Quadratic equations is one of the most predictable chapters in the Class 10 paper. The questions below are the ones that come back year after year, in the form they come back in. Do each one on paper before reading the answer.
How to use this list
Solve first, check second. If you got it right, move on. If not, write the question into your Mistake Book with the step that went wrong — that is the list you revise the night before.
Short answer (1–2 marks)
Q1. Find the discriminant of 2x² − 4x + 3 = 0 and state the nature of the roots.
D = b² − 4ac = (−4)² − 4(2)(3) = 16 − 24 = −8. Since D < 0, there are no real roots.
Q2. For what value of k does kx² + 4x + 1 = 0 have equal roots?
Equal roots need D = 0: 16 − 4k = 0, so k = 4.
Q3. Is x = −1 a root of x² + 3x + 2 = 0?
Substitute: 1 − 3 + 2 = 0. Yes.
Q4. Write 3x² = 5x − 2 in standard form and state a, b, c.
3x² − 5x + 2 = 0, so a = 3, b = −5, c = 2. (The sign of b is where marks are lost.)
Standard problems (3 marks)
Q5. Solve by factorisation: x² − 7x + 12 = 0.
Two numbers that multiply to 12 and add to −7: −3 and −4. So (x − 3)(x − 4) = 0, giving x = 3 or x = 4.
Q6. Solve 2x² + x − 6 = 0 by factorisation.
Product ac = −12, sum b = 1: the pair is 4 and −3. Split: 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2) = 0. So x = 3/2 or x = −2.
Q7. Solve x² − 2x − 2 = 0 using the quadratic formula.
x = [2 ± √(4 + 8)]/2 = [2 ± √12]/2 = [2 ± 2√3]/2 = 1 ± √3.
Q8. Find k so that x² + kx + 9 = 0 has real roots.
D ≥ 0: k² − 36 ≥ 0, so k ≤ −6 or k ≥ 6. Write both parts — "k ≥ 6" alone loses a mark.
Long answer (4–5 marks): word problems
The recipe is always the same: name the unknown, write the equation from the sentence, solve, and reject the root that makes no sense (a negative length, a fractional number of people).
Q9. The sum of the squares of two consecutive natural numbers is 313. Find them.
Let the numbers be n and n + 1. Then n² + (n + 1)² = 313, so 2n² + 2n + 1 = 313, so n² + n − 156 = 0, so (n + 13)(n − 12) = 0. Reject n = −13. The numbers are 12 and 13.
Q10. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed.
Let speed be x km/h. Time difference: 360/x − 360/(x + 5) = 1. Multiply out: 360(x + 5) − 360x = x(x + 5), so 1800 = x² + 5x, so x² + 5x − 1800 = 0, so (x + 45)(x − 40) = 0. Speed = 40 km/h.
Q11. The hypotenuse of a right triangle is 13 cm and one side is 7 cm longer than the other. Find the sides.
Let the shorter side be x. Then x² + (x + 7)² = 169, so 2x² + 14x − 120 = 0, so x² + 7x − 60 = 0, so (x + 12)(x − 5) = 0. Sides: 5 cm and 12 cm.
Q12. Two pipes together fill a tank in 6 hours. The larger pipe alone takes 5 hours less than the smaller. How long does each take alone?
Let the smaller take x hours; the larger takes x − 5. In one hour they fill 1/x + 1/(x − 5) = 1/6. So 6(x − 5) + 6x = x(x − 5), so 12x − 30 = x² − 5x, so x² − 17x + 30 = 0, so (x − 15)(x − 2) = 0. x = 2 makes the larger pipe take −3 hours, so reject it. Smaller: 15 h, larger: 10 h.
What to revise if you got these wrong
- Q1–Q2, Q8: the discriminant and what each sign means. Ten minutes.
- Q5–Q6: splitting the middle term. Do five more with ac negative.
- Q7: the formula, including simplifying √12 to 2√3 — that simplification is a mark.
- Q9–Q12: translating sentences into equations. Write the equation before touching any algebra, and always state which root you reject and why.