Quadratic Equations Questions with Solutions: 12 Solved MCQs

Remembering the quadratic formula is not enough when a question asks about transformed roots, a parameter range, or an equation hidden inside a word problem. Attempt each MCQ before reading its solution, and keep your own steps visible. These 12 published question-bank questions move from direct root calculation to discriminants, Vieta's relations, graphs, word problems, and an unknown-base problem. They are a focused practice set, not a representation of any official exam pattern.

Quadratic equations: four tools to keep beside the set

Write a quadratic equation in the standard form ax² + bx + c = 0, where a ≠ 0. Its roots are

x = (-b ± √(b² - 4ac))/(2a).

The discriminant is D = b² - 4ac. If D > 0, there are two distinct real roots. If D = 0, the real roots are equal. If D < 0, there are no real roots.

For roots α and β, Vieta's relations give α + β = -b/a and αβ = c/a. For reciprocal roots 1/α and 1/β, the sum becomes (α + β)/(αβ) and the product becomes 1/(αβ).

Use the Engineering Mathematics course for structured theory and the Aptitude category for broader practice. The CAT Quant: Arithmetic and Algebra First guide shows where algebra fits in a sensible study order.

Questions 1-3: calculate roots and read the discriminant

Question 1: direct quadratic formula

Solve using the quadratic formula: x2 - 2x - 24 = 0

• (a) x = -6; x = 4

• (b) x = 4; x = 6

• (c) x = 6; x = -4

• (d) x = -4; x = -6

Solution. Typeset the equation as x² - 2x - 24 = 0, so a = 1, b = -2, and c = -24.

D = (-2)² - 4(1)(-24) = 4 + 96 = 100

Therefore, x = (2 ± 10)/2, which gives x = 6 or x = -4. Answer: (c).

Question 2: irrational coefficients

Find the zeros of the quadratic polynomial √3x² − 8x + 4√3.

• (a) −2√3, 2/√3

• (b) −2√3, −2/√3

• (c) 2√3, 2/√3

• (d) 2√3, −2/√3

Solution. Here a = √3, b = -8, and c = 4√3.

D = 64 - 4(√3)(4√3) = 64 - 48 = 16

Thus x = (8 ± 4)/(2√3). The roots are 12/(2√3) = 2√3 and 4/(2√3) = 2/√3. Answer: (c).

Question 3: condition for distinct real roots

The quadratic equation ax^2 + bx + c = 0 has two distinct real roots if:

• (a) b^2 - 4ac < 0

• (b) b^2 - 4ac > 0

• (c) b^2 - 4ac = 0

• (d) More than one of the above

• (e) None of the above

Solution. Let D = b² - 4ac. A negative D gives no real roots, zero gives equal real roots, and a positive D gives two distinct real roots. Answer: (b).

Questions 4-6: parameter ranges, the vertex, and root spacing

Question 4: count integer parameters that keep roots real

x² - 9x + |k| = 0 has real roots. How many integer values can k take?

• (a) 20

• (b) 41

• (c) 40

• (d) 21

Solution. Real roots require

D = 81 - 4|k| ≥ 0, so |k| ≤ 81/4 = 20.25.

For integer k, -20 ≤ k ≤ 20: 20 negative values, zero, and 20 positive values, giving 41. At the boundary, k = 20 gives D = 1, while k = 21 gives D = -3. Answer: (b).

Question 5: vertex of a parabola

The graph of y = x² + 4x + 3 has its vertex at:

• (a) (−2, −1)

• (b) (2, 1)

• (c) (−1, 3)

• (d) (1, 3)

Solution. The vertex has

x_v = -b/(2a) = -4/2 = -2.

Substitution gives y_v = (-2)² + 4(-2) + 3 = -1. Completing the square confirms it: y = (x + 2)² - 1. Answer: (a).

Question 6: consecutive integer roots

If the roots of the equation x² − bx + c = 0 are two consecutive integers, then b² − 4c is

• (a) 1

• (b) -1

• (c) 0

• (d) 2

Solution. Let the roots be n and n + 1. Vieta gives b = 2n + 1 and c = n(n + 1). Hence

(2n + 1)² - 4n(n + 1) = 4n² + 4n + 1 - 4n² - 4n = 1.

The faster check is D = (α - β)² = 1² = 1. Answer: (a).

Questions 7-9: Vieta's relations and transformed roots

Question 7: recover a coefficient from a relation between roots

If α, β are zeros of x² − 6x + k, what is the value of k if 3α + 2β = 20?

• (a) 8

• (b) -8

• (c) -2

• (d) -16

Solution. Vieta gives α + β = 6 and αβ = k. Put β = 6 - α into the given relation:

3α + 2(6 - α) = 20

3α + 12 - 2α = 20, so α = 8, β = -2, and k = αβ = -16. Answer: (d).

Question 8: equation whose roots are reciprocals

If x₁ and x₂ are the roots of the equation x² + 2x − 15 = 0 then the quadratic equation which has the roots 1/x₁ and 1/x₂ is:

• (a) −15x² − 2x − 1 = 0

• (b) 15x² − 2x − 1 = 0

• (c) 15x² − 2x + 1 = 0

• (d) 15x² + 2x − 1 = 0

Solution. Here x₁ + x₂ = -2 and x₁x₂ = -15. The reciprocal roots have sum (-2)/(-15) = 2/15 and product 1/(-15) = -1/15. Their equation is

x² - (2/15)x - 1/15 = 0.

Multiplying by 15 gives 15x² - 2x - 1 = 0. Answer: (b).

Question 9: reduce an expression using the root equation

Let r be a root of the equation x² + 2x + 6 = 0. Then the value of the expression (r + 2)(r + 3)(r + 4)(r + 5) is

• (a) 51

• (b) -51

• (c) 126

• (d) -126

Solution. The root equation gives r² = -2r - 6. Pair the factors:

(r + 2)(r + 5) = r² + 7r + 10 = 5r + 4

(r + 3)(r + 4) = r² + 7r + 12 = 5r + 6

Their product is 25r² + 50r + 24. Substitution gives 25(-2r - 6) + 50r + 24 = -126. Answer: (d).

Questions 10-12: build the quadratic hidden in the wording

Question 10: rectangle equation formation

A rectangle has area 154 cm². Its length is 4 times the breadth plus 4 cm. What would be the quadratic equation?

• (a) 4b²+4b−154=0.

• (b) 4b²+4b−1=0.

• (c) 4b²+4b−4=0.

• (d) 4b²+4b−14=0.

Solution. Let breadth be b cm, so length is 4b + 4 cm. Area gives b(4b + 4) = 154, hence 4b² + 4b - 154 = 0. The stem asks only for the equation, so stop here. Answer: (a).

Question 11: benches and a fixed total

In a school, 600 students are to be seated on benches, with an equal number of students seated on each bench. If the number of benches available is reduced by 10, each bench would need to seat 2 more students to accommodate all 600 students. Find the original number of benches.

• (a) 65

• (b) 120

• (c) 60

• (d) 70

Solution. Let x be the original benches and y the students per bench. Then xy = 600 and (x - 10)(y + 2) = 600. Expanding and using xy = 600 gives

600 + 2x - 10y - 20 = 600, so x - 5y = 10 and x = 5y + 10.

Now xy = 600 becomes y² + 2y - 120 = 0 = (y + 12)(y - 10). Reject y = -12; with y = 10, x = 60. Check: 60 × 10 = 600 and 50 × 12 = 600. Answer: (c).

Question 12: unknown radix

What is the radix of the numbers if the solution to the quadratic equation x² - 10x + 26 = 0 is x = 4 and x = 7?

• (a) 8

• (b) 9

• (c) 10

• (d) 11

Solution. Let the unknown base be R. Then 10_R means decimal R, while 26_R means decimal 2R + 6. The equation is x² - Rx + (2R + 6) = 0.

The roots sum to 11, so Vieta gives R = 11. Their product is 4 × 7 = 28 = 2(11) + 6, confirming the constant term. Digits 0, 1, 2, and 6 are valid in base 11. Answer: (d).

Score the set and diagnose the method to revise

Give yourself one point for each correct response. These are editorial practice bands, not official cutoffs:

• 10-12: ready for a mixed timed set

• 7-9: revise the one method that caused most errors

• 4-6: rebuild discriminants and Vieta's relations

• 0-3: restart with direct factorisation and the quadratic formula

Use question numbers to find the gap. Q1-Q2 test formula use and arithmetic; Q3-Q6 test the discriminant and graph meaning; Q7-Q9 test Vieta and transformed roots; Q10-Q12 test modelling and interpretation. The MCA Entrance Mathematics: Algebra to Calculus Order guide can help you decide what should follow algebra.

Answer key, short revision card, and next practice step

Keep this four-step revision card beside your next set:

1. Standardise the equation.

2. Choose factorisation, the formula, the discriminant, or Vieta.

3. Keep the required condition visible.

4. Substitute back or cross-check the roots.

For structured drills, continue with the Aptitude for Placement course. For the wider algebra study order, return to the CAT Quant guide linked above. The useful next step is simple: redo every missed question without looking at its solution, then explain why the other options fail.