Course Notes Practice Past Paper Archive

Solving Quadratic Equations0%

Algebra · Topic 12 of 12

Solving Quadratic Equations

Video lesson6 worked examples

Theory

Solving

To calculate the roots, set the equation to zero. Try to solve by factorising first. If that isn't possible, use the Quadratic Formula:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The Discriminant

The expression $b^2 - 4ac$ is the discriminant. It dictates the nature of the roots:

$b^2 - 4ac > 0$ : 2 real distinct roots.
$b^2 - 4ac = 0$ : 1 repeated real root (equal roots).
$b^2 - 4ac < 0$ : No real roots.

The Golden Rule: to solve, always make one side zero first, then factorise if you can and use the formula only if you cannot. To describe the roots without solving, evaluate the discriminant $b^2 - 4ac$ .

⚠️ Common Examiner Traps

Not setting to zero: $16t - t^2 = 60$ must be rearranged to $t^2 - 16t + 60 = 0$ before factorising.
Formula sign errors: substitute $a, b, c$ with their signs. If $c = -4$ then $-4ac$ becomes $+$ — mishandling this is the most common formula slip.
Discriminant vs roots: $b^2 - 4ac$ tells you the nature of the roots, not the roots themselves.
Rejecting impossible answers: in a context (a length, a number of items), discard any negative or otherwise impossible solution — and say why.
Use the formula when asked for decimals: “correct to 2 d.p.” signals the quadratic does not factorise — go straight to the formula.

Worked examples

Example 1

Solve by Factorising (unitary)

Solve $x^2 - 8x + 15 = 0$ .

Step 1: Factorise the trinomial: $(x - 3)(x - 5) = 0$ .

Step 2: A product is zero when a factor is zero, so set each bracket to zero. Answer: $x = 3$ or $x = 5$ .

Example 2

Solve by Factorising (non-unitary)

Solve $3x^2 - 5x - 2 = 0$ .

Step 1: Factorise, with first terms multiplying to $3x^2$ and last terms to $-2$ : $(3x + 1)(x - 2) = 0$ .

Step 2: Set each bracket to zero: $3x + 1 = 0$ gives $x = -\tfrac{1}{3}$ , and $x - 2 = 0$ gives $x = 2$ .

Answer: $x = -\tfrac{1}{3}$ or $x = 2$ .

Example 3

Rearranging First

Solve $x^2 = 3x + 10$ .

Step 1: Make one side zero before doing anything else: $x^2 - 3x - 10 = 0$ .

Step 2: Factorise: $(x - 5)(x + 2) = 0$ .

Answer: $x = 5$ or $x = -2$ .

Example 4

Using the Quadratic Formula

Solve $2x^2 + 3x - 4 = 0$ , giving the solutions correct to 2 decimal places.

Step 1: It does not factorise, so use the formula with $a = 2,\ b = 3,\ c = -4$ :

x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4}

Step 2: Since $\sqrt{41} \approx 6.40$ , work out both values:

x = \frac{-3 + 6.40}{4} = 0.85 \qquad \text{or} \qquad x = \frac{-3 - 6.40}{4} = -2.35

Answer: $x = 0.85$ or $x = -2.35$ (2 d.p.).

Example 5

Nature of the Roots (discriminant)

Determine the nature of the roots of $2x^2 + 3x + 4 = 0$ .

Step 1: Identify $a = 2,\ b = 3,\ c = 4$ and evaluate the discriminant $b^2 - 4ac$ : $3^2 - 4(2)(4) = 9 - 32 = -23$ .

Answer: Since $-23 < 0$ , the equation has no real roots.

Example 6

🎯 Exam-style (word problem)

A rectangle has length $(x + 3)$ cm and breadth $x$ cm. Its area is $40$ cm². Show that $x^2 + 3x - 40 = 0$ , and find the breadth.

Step 1: Area is length × breadth, so $x(x + 3) = 40$ . Expand and set to zero:

x^2 + 3x = 40 \implies x^2 + 3x - 40 = 0

Step 2: Factorise: $(x + 8)(x - 5) = 0$ , giving $x = -8$ or $x = 5$ .

Step 3: A breadth cannot be negative, so reject $x = -8$ .

Answer: the breadth is $5$ cm.

Ready to practise?

22 Quadratics questions with answers and video solutions