Quadratic Equations Aptitude Questions and Answers (Solved MCQs)

c) 8 — Factorising gives x²−11x+24 = (x−3)(x−8). Therefore, the roots are 3 and 8, and the larger root is 8.

a) 3 and 1/2 — Splitting the middle term gives 2x²−6x−x+3 = 0. Therefore, 2x(x−3)−1(x−3) = 0, so (2x−1)(x−3) = 0. Hence, x = 1/2 or x = 3.

a) 1 and −7 — We need two numbers whose product is −7 and sum is 6. These numbers are 7 and −1. Therefore, x²+6x−7 = (x−1)(x+7), giving the roots x = 1 and x = −7.

c) 6 — Factorising gives x²−8x+7 = (x−1)(x−7). Thus, the roots are 1 and 7. Their positive difference is 7−1 = 6.

b) 4/3 and −2 — Splitting the middle term gives 3x²+6x−4x−8 = 0. Therefore, 3x(x+2)−4(x+2) = 0, so (3x−4)(x+2) = 0. Hence, x = 4/3 or x = −2.

c) 6 — Multiplying the equation by 2 gives x²−5x+6 = 0. Factorising gives (x−2)(x−3) = 0, so the roots are 2 and 3. Their product is 2×3 = 6.

b) 2 and 7 — Expanding gives x²−4x+4 = 5x−10. Bringing all terms to one side gives x²−9x+14 = 0. Factorising gives (x−2)(x−7) = 0. Therefore, the roots are 2 and 7.

b) 2 and 3 — Multiplying by x gives x²+6 = 5x. Therefore, x²−5x+6 = 0, which factors as (x−2)(x−3) = 0. Hence, x = 2 or x = 3, and both satisfy x ≠ 0.

d) 4 — Let y = x². The equation becomes y²−13y+36 = 0, which factors as (y−4)(y−9) = 0. Thus, x² = 4 or x² = 9. Therefore, x = ±2 or x = ±3, giving four distinct real roots.

c) 9 — Let y = √x, where y ≥ 0. Then x = y², and the equation becomes y²−y−6 = 0. Factorising gives (y−3)(y+2) = 0. Since y cannot be negative, y = 3. Therefore, x = y² = 9.

b) 4 — Factorising gives x²−10x+21 = (x−3)(x−7). Thus, the roots are 3 and 7, and their positive difference is 7−3 = 4.

b) 5 — Substituting x = 2 gives 2²−2k+6 = 0. Therefore, 10−2k = 0, so k = 5. Alternatively, the other root is 3 because the product of the roots is 6, making their sum 5.

b) m = ±10 — A perfect-square expression with first term x² and last term 25 must be either (x+5)² or (x−5)². Expanding gives x²+10x+25 or x²−10x+25. Therefore, m = 10 or m = −10.

a) (5±√89)/4 — Expanding gives 2x²−5x−3 = 5. Therefore, 2x²−5x−8 = 0. Using the quadratic formula, x = [5±√{(−5)²−4(2)(−8)}]/4 = [5±√89]/4.

b) 5 — Completing the square gives x²−12x+41 = (x−6)²+5. Since (x−6)² is always non-negative, its minimum value is 0 at x = 6. Therefore, the minimum value of the expression is 5.

c) Two equal real roots — The discriminant is D = b²−4ac = (−6)²−4(1)(9) = 36−36 = 0. Therefore, the equation has two equal real roots. In fact, x²−6x+9 = (x−3)², so both roots are 3.

c) k = 8 or −4 — For equal roots, the discriminant must be zero. Therefore, (k−2)²−4(1)(9) = 0. This gives (k−2)² = 36, so k−2 = ±6. Hence, k = 8 or k = −4.

c) Two non-real conjugate roots — The discriminant is D = 4²−4(2)(5) = 16−40 = −24. Since the discriminant is negative, the equation has two non-real conjugate roots.

c) 37/9 — By the relationships between roots and coefficients, α+β = 7/3 and αβ = 2/3. Therefore, α²+β² = (α+β)²−2αβ = 49/9−4/3 = 49/9−12/9 = 37/9.

b) x²−4x+1 = 0 — The sum of the roots is (2+√3)+(2−√3) = 4. Their product is (2+√3)(2−√3) = 4−3 = 1. Therefore, the required equation is x²−4x+1 = 0.

b) 5 — If the roots are reciprocals, their product is 1. For 5x²−9x+k = 0, the product of the roots is k/5. Therefore, k/5 = 1, giving k = 5.

c) 3 — Opposite roots have a sum of zero. The sum of the roots is −(k−3) = 3−k. Therefore, 3−k = 0, giving k = 3. The equation then becomes x²−16 = 0, whose roots are 4 and −4.

c) 32 — Let the roots be r and 2r. Their sum is 3r = 12, so r = 4. Therefore, the roots are 4 and 8. Their product is k = 4×8 = 32.

c) x²−7x+12 = 0 — We have α+β = 5 and αβ = 6. The sum of the new roots is α+β+2 = 7. Their product is (α+1)(β+1) = αβ+α+β+1 = 6+5+1 = 12. Therefore, the required equation is x²−7x+12 = 0.

b) 3x²−5x+2 = 0 — To form the equation whose roots are reciprocals, replace x by 1/x in 2x²−5x+3 = 0. This gives 2/x²−5/x+3 = 0. Multiplying by x² gives 3x²−5x+2 = 0.

c) 9 — The first equation factors as (x−2)(x−3) = 0, so its roots are 2 and 3. Substituting x = 2 into the second equation gives 4−2k+2k = 4, which can never be zero. Substituting x = 3 gives 9−3k+2k = 0, so 9−k = 0. Therefore, k = 9.

b) 3 — We have α+β = 5 and αβ = 5. Now, α/β+β/α = (α²+β²)/(αβ). Also, α²+β² = (α+β)²−2αβ = 25−10 = 15. Therefore, the required value is 15/5 = 3.

c) 21 — Let the roots be α and β. Then α+β = 10 and αβ = k. Using (α−β)² = (α+β)²−4αβ, we get 4² = 10²−4k. Therefore, 16 = 100−4k, giving 4k = 84 and k = 21.

b) k > 0 and k ≠ 2 — The equation factors as (x−2)(x−k) = 0, so its roots are 2 and k. Both roots are positive when k > 0. They are distinct when k ≠ 2. Therefore, the required condition is k > 0 and k ≠ 2.

c) 8 — Let α and β be the roots of x²−3x+1 = 0. Then α+β = 3 and αβ = 1. The new roots are α² and β². Their sum is α²+β² = (α+β)²−2αβ = 9−2 = 7, so p = 7. Their product is α²β² = (αβ)² = 1, so q = 1. Therefore, p+q = 8.

c) 18 — Let the smaller integer be x, so the larger integer is x+1. Then x(x+1) = 306, giving x²+x−306 = 0. Factorising gives (x−17)(x+18) = 0. Since the integers are positive, x = 17. Therefore, the larger integer is 18.

c) 28 — Let the smaller odd integer be x, so the next odd integer is x+2. Then x(x+2) = 195, giving x²+2x−195 = 0. Factorising gives (x−13)(x+15) = 0. Since the integers are positive, x = 13. The numbers are 13 and 15, and their sum is 28.

c) 20 metres — Let the breadth be x metres. Then the length is x+5 metres. Therefore, x(x+5) = 300, giving x²+5x−300 = 0. Factorising gives (x−15)(x+20) = 0. Since a dimension must be positive, x = 15. Hence, the length is 20 metres.

b) 16 years — Let the daughter’s present age be x years. The mother’s present age is x+24. Four years ago, their ages were x−4 and x+20. Therefore, (x−4)(x+20) = 432. Expanding gives x²+16x−80 = 432, or x²+16x−512 = 0. Factorising gives (x−16)(x+32) = 0. Since age cannot be negative, x = 16 years.

b) 40 km/h — Let the original speed be x km/h. The original and reduced journey times are 240/x and 240/(x+20) hours. Therefore, 240/x−240/(x+20) = 2. Simplifying gives 4,800 = 2x(x+20), or x²+20x−2,400 = 0. Factorising gives (x−40)(x+60) = 0. Since speed must be positive, x = 40 km/h.

b) 2 metres — Let the width of the path be x metres. The dimensions of the central area are 30−2x and 20−2x. Therefore, (30−2x)(20−2x) = 416. Expanding gives 4x²−100x+184 = 0, or x²−25x+46 = 0. Factorising gives (x−2)(x−23) = 0. Since x = 23 is physically impossible, the path is 2 metres wide.

c) 13 — Let the number be x. According to the condition, x²−9x = 52. Therefore, x²−9x−52 = 0. Factorising gives (x−13)(x+4) = 0. Since the number is positive, x = 13.

b) ₹25 — At ₹x per unit, the number of units sold is 1,500/x. After the price reduction, (x−5)(1,500/x+20) = 1,600. Multiplying by x and simplifying gives 20x²−200x−7,500 = 0, or x²−10x−375 = 0. Factorising gives (x−25)(x+15) = 0. Since price must be positive, x = ₹25.

c) 7 seconds — The object reaches the ground when h = 0. Therefore, −5t²+30t+35 = 0. Dividing by −5 gives t²−6t−7 = 0. Factorising gives (t−7)(t+1) = 0. Since time cannot be negative, t = 7 seconds.

c) 17 cm — Let the shorter perpendicular side be x cm. The other side is x+7 cm. Since the area is 60 cm², x(x+7)/2 = 60. Thus, x²+7x−120 = 0, which factors as (x−8)(x+15) = 0. Hence, the perpendicular sides are 8 cm and 15 cm. The hypotenuse is √(8²+15²) = √289 = 17 cm.

b) 6 days — Let A take x days, so B takes x+6 days. Their combined one-day work is 1/x+1/(x+6) = 1/4. Simplifying gives 4(2x+6) = x(x+6), or x²−2x−24 = 0. Factorising gives (x−6)(x+4) = 0. Since time must be positive, A takes 6 days.

b) 10 — Let the number of rows be x. Then each row contains x+2 students. Therefore, x(x+2) = 120, giving x²+2x−120 = 0. Factorising gives (x−10)(x+12) = 0. Since the number of rows must be positive, x = 10.

b) 64 — Let the tens digit be x and the units digit be y. Since the reversed number is 18 less, (10x+y)−(10y+x) = 18. Thus, x−y = 2, so x = y+2. Also, xy = 24. Therefore, y(y+2) = 24, giving y²+2y−24 = 0. Factorising gives (y−4)(y+6) = 0. As a digit cannot be negative, y = 4 and x = 6. Hence, the number is 64.

c) 40 or 60 — Revenue equals quantity multiplied by price. Therefore, x(100−x) = 2,400. This gives x²−100x+2,400 = 0. Factorising gives (x−40)(x−60) = 0. Hence, the possible quantities are 40 or 60 units.

c) 12 metres — Let the distance of the foot of the ladder from the wall be x metres. Then the height reached is x+7 metres. By the Pythagorean theorem, x²+(x+7)² = 13². Simplifying gives 2x²+14x−120 = 0, or x²+7x−60 = 0. Factorising gives (x−5)(x+12) = 0. Thus, x = 5 and the height is 5+7 = 12 metres.

b) 2 < x < 5 — Factorising gives x²−7x+10 = (x−2)(x−5). Since the coefficient of x² is positive, the quadratic expression is negative between its two roots. Therefore, the solution is 2 < x < 5.

d) 4 — The equation gives two cases. First, x²−5x+4 = 2, so x²−5x+2 = 0, which has two distinct real roots because its discriminant is 17. Second, x²−5x+4 = −2, so x²−5x+6 = 0, whose roots are 2 and 3. None of these roots coincide, so there are four distinct real solutions.

d) 4 — Adding 2 to both sides gives x²+2+1/x² = 16. Therefore, (x+1/x)² = 16, so x+1/x = 4 or x+1/x = −4. These lead to x²−4x+1 = 0 and x²+4x+1 = 0. Each equation has two distinct real roots, giving four real non-zero solutions.

b) 0 — For equal roots, the discriminant must be zero. Therefore, [−2(k+1)]²−4(k²+1) = 0. Simplifying gives 4(k+1)²−4(k²+1) = 0, or (k+1)²−k²−1 = 0. Thus, 2k = 0 and k = 0.

c) 5 — Completing the square gives 2x²−12x+23 = 2(x²−6x)+23 = 2[(x−3)²−9]+23 = 2(x−3)²+5. Since the square term is non-negative, the minimum value is 5, attained at x = 3.

b) 7 — The positive factor pairs of 12 are 1 and 12, 2 and 6, and 3 and 4. The only consecutive pair is 3 and 4. Their sum is 7, so k = 7.

c) 52 — We have α+β = 4 and αβ = 1. Using α³+β³ = (α+β)³−3αβ(α+β), the required value is 4³−3(1)(4) = 64−12 = 52.

a) 2x²−8x+3 = 0 — Replacing x by 1/x in 3x²−8x+2 = 0 gives 3/x²−8/x+2 = 0. Multiplying by x² gives 2x²−8x+3 = 0. Its roots are the reciprocals of the original roots.

b) [−2,−1] ∪ [1,2] — Let y = x². The inequality becomes y²−5y+4 ≤ 0, or (y−1)(y−4) ≤ 0. Therefore, 1 ≤ y ≤ 4. Since y = x², we need 1 ≤ x² ≤ 4, giving −2 ≤ x ≤ −1 or 1 ≤ x ≤ 2.

c) 3 — Since the square root is non-negative, x must satisfy x ≥ 0. Squaring gives x+6 = x², or x²−x−6 = 0. Factorising gives (x−3)(x+2) = 0. The algebraic roots are 3 and −2, but −2 violates x ≥ 0. Therefore, the only valid solution is x = 3.

c) k > 0 — The product of the roots is 2k, which must be positive, so k > 0. Their sum is k+3, which is also positive for k > 0. The discriminant is (k+3)²−8k = k²−2k+9 = (k−1)²+8, which is always positive. Therefore, for every k > 0, the equation has two distinct positive real roots.

c) 21 — We have α+β = 5/2 and αβ = 1/2. Therefore, α²+β² = (5/2)²−2(1/2) = 25/4−1 = 21/4. Now, 1/α²+1/β² = (α²+β²)/(αβ)² = (21/4)/(1/4) = 21.

b) 26/5 — The equation factors as (x−1)(x−k) = 0, so its roots are 1 and k. If k is the larger root, k = 5. If 1 is the larger root, then 1 = 5k, giving k = 1/5. Therefore, the sum of all possible values is 5+1/5 = 26/5.

c) 11 — Completing the square gives −2x²+8x+3 = −2(x²−4x)+3 = −2[(x−2)²−4]+3 = −2(x−2)²+11. Since −2(x−2)² ≤ 0, the maximum value is 11, attained at x = 2.

c) 10 — The equation gives x²−5x+5 = 1 or x²−5x+5 = −1. The first case gives x²−5x+4 = 0, whose roots are 1 and 4. The second gives x²−5x+6 = 0, whose roots are 2 and 3. Therefore, the sum of all distinct real solutions is 1+2+3+4 = 10.