Solving Quadratic Equations: Completing the Square & The Formula
What happens when a quadratic equation won’t factorise? This guide will teach you two powerful, universal methods—completing the square and the quadratic formula—that can solve *any* quadratic equation, a key skill for higher-tier GCSE Maths.
Two Powerful Methods to Solve Any Quadratic
Both methods are used to find the roots (solutions) for equations in the form $ax^2 + bx + c = 0$.
1. Completing the Square
This method rewrites the equation into the form $(x+p)^2=q$, making it easy to isolate $x$. The key steps are:
- Move the constant term ($c$) to the other side.
- Halve the coefficient of $x$, square it, and add it to both sides.
- Rewrite the left side as a perfect square $(x+p)^2$.
- Square root both sides (remembering the $\pm$) and solve for $x$.
2. The Quadratic Formula
This formula is a shortcut derived from completing the square. You just need to identify $a$, $b$, and $c$ from your equation and substitute them in.
The Discriminant: How Many Solutions?
The part of the quadratic formula under the square root, $b^2 – 4ac$, is called the discriminant. It quickly tells you how many real solutions an equation has.
$b^2 – 4ac > 0$
Two distinct real solutions.
$b^2 – 4ac = 0$
One repeated real solution.
$b^2 – 4ac < 0$
No real solutions.
Worked Examples
Example 1: Completing the Square
Solve $x^2 + 6x + 5 = 0$.
- Move the constant: $x^2 + 6x = -5$.
- Half of 6 is 3. $3^2=9$. Add 9 to both sides:
$x^2 + 6x + 9 = -5 + 9$. - Factorise the left side: $(x+3)^2 = 4$.
- Square root both sides: $x+3 = \pm 2$.
- Solve for both cases:
$x = -3 + 2 = -1$
$x = -3 – 2 = -5$
Solutions: $x = -1, x = -5$
Example 2: The Quadratic Formula
Solve $3x^2 + 5x – 2 = 0$.
- Identify $a=3, b=5, c=-2$.
- Substitute into the formula:
$x = \frac{-5 \pm \sqrt{5^2 – 4(3)(-2)}}{2(3)}$ - Calculate the discriminant: $25 – (-24) = 49$.
- Simplify: $x = \frac{-5 \pm \sqrt{49}}{6} = \frac{-5 \pm 7}{6}$.
- Solve for both cases:
$x = \frac{-5+7}{6} = \frac{2}{6} = \frac{1}{3}$
$x = \frac{-5-7}{6} = \frac{-12}{6} = -2$
Solutions: $x = \frac{1}{3}, x = -2$
Tutor Insights
🤔 Common Misunderstandings
- Completing the Square: Forgetting to add the balancing term to *both sides* of the equation when solving.
- Quadratic Formula: Sign errors are the biggest issue! Especially with $-b$ when $b$ is already negative, or in $-4ac$ when $c$ is negative. Use brackets around negative numbers when squaring on a calculator, e.g., $(-5)^2$.
📝 Common Exam Mistakes
- Not rearranging the equation first. It must be in the form $ax^2+bx+c=0$ before you start.
- Forgetting the $\pm$ symbol after square rooting, leading to only one solution.
- Not showing your substitution into the formula, which can cost method marks.
- Rounding too early. Keep the full value from the square root until the very final step.
Practice Questions
- Solve $x^2 – 6x + 2 = 0$ by completing the square. Give your answer in simplest surd form.
- Solve $2x^2 – 11x + 5 = 0$ using the quadratic formula.
- Solve $3x^2 + 2x = 7$ using the quadratic formula. Give your answers to 2 decimal places.
Show Answers
- Working: $x^2 – 6x = -2 \implies (x-3)^2 – 9 = -2 \implies (x-3)^2 = 7 \implies x-3 = \pm\sqrt{7}$.
Answer: $x = 3 \pm \sqrt{7}$. - Working: $a=2, b=-11, c=5$. $x = \frac{11 \pm \sqrt{(-11)^2 – 4(2)(5)}}{4} = \frac{11 \pm \sqrt{81}}{4} = \frac{11 \pm 9}{4}$.
Answers: $x=5, x=0.5$. - Working: Rearrange to $3x^2+2x-7=0$. $a=3, b=2, c=-7$. $x = \frac{-2 \pm \sqrt{2^2 – 4(3)(-7)}}{6} = \frac{-2 \pm \sqrt{88}}{6}$.
Answers: $x \approx 1.23$ and $x \approx -1.90$.
