Solution by Completing the Square

Why Completing the Square?

Factorisation is quick, but many quadratics don't factor over rationals. Completing the square is a method that works for every quadratic equation. It turns the equation into the form $(x + p)^2 = q$, which you can solve by taking square roots.

The idea: Rewrite $ax^2 + bx + c = 0$ so the $x$-terms become a perfect square $(x + p)^2$, then isolate it and take the square root of both sides.

Think of it this way: we 'force' the left side into a perfect square by adding and subtracting the right constant.

[Board Important] This method is the foundation of the quadratic formula. Even where it's not on your exact syllabus, understanding it makes the formula far less mysterious.

The Key Algebraic Identity

Completing the square relies on:

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

So to complete the square for $x^2 + bx$, we need the constant $\left(\dfrac{b}{2}\right)^2$:

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

Quick demo

$x^2 + 6x = (x + 3)^2 - 9$ (since half of 6 is 3, and $3^2 = 9$).

Key Point: Take half the coefficient of $x$, square it, and add-and-subtract it. The first three terms then form a perfect square.

[Board Important] The magic number is always $\left(\dfrac{\text{coefficient of }x}{2}\right)^2$. Memorise this — it's the core of the method.

The Steps (with $a = 1$)

Let's solve $x^2 + 4x - 5 = 0$.

1. Move the constant to the right: $x^2 + 4x = 5$.
2. Complete the square: half of 4 is 2, $2^2 = 4$. Add 4 to both sides: $x^2 + 4x + 4 = 5 + 4 = 9$.
3. Write as a square: $(x + 2)^2 = 9$.
4. Take square roots: $x + 2 = \pm 3$.
5. Solve: $x = 3 - 2 = 1$ or $x = -3 - 2 = -5$.

So the roots are $x = 1$ and $x = -5$.

Key Point: Don't forget the $\pm$ when taking the square root — it gives both roots.

[Board Important] Always add the completing constant to both sides to keep the equation balanced.

When $a \neq 1$

If the coefficient of $x^2$ is not 1, divide the whole equation by $a$ first, then complete the square.

Worked outline: $2x^2 - 8x + 6 = 0$

1. Divide by 2: $x^2 - 4x + 3 = 0$.
2. Move constant: $x^2 - 4x = -3$.
3. Half of $-4$ is $-2$, square is 4. Add 4: $x^2 - 4x + 4 = 1 \Rightarrow (x - 2)^2 = 1$.
4. $x - 2 = \pm 1 \Rightarrow x = 3$ or $x = 1$.

Key Point: Divide by $a$ before completing the square, so the $x^2$ coefficient becomes 1.

[Board Important] Forgetting to divide by $a$ first is the most common mistake here. Always reduce the $x^2$ coefficient to 1 at the start.

Solved Examples

Example 1: Basic completing the square

Solve $x^2 + 6x + 5 = 0$.

Solution:

1. $x^2 + 6x = -5$.
2. Half of 6 is 3, $3^2 = 9$: $x^2 + 6x + 9 = -5 + 9 = 4$.
3. $(x + 3)^2 = 4 \Rightarrow x + 3 = \pm 2$.
4. $x = -1$ or $x = -5$.

Final Answer: $x = -1, -5$.

Takeaway: Add $(b/2)^2$ to both sides, then square-root.

Example 2: Roots with surds

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

Solution:

1. $x^2 - 4x = 1$.
2. Half of $-4$ is $-2$, square 4: $x^2 - 4x + 4 = 5 \Rightarrow (x - 2)^2 = 5$.
3. $x - 2 = \pm\sqrt5 \Rightarrow x = 2 \pm \sqrt5$.

Final Answer: $x = 2 + \sqrt5, 2 - \sqrt5$.

Takeaway: Completing the square handles surd roots that factorisation can't.

Example 3: With $a \neq 1$

Solve $2x^2 - 8x + 6 = 0$.

Solution:

1. Divide by 2: $x^2 - 4x + 3 = 0 \Rightarrow x^2 - 4x = -3$.
2. Add 4: $(x - 2)^2 = 1$.
3. $x - 2 = \pm 1 \Rightarrow x = 3, 1$.

Final Answer: $x = 1, 3$.

Takeaway: Divide by $a$ first so the $x^2$ coefficient is 1.

Example 4: Negative middle term

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

Solution:

1. $x^2 - 6x = -7$.
2. Half of $-6$ is $-3$, square 9: $(x - 3)^2 = -7 + 9 = 2$.
3. $x - 3 = \pm\sqrt2 \Rightarrow x = 3 \pm \sqrt2$.

Final Answer: $x = 3 + \sqrt2, 3 - \sqrt2$.

Takeaway: Half of a negative coefficient is negative, but its square is positive.

Example 5: Fractional half

Solve $x^2 + 3x + 1 = 0$.

Solution:

1. $x^2 + 3x = -1$.
2. Half of 3 is $\dfrac{3}{2}$, square $\dfrac{9}{4}$: $\left(x + \dfrac{3}{2}\right)^2 = -1 + \dfrac{9}{4} = \dfrac{5}{4}$.
3. $x + \dfrac{3}{2} = \pm\dfrac{\sqrt5}{2} \Rightarrow x = \dfrac{-3 \pm \sqrt5}{2}$.

Final Answer: $x = \dfrac{-3 \pm \sqrt5}{2}$.

Takeaway: Odd middle coefficients give fractional halves — keep them as fractions.

Example 6: No real roots revealed

Solve $x^2 + 2x + 5 = 0$ by completing the square.

Solution:

1. $x^2 + 2x = -5$.
2. Add 1: $(x + 1)^2 = -4$.
3. The right side is negative; no real number squares to a negative.

Final Answer: No real roots.

Takeaway: A negative value after completing the square means no real roots.

Example 7: With $a \neq 1$ and surds

Solve $3x^2 - 6x - 2 = 0$.

Solution:

1. Divide by 3: $x^2 - 2x - \dfrac{2}{3} = 0 \Rightarrow x^2 - 2x = \dfrac{2}{3}$.
2. Add 1: $(x - 1)^2 = \dfrac{2}{3} + 1 = \dfrac{5}{3}$.
3. $x - 1 = \pm\sqrt{\dfrac{5}{3}} \Rightarrow x = 1 \pm \dfrac{\sqrt{15}}{3}$.

Final Answer: $x = 1 \pm \dfrac{\sqrt{15}}{3}$.

Takeaway: Rationalise $\sqrt{5/3}$ to $\dfrac{\sqrt{15}}{3}$ for the final answer.

Example 8: Equal roots case

Solve $x^2 - 8x + 16 = 0$ by completing the square.

Solution:

1. $x^2 - 8x = -16$.
2. Add 16: $(x - 4)^2 = 0$.
3. $x - 4 = 0 \Rightarrow x = 4$ (repeated).

Final Answer: $x = 4$ (equal roots).

Takeaway: $(x - p)^2 = 0$ gives a single repeated root.

Example 9: Verify the method matches factorisation

Solve $x^2 - 5x + 6 = 0$ by completing the square and confirm with factorisation.

Solution:

1. $x^2 - 5x = -6$. Half of $-5$ is $-\dfrac{5}{2}$, square $\dfrac{25}{4}$: $\left(x - \dfrac{5}{2}\right)^2 = -6 + \dfrac{25}{4} = \dfrac{1}{4}$.
2. $x - \dfrac{5}{2} = \pm\dfrac{1}{2} \Rightarrow x = 3$ or $x = 2$.
3. Factorisation: $(x-2)(x-3) = 0$ gives the same. ✓

Final Answer: $x = 2, 3$.

Takeaway: Both methods give identical roots — a good self-check.

Example 10: Word problem via completing the square

The sum of a number and its reciprocal is $\dfrac{10}{3}$. Find the number.

Solution:

1. Let the number be $x$: $x + \dfrac{1}{x} = \dfrac{10}{3} \Rightarrow 3x^2 - 10x + 3 = 0$.
2. Divide by 3: $x^2 - \dfrac{10}{3}x + 1 = 0 \Rightarrow x^2 - \dfrac{10}{3}x = -1$.
3. Half of $-\dfrac{10}{3}$ is $-\dfrac{5}{3}$, square $\dfrac{25}{9}$: $\left(x - \dfrac{5}{3}\right)^2 = -1 + \dfrac{25}{9} = \dfrac{16}{9}$.
4. $x - \dfrac{5}{3} = \pm\dfrac{4}{3} \Rightarrow x = 3$ or $x = \dfrac{1}{3}$.

Final Answer: The number is 3 (or $\dfrac{1}{3}$).

Takeaway: Reciprocal problems lead to a quadratic; completing the square solves it cleanly.