Question 1

If the discriminant of the equation 3x² - 6x + k = 0 is 0, find the value of k.

Accepted Answer

Step 1: For a quadratic ax² + bx + c = 0, discriminant Δ = b² - 4ac. Step 2: Here a = 3, b = -6, c = k. Step 3: Δ = (-6)² - 4(3)(k) = 0. Step 4: 36 - 12k = 0. Step 5: 12k = 36, so k = 3. Therefore, the answer is 3.

Question 2

Solve: x² - 5x + 6 = 0. What are the roots?

Accepted Answer

Step 1: Factorize x² - 5x + 6 = 0. Step 2: Find two numbers that multiply to 6 and add to -5: these are -2 and -3. Step 3: (x - 2)(x - 3) = 0. Step 4: x = 2 or x = 3. Therefore, the roots are 2 and 3.

Question 3

If one root of the equation x² + px - 12 = 0 is 3, find the value of p.

Accepted Answer

Step 1: If x = 3 is a root, substitute it into the equation: 3² + p(3) - 12 = 0. Step 2: 9 + 3p - 12 = 0. Step 3: 3p - 3 = 0. Step 4: 3p = 3, so p = 1. Therefore, the answer is 1.

Question 4

The sum of two consecutive integers is 15. Form a quadratic equation and find the integers.

Accepted Answer

Step 1: Let the two consecutive integers be x and x+1. Step 2: x + (x+1) = 15. Step 3: 2x + 1 = 15, so 2x = 14, x = 7. Step 4: The integers are 7 and 8. Step 5: Verify: 7 + 8 = 15 ✓. Therefore, the integers are 7 and 8.

Question 5

If x² - 7x + 12 = 0, find the sum of the roots of the equation.

Accepted Answer

Step 1: For a quadratic equation ax² + bx + c = 0, the sum of roots = -b/a. Step 2: Here a = 1, b = -7, c = 12. Step 3: Sum of roots = -(-7)/1 = 7. Therefore, the answer is 7.

Question 6

Find the product of the roots of the equation 2x² - 8x + 6 = 0.

Accepted Answer

Step 1: For a quadratic equation ax² + bx + c = 0, the product of roots = c/a. Step 2: Here a = 2, b = -8, c = 6. Step 3: Product of roots = 6/2 = 3. Therefore, the answer is 3.

Question 7

If α and β are roots of x² - 5x + 6 = 0, find the value of α³ + β³.

Accepted Answer

Step 1: From the equation x² - 5x + 6 = 0, by Vieta's formulas: α + β = 5 and αβ = 6. Step 2: Factor the equation: (x-2)(x-3) = 0, so α = 2, β = 3 (or vice versa). Step 3: α³ + β³ = 2³ + 3³ = 8 + 27 = 35. Alternatively, using the identity: α³ + β³ = (α + β)³ - 3αβ(α + β) = 5³ - 3(6)(5) = 125 - 90 = 35.

Question 8

The quadratic equation x² - (a+b)x + ab = 0 has roots α and β. If α = 2 and β = 3, find the value of a² + b².

Accepted Answer

Step 1: The equation x² - (a+b)x + ab = 0 has roots α and β. Step 2: By Vieta's formulas: α + β = a + b and αβ = ab. Step 3: Given α = 2 and β = 3: sum = 2 + 3 = 5 = a + b, and product = 2 × 3 = 6 = ab. Step 4: So a + b = 5 and ab = 6. Step 5: Using the identity (a+b)² = a² + 2ab + b², we get a² + b² = (a+b)² - 2ab = 25 - 12 = 13.

Question 9

The sum of the squares of two consecutive integers is 145. Find the larger integer.

Accepted Answer

Step 1: Let the two consecutive integers be n and n+1. Step 2: n² + (n+1)² = 145. Step 3: n² + n² + 2n + 1 = 145 → 2n² + 2n + 1 = 145 → 2n² + 2n - 144 = 0 → n² + n - 72 = 0. Step 4: Using quadratic formula: n = (-1 ± √(1 + 288))/2 = (-1 ± √289)/2 = (-1 ± 17)/2. Step 5: n = 8 or n = -9. Step 6: If n = 8, larger integer = 9. If n = -9, larger integer = -8. For positive integers, answer is 9.

Question 10

A quadratic equation with integer coefficients has roots (2 + √3) and (2 - √3). If the equation is written as x² + px + q = 0, find p + q.

Accepted Answer

Step 1: Roots are (2 + √3) and (2 - √3). Step 2: Sum of roots = (2 + √3) + (2 - √3) = 4, so -p = 4, giving p = -4. Step 3: Product of roots = (2 + √3)(2 - √3) = 4 - 3 = 1, so q = 1. Step 4: Therefore, p + q = -4 + 1 = -3.

Question 11

A quadratic equation x² - (a + b)x + ab = 0 has roots that are 3 more than the roots of x² - 7x + 10 = 0. Find the value of (a + b).

Accepted Answer

Step 1: Find roots of x² - 7x + 10 = 0. Factoring: (x - 2)(x - 5) = 0, so roots are 2 and 5. Step 2: The new roots are 2 + 3 = 5 and 5 + 3 = 8. Step 3: For equation x² - (a + b)x + ab = 0 with roots 5 and 8, by Vieta's formulas: sum of roots = a + b = 5 + 8 = 13. Step 4: Therefore, a + b = 13.

Question 12

If one root of the equation 2x² - 5x + c = 0 is the reciprocal of the other, find the value of c.

Accepted Answer

Step 1: Let the roots be r and 1/r (reciprocals). Step 2: By Vieta's formulas, product of roots = c/2. Step 3: Since roots are r and 1/r, their product is r × (1/r) = 1. Step 4: Therefore, c/2 = 1, which gives c = 2.

Question 13

The quadratic equation x² - 6x + k = 0 has two distinct real roots. If the difference between the roots is 4, find the value of k.

Accepted Answer

Step 1: Let roots be α and β. Given: α - β = 4 (or β - α = 4). Step 2: By Vieta's formulas: α + β = 6 and αβ = k. Step 3: We know (α - β)² = (α + β)² - 4αβ. Step 4: Substituting: 16 = 36 - 4k. Step 5: 4k = 20, so k = 5.

Question 14

If α and β are roots of x² + 3x + 5 = 0, find the value of (α/β) + (β/α).

Accepted Answer

Step 1: By Vieta's formulas: α + β = -3 and αβ = 5. Step 2: (α/β) + (β/α) = (α² + β²)/(αβ). Step 3: α² + β² = (α + β)² - 2αβ = (-3)² - 2(5) = 9 - 10 = -1. Step 4: Therefore, (α/β) + (β/α) = -1/5.

Question 15

If the roots of the quadratic equation x² + px + q = 0 are α and β, and the roots of x² + mx + n = 0 are α² and β², then which of the following is true?

Accepted Answer

Step 1: For equation x² + px + q = 0, by Vieta's formulas: α + β = -p and αβ = q. Step 2: For the second equation with roots α² and β², the sum is α² + β² = (α + β)² - 2αβ = p² - 2q, so m = -(p² - 2q) = 2q - p². Step 3: The product is α²β² = (αβ)² = q², so n = q². Step 4: Therefore, m = 2q - p² and n = q².

SBI Clerk Quadratic Equations

SBI Clerk Quadratic Equations — Past Exam Questions

Quadratic Equations— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Quadratic Equations