Algebra & Equations

Algebra covers solving linear, quadratic, simultaneous, exponential and surd equations, plus working with inequalities and the nature of roots.

Algebra covers solving linear, quadratic, simultaneous, exponential and surd equations, plus working with inequalities and the nature of roots.

Key concepts

  • Quadratic formula: x = (−b ± √(b²−4ac)) / 2a
  • Discriminant: Δ = b² − 4ac determines nature of roots
  • Simultaneous equations by substitution or elimination
  • Exponential equations: equate bases or take logs

How it works

Always write the equation in standard form first. Try factoring before using the formula. Check solutions in the original equation, especially with surds and rational equations. Before solving, look at the form of the equation: a standard quadratic, an equation with a surd, a fraction with x in the denominator, or a pair of simultaneous equations. Each has a standard first move: factorise or use the formula, isolate and square the surd, multiply by the lowest common denominator, or substitute from the linear equation.

Step by step

1. Standard form

Move everything to one side equal to 0.

2. Try factoring

Factoring is faster than the formula when possible.

3. Use the formula

When factoring fails, substitute into the quadratic formula carefully.

4. Verify

Substitute solutions back into the original equation.

Worked examples

Solve x² − 5x + 6 = 0

  1. Factor: (x − 2)(x − 3) = 0
  2. x = 2 or x = 3

Answer: x = 2 or x = 3

Solve x² − 2x − 5 = 0, correct to two decimal places

  1. It does not factorise, so use the formula with a = 1, b = −2, c = −5
  2. x = [2 ± √(4 + 20)] / 2 = [2 ± √24] / 2
  3. x = 1 ± √6

Answer: x ≈ 3,45 or x ≈ −1,45

Solve x − √(x + 2) = 0

  1. Isolate the root: x = √(x + 2)
  2. Square both sides: x² = x + 2, so x² − x − 2 = 0
  3. (x − 2)(x + 1) = 0, so x = 2 or x = −1
  4. Check x = −1: −1 − √1 = −2 ≠ 0, so reject it

Answer: x = 2

Solve simultaneously: x + y = 5 and xy = 6

  1. From the first equation, y = 5 − x
  2. x(5 − x) = 6, so x² − 5x + 6 = 0
  3. (x − 2)(x − 3) = 0

Answer: x = 2, y = 3 or x = 3, y = 2

Solve x² + 3x ≤ 10

  1. x² + 3x − 10 ≤ 0
  2. (x + 5)(x − 2) ≤ 0
  3. The parabola opens upwards, so it is negative between its roots

Answer: −5 ≤ x ≤ 2

Common mistakes

  • Forgetting ± when taking square roots
  • Dropping a solution after factoring
  • Sign errors with the discriminant
  • Not checking for extraneous roots after squaring both sides

Frequently asked questions

What's the nature of roots?

Δ > 0: real & unequal; Δ = 0: real & equal; Δ < 0: non-real.

What are extraneous roots?

Roots that appear when you square both sides of an equation but do not satisfy the original equation. Always substitute your answers back to check.

How do I write the solution of a quadratic inequality?

Either as one interval between the roots, such as −5 ≤ x ≤ 2, or as two regions, such as x < −5 or x > 2, depending on the sign the inequality needs.

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