, Khan Academy, embedded from YouTube.

The Rwandan worked examples, the diagrams, the practice items and the quiz questions were written by Ganzaa.

1When factorising fails

2Completing the square: the picture

3The method (when a=1 a = 1 a=1)

4What completing the square shows about the graph

When factorising fails

Completing the square: the picture

The method (when $a = 1$ )

What completing the square shows about the graph