Quadratic Equation: A Must-know GCSE Maths Topic

What is a quadratic equation?

Quadratic equation is one of the most important algebra topics in the GCSE Maths syllabus. They appear frequently in both Foundation and Higher Tier exams, often in questions worth several marks. Mastering quadratics not only helps with algebra but also supports topics like graphs, coordinate geometry, and even problem-solving questions.—A quadratic equation is an equation that includes a variable raised to the power of two.It usually takes the form:\[ ax² + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \).The word quadratic comes from “quadratus,” meaning “square” — because the variable is squared!Quadratic equations appear everywhere in maths — from physics and engineering to economics and geometry. Solving them helps us find the values of \( x \) that make the equation true, known as the roots or solutions.

How to Solve a Quadratic Equation

There are three main methods to solve quadratic equations:

1. Factorisation

Factorisation is the process of breaking down a quadratic expression into two linear brackets that multiply to give the original equation. It’s the simplest method when the numbers are easy to work with.If the quadratic can be neatly factorised, this is often the quickest method.

Example:

\[ x² + 5x + 6 = 0 \]Factorise into brackets:\[(x + 2)(x + 3) = 0 \]Set each bracket equal to zero:\[ x + 2 = 0 → x = -2, x + 3 = 0 → x = -3 \]Tip: This works best when the coefficients are simple integers.

2. Completing the Square

Completing the Square is a powerful technique that rewrites a quadratic into a perfect square form, making it easier to understand the shape and position of its graph.

Example:

\[ x^2 + 2x – 5 = 0 \]

Rewrite as:

\[ (x + 1)^2 – 6 = 0 \]

Taking square roots:

\[ x + 1 = ±\sqrt{6} \]So, \( x = -1 ± \sqrt{6} \).

Tip: This method is especially useful when the quadratic does not factorise nicely.

3. Quadratic Formula

When a quadratic equation cannot be easily factorised, or when the numbers are not nice and tidy, we use the quadratic formula — a powerful tool that works for any quadratic equation:

\(x = \dfrac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

Example:\[ 2x^2 + 3x – 2 = 0 \]Coefficients: \( a = 2, b = 3, c = -2 \)

Substitute into the formula:

\[ x = \dfrac{-3 ± \sqrt{3^2 – 4(2)(-2)}}{2(2)} \]Simplify:\[ x = \dfrac{-3 ± \sqrt{9 + 16}}{4} \]\[ x = \dfrac{-3 ± \sqrt{25}}{4} \]Final solutions:\[ x = \dfrac{-3 + 5}{4} = \dfrac{1}{2}, \quad x = \dfrac{-3 – 5}{4} = -2 \]

What Is the Discriminant?

The expression inside the square root — b² – 4ac — is called the discriminant.The discriminant tells us the nature of the roots of the quadratic:

Value of Discriminant	Meaning	Type of Roots
b² – 4ac > 0	Two distinct real roots	The graph cuts the x-axis twice
b² – 4ac = 0	One repeated real root	The graph touches the x-axis once
b² – 4ac < 0	No real roots	The graph does not cross the x-axis

In short: The discriminant helps us predict what kind of solutions we’ll get — even before solving!

Examiner Tips & Tricks

• Make sure the equation has “= 0” on the right-hand side. If not, rearrange first.
• Try to have the ax² term on the positive side (i.e., a > 0) — it’s clearer for completing the square and for checking signs.
• Always check how the question asks you to present answers: exact form, 2 decimal places, 3 significant figures, or simplified fractions.
• If the discriminant is a perfect square, expect rational/surd-free answers; if not, give exact surd form or round as instructed.
• After solving, substitute one root back into the original equation to double check (optional in exam if time permits).

Worked Example (All Three Methods)

Method A — Factorisation

1. Start with:\[ 2x² + 3x – 2 = 0 \]
2. Multiply\[ a × c : 2 × (-2) = -4 \]
3. Find two integers that multiply to \( -4 \) and add to \( b = 3 \): those are \( 4 \) and \( -1 \).
4. Split the middle term:\[ 2x² + 4x – x – 2 = 0 \]
5. Factorise by grouping:\[ 2x(x+2) -1(x+2) = 0 → (2x-1)(x+2)=0 \]
6. Set each factor equal to zero:\[ 2x – 1 = 0 → x = \dfrac{1}{2} , x + 2 = 0 → x = -2 \]
Solutions: \( x = \dfrac{1}{2}, x = -2 \)

Method B — Completing the Square

1. Start with the quadratic:

   \[ 2x² + 3x – 2 = 0 \]

2. Divide through by 2:

   \[ x² + \dfrac{3}{2}x – 1 = 0 \]

3. Move the constant:

   \[ x² + \dfrac{3}{2}x = 1 \]

4. Take half of the coefficient of \( x \), square it:Half of \(\dfrac{3}{2}\) is \(\dfrac{3}{4}\); square → \(\dfrac{9}{16}\).
5. Add \(\dfrac{9}{16}\) to both sides:

   \[ x² + \dfrac{3}{2}x + \dfrac{9}{16} = 1 + \dfrac{9}{16} \]

6. Rewrite as a perfect square:

   \[ (x + \dfrac{3}{4})^2 = \dfrac{25}{16} \]

7. Take square roots:

   \[ x + \dfrac{3}{4} = ± \dfrac{5}{4} \]

8. Solve for \(x\):

   \[ x = -\dfrac{3}{4} + \dfrac{5}{4} = \dfrac{1}{2}, x = -\dfrac{3}{4} – \dfrac{5}{4} = -2 \]

Solutions: \( x = \dfrac{1}{2}, x = -2 \)

Method C — Quadratic Formula

1. Identify coefficients: \( a = 2, b = 3, c = -2 \)
2. Compute discriminant:

   \[ b² – 4ac = 3² – 4(2)(-2) = 9 + 16 = 25\]

3. Substitute into the formula:

   \[ x = \dfrac{-3 ± \sqrt{25}}{4}\]

4. Split into cases:

   \[ x = \dfrac{-3 + 5}{4} = \dfrac{1}{2}, x = \dfrac{-3 – 5}{4} = -2 \]

   Solutions: \( x = \dfrac{1}{2}, x = -2 \)

Summary

• A quadratic equation is of the form \( ax² + bx + c = 0 \).
• The three solving methods: Factorising, Completing the Square, and Quadratic Formula.
• The discriminant \( b² – 4ac \) determines the type of roots.

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