Linear and Quadratic Equations

Linear and Quadratic Equations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A linear equation in one variable has the form ax + b = 0 with a ≠ 0 and gives the single solution x = −b/a. A quadratic equation has the form ax² + bx + c = 0 with a ≠ 0, degree 2 in the variable, and up to two roots. The quadratic formula x = (−b ± √(b² − 4ac)) / 2a solves any quadratic, where the discriminant D = b² − 4ac tells you the nature of the roots: D > 0 gives two distinct real roots, D = 0 gives a repeated root, D < 0 gives no real roots. Sum of roots = −b/a and product of roots = c/a. NECO SSCE almost always tests factorization, the formula, and one word problem — practise all three.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Forms and Core Definitions

A linear equation in x is any equation reducible to ax + b = 0, a ≠ 0. Solving means isolating x by transposition (moving a term across the equality sign reverses its sign) and simplifying. A quadratic equation in x is ax² + bx + c = 0, a ≠ 0; a, b, c are the coefficients and c the constant term.

Four Methods of Solving Quadratics

1. Factorization — rewrite the left side as a product of two linear factors, then set each to zero. Works for trinomials and difference of two squares (x² − k² = (x − k)(x + k)).
2. Completing the square — rewrite as (x + p)² = q, then take square roots. Useful when a = 1 and factorization is awkward.
3. Quadratic formula — universal; substitute a, b, c into x = (−b ± √(b² − 4ac)) / 2a. Always keep the ± sign.
4. Graphical method — the parabola y = ax² + bx + c crosses the x-axis where y = 0, so the roots are the x-intercepts.

Discriminant at a Glance

D = b² − 4ac	Nature of roots
D > 0	Two distinct real roots
D = 0	One repeated (equal) real root
D < 0	No real roots (complex conjugates)

Forming Equations from Roots

If α and β are the roots, the equation is x² − (α + β)x + αβ = 0. This is faster than re-deriving from coefficients.

Simultaneous Linear–Quadratic Pairs

A typical NECO pair is y = mx + c together with y = ax² + bx + c. Equate the two expressions for y to get one quadratic in x, solve it, then substitute back to find y. Expect two ordered pairs, sometimes a repeated one.

Exam Pointers

NECO Paper 1 carries 60 objective items; expect 4–6 direct questions on linear/quadratic solving plus a discriminant item. Paper 2 (essay) almost always allocates 15–25 marks to a word problem, simultaneous equations, or “form the equation whose roots are…”.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Common Traps

• a = 0 collapses the quadratic to a linear equation. If a question gives a “quadratic” with leading coefficient 0 after simplification, reclassify it — there is only one root.
• Sign discipline with the formula. The numerator is −b, not b. A slip here produces two wrong roots, no partial credit on objective items.
• D < 0 ≠ no solution — the equation still has two complex roots x = (−b ± i√(4ac − b²)) / 2a. In NECO SSCE (real-number syllabus) the safe wording is “no real roots”.
• Product of roots = c/a, not −c/a. Memorise via the factored form a(x − α)(x − β) = 0 and compare.
• Completing the square sign. Rewrite x² + bx as (x + b/2)² − (b/2)²; the sign inside the bracket mirrors the coefficient, not its opposite.

Worked Example

Solve 2x² − 5x − 3 = 0. Here a = 2, b = −5, c = −3, so D = 25 + 24 = 49, √D = 7. Then x = (5 ± 7) / 4, giving x = 3 or x = −1/2. Check by factorization: 2x² − 5x − 3 = (2x + 1)(x − 3) — confirmed.

Connection to Other Topics

Quadratics link directly to coordinate geometry (parabola sketches, axis of symmetry x = −b/2a, vertex), inequalities (sign of ax² + bx + c), and sequences (quadratic nth-term problems). Word problems on ages, areas, and uniform motion typically reduce to a linear or quadratic once you translate “is twice as old”, “exceeds by”, or “product equals” into algebra.

Two Practice Prompts

1. Find k for which x² − 3kx + 9 = 0 has equal roots, hence solve.
2. The sum of two numbers is 11 and the product is 28; form the quadratic equation whose roots are the numbers, then solve it.

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