Introduction to Quadratics
A quadratic equation is any single-variable equation that can be rearranged into the standard form:
Where $a$, $b$, and $c$ are real numbers, and $a \neq 0$. To solve a quadratic equation means to find the values of $x$ (often called the **roots**) that make the equation true.
At the GCSE Higher Tier, you are expected to master three distinct mathematical methods to find these roots. Selecting the most efficient method will save you valuable exam time.
Method 1: Factorisation (Simplifying into Brackets)
Factorising is the fastest method to solve a quadratic equation when the roots are integers. It involves rewriting the expression as the product of two linear brackets.
Ensure the equation equals zero
Rearrange your quadratic so all terms are on the left side: $ax^2 + bx + c = 0$.
Find two numbers that multiply to $c$ and add to $b$
When $a = 1$, search for factors of the constant term $c$ that sum to the linear coefficient $b$.
Worked GCSE Example
Question: Solve $x^2 - 5x + 6 = 0$ by factorisation.
Solution:
- We need two numbers that multiply to $+6$ and add to $-5$. These numbers are $-2$ and $-3$.
- Write the quadratic in bracket form: $(x - 2)(x - 3) = 0$.
- Set each bracket equal to zero to find the roots:
$x - 2 = 0 \implies x = 2$
$x - 3 = 0 \implies x = 3$
Final Roots: $x = 2$ or $x = 3$.
Method 2: The Quadratic Formula
When a quadratic equation does not factorise easily, or when the question asks for decimal rounding (e.g. to 3 significant figures), you must use the quadratic formula:
Worked GCSE Example
Question: Solve $2x^2 + 5x - 4 = 0$ using the quadratic formula. Give your answers to 2 decimal places.
Solution:
- Identify coefficients: $a = 2$, $b = 5$, $c = -4$.
- Substitute into the formula: $$x = \frac{-5 \pm \sqrt{5^2 - 4(2)(-4)}}{2(2)}$$
- Simplify the discriminant ($b^2 - 4ac$): $$x = \frac{-5 \pm \sqrt{25 - (-32)}}{4} \implies x = \frac{-5 \pm \sqrt{57}}{4}$$
- Split the plus and minus calculations:
$x_1 = \frac{-5 + 7.5498}{4} = 0.64$ (to 2 d.p.)
$x_2 = \frac{-5 - 7.5498}{4} = -3.14$ (to 2 d.p.)
Final Roots: $x = 0.64$ or $x = -3.14$.
Method 3: Completing the Square
Completing the square rearranges $x^2 + bx + c$ into the form $(x + p)^2 + q$. This is highly useful for identifying coordinate turning points of quadratic graphs, as well as solving algebraic fractions.
Worked GCSE Example
Question: Solve $x^2 + 6x - 1 = 0$ by completing the square. Leave your answer in surd form.
Solution:
- Halve the coefficient of $x$: $\frac{6}{2} = 3$.
- Write in bracket form and subtract the square of that number: $$(x + 3)^2 - 3^2 - 1 = 0 \implies (x + 3)^2 - 9 - 1 = 0$$
- Simplify: $(x + 3)^2 - 10 = 0 \implies (x + 3)^2 = 10$.
- Take the square root of both sides: $x + 3 = \pm\sqrt{10}$.
- Solve for $x$: $x = -3 \pm\sqrt{10}$.
Final Roots: $x = -3 + \sqrt{10}$ or $x = -3 - \sqrt{10}$.
Frequently Asked Questions
You should use the quadratic formula if the quadratic does not factorise easily into integers, or if the exam question specifically asks you to give your answers to a certain number of decimal places (e.g. 3 significant figures) or in surd form.
Always list the values of $a$, $b$, and $c$ separately first. If $b$ is negative, $-b$ becomes positive. When calculating $b^2 - 4ac$, double check if $c$ is negative, as subtracting a negative term will change to addition ($b^2 + 4ac$).
The discriminant is the expression inside the square root: $b^2 - 4ac$. If it is positive, the equation has two distinct real roots (crosses the x-axis twice). If it is zero, it has one repeated root (touches the x-axis once). If it is negative, it has no real roots (never touches the x-axis).
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