12.1 Algebra

12.1 Algebra

Quadratic Equations

A quadratic equation in one variable is any equation that can be written in the standard form ax² + bx + c = 0, where a ≠ 0. The highest power of x is 2, which is what makes it quadratic. On the PAF GD Pilot Initial Test, quadratics appear in factorisation, root-finding, and word problems on areas and projectiles.

The two roots are given by the quadratic formula:

• x = (−b ± √(b² − 4ac)) / 2a

The expression under the radical, Δ = b² − 4ac, is called the discriminant. It tells you the nature of the roots before you solve:

• Δ > 0 → two distinct real roots
• Δ = 0 → one repeated real root (the parabola just touches the x-axis)
• Δ < 0 → no real roots (two complex conjugate roots)

Worked example — using the discriminant: For 2x² − 4x + 3 = 0, Δ = (−4)² − 4(2)(3) = 16 − 24 = −8. Since Δ < 0, the equation has no real roots — the parabola sits entirely above the x-axis.

Worked example — using the formula: Solve x² − 5x + 6 = 0. Here a = 1, b = −5, c = 6, so x = (5 ± √(25 − 24)) / 2 = (5 ± 1) / 2, giving x = 3 or x = 2.

Factorisation

When the coefficients are friendly, factorisation is faster than the formula. You look for two numbers that multiply to a·c and add to b.

Worked example — factorising x² − 5x + 6:

Here a = 1, b = −5, c = 6. We need two numbers that multiply to +6 and add to −5. Those numbers are −3 and −2, so:

• x² − 5x + 6 = (x − 3)(x − 2)

Setting each factor to zero gives the roots x = 3 and x = 2. Verify by expanding: (x − 3)(x − 2) = x² − 2x − 3x + 6 = x² − 5x + 6. ✓

Worked example — non-monic quadratic: Factorise 3x² − 5x − 2. We need two numbers that multiply to (3)(−2) = −6 and add to −5. Those are −6 and +1. Split the middle term:

• 3x² − 6x + x − 2 = 3x(x − 2) + 1(x − 2) = (3x + 1)(x − 2)

Roots: x = 2 and x = −1/3. Check with the formula: x = (5 ± √(25 + 24)) / 6 = (5 ± 7) / 6, so x = 2 or x = −1/3. ✓

Sum and Product of Roots (Vieta's Formulas)

For any quadratic ax² + bx + c = 0 with roots α and β, Vieta's formulas give:

• Sum of roots: α + β = −b/a
• Product of roots: αβ = c/a

These let you answer questions about roots without solving the equation — a frequent shortcut on the test.

Worked example: For 3x² − 5x − 2 = 0, the sum of roots is α + β = −(−5)/3 = 5/3, and the product is αβ = (−2)/3. Verify with the roots found above: 2 + (−1/3) = 5/3 ✓ and 2 × (−1/3) = −2/3 ✓.

Simultaneous Equations

A system of two linear equations in two unknowns can be solved by substitution or elimination. Elimination is usually faster on the test.

Worked example: Solve

• 2x + 3y = 13
• 3x − y = 3

Multiply the second equation by 3: 9x − 3y = 9. Add to the first equation: 11x = 22, so x = 2. Substitute back: 3(2) − y = 3 ⇒ y = 3. Solution: (2, 3). Always substitute back into the other equation to check: 2(2) + 3(3) = 4 + 9 = 13. ✓

Polynomials and HCF/LCM

A polynomial is a sum of terms of the form axⁿ. The degree is the highest power of x. Division of polynomials uses the identity Dividend = Divisor × Quotient + Remainder.

The HCF (highest common factor) of two algebraic expressions is the largest expression that divides both exactly; the LCM is the smallest expression both divide exactly. Use factorisation to find them.

Worked example — HCF of x² − 1 and x³ − 1:

• x² − 1 = (x − 1)(x + 1)
• x³ − 1 = (x − 1)(x² + x + 1)

The only common factor is (x − 1), so the HCF is (x − 1). The LCM is (x − 1)(x + 1)(x² + x + 1) — take each distinct factor at its highest power.

Extraneous Roots and Squaring Both Sides

A common PAF trap arises when an equation contains a square root and you "square both sides" to remove it. Squaring is not a reversible operation — it can introduce extraneous roots that satisfy the squared equation but not the original. Always substitute every candidate back into the original equation before accepting it.

Worked example — extraneous root: Solve √(2x + 1) = x − 2.

• Square both sides: 2x + 1 = (x − 2)² = x² − 4x + 4
• Rearrange: x² − 6x + 3 = 0
• Quadratic formula: x = (6 ± √(36 − 12)) / 2 = (6 ± √24) / 2 = 3 ± √6
• Candidates: x ≈ 5.449 or x ≈ 0.551
• Check x ≈ 5.449: √(2(5.449) + 1) = √11.898 ≈ 3.449, and x − 2 ≈ 3.449. ✓
• Check x ≈ 0.551: √(2(0.551) + 1) = √2.102 ≈ 1.450, but x − 2 ≈ −1.449. ✗ (a square root is non-negative, so it cannot equal a negative number)

So the only valid solution is x = 3 + √6. The other candidate is extraneous — introduced by squaring, not by the original equation. The same trap appears whenever you raise both sides to any even power.

Linear–Quadratic Systems

A simultaneous system can also mix a linear equation with a quadratic. Substitute the linear expression into the quadratic, then solve the resulting quadratic.

Worked example: Solve

• y = x²
• y = x + 2

Substitute: x² = x + 2 ⇒ x² − x − 2 = 0 ⇒ (x − 2)(x + 1) = 0, so x = 2 or x = −1. Corresponding y values: y = 4 or y = 1. Two solution pairs: (2, 4) and (−1, 1). Geometrically these are the intersection points of the parabola y = x² and the line y = x + 2. A linear–quadratic system can have zero, one, or two solution pairs — count the intersections of the line and the parabola, and use the discriminant of the resulting quadratic to predict how many to expect.