Question 1

If the roots of x² + px + 12 = 0 are 3 and 4, find the value of p.

Accepted Answer

Step 1: For a quadratic with roots α and β, the equation is x² - (α + β)x + αβ = 0. Step 2: Sum of roots = 3 + 4 = 7. Step 3: The equation becomes x² - 7x + 12 = 0. Step 4: Comparing with x² + px + 12 = 0, we get p = -7. Therefore, the answer is -7.

Question 2

If α and β are roots of x² - 6x + 8 = 0, find the value of α² + β².

Accepted Answer

Step 1: From the equation, sum of roots α + β = 6 and product αβ = 8. Step 2: Use the identity α² + β² = (α + β)² - 2αβ. Step 3: α² + β² = (6)² - 2(8) = 36 - 16 = 20. Therefore, the answer is 20.

Question 3

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 4

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

Accepted Answer

Step 1: If x = 2 is a root, it must satisfy the equation. Step 2: Substitute x = 2: (2)² - 5(2) + k = 0. Step 3: 4 - 10 + k = 0. Step 4: -6 + k = 0, so k = 6. Therefore, the answer is 6.

Question 5

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 6

Solve: x² - 9 = 0. What are the roots?

Accepted Answer

Step 1: Recognize this as a difference of squares: x² - 9 = (x - 3)(x + 3) = 0. Step 2: Set each factor to zero: x - 3 = 0 or x + 3 = 0. Step 3: x = 3 or x = -3. Therefore, the roots are 3 and -3.

Question 7

A quadratic equation x² + px + q = 0 has roots that are reciprocals of each other. If one root is 4, find the value of q.

Accepted Answer

Step 1: If roots are reciprocals and one root is 4, the other root is 1/4. Step 2: Product of roots = 4 × (1/4) = 1. Step 3: By Vieta's formula, product of roots = q, so q = 1.

Question 8

The quadratic equation x² - (a+b)x + ab = 0 has roots that are 3 more than the roots of x² - 5x + 6 = 0. Find the value of (a+b).

Accepted Answer

Step 1: Find roots of x² - 5x + 6 = 0. Factoring: (x-2)(x-3) = 0, so roots are 2 and 3. Step 2: New roots are 2+3 = 5 and 3+3 = 6. Step 3: For equation x² - (a+b)x + ab = 0 with roots 5 and 6, sum of roots = a+b = 5+6 = 11.

Question 9

If α and β are roots of x² - 6x + 8 = 0, find the value of (α² + β²).

Accepted Answer

Step 1: Find roots of x² - 6x + 8 = 0. Factoring: (x-2)(x-4) = 0, so α = 2 and β = 4. Step 2: α² + β² = 4 + 16 = 20. Alternatively, use identity: α² + β² = (α+β)² - 2αβ. Step 3: From Vieta's: α+β = 6, αβ = 8. Step 4: α² + β² = 36 - 16 = 20.

Question 10

If the roots of 2x² - 8x + k = 0 are equal, find the value of k.

Accepted Answer

Step 1: For a quadratic ax² + bx + c = 0 to have equal roots, discriminant Δ = b² - 4ac = 0. Step 2: Here, a = 2, b = -8, c = k. Step 3: Δ = (-8)² - 4(2)(k) = 64 - 8k = 0. Step 4: 8k = 64, so k = 8.

Question 11

The sum of the squares of the roots of x² - 4x + 3 = 0 exceeds the product of the roots by how much?

Accepted Answer

Step 1: From x² - 4x + 3 = 0, by Vieta's: sum of roots = 4, product of roots = 3. Step 2: Calculate α² + β² = (α+β)² - 2αβ = 16 - 6 = 10. Step 3: Difference = (α² + β²) - (αβ) = 10 - 3 = 7.

Question 12

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, we have α + β = -p and αβ = q. Step 2: For the second equation, the sum of roots is α² + β² = (α + β)² - 2αβ = p² - 2q, so m = -(p² - 2q) = 2q - p². Step 3: The product of roots is α²β² = (αβ)² = q², so n = q². Step 4: Therefore, m = 2q - p² and n = q².

Question 13

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 the 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 14

The quadratic equation x² + bx + c = 0 has roots r and s. If r² + s² = 13 and rs = 3, then the value of b² - 4c is:

Accepted Answer

Step 1: From Vieta's formulas, r + s = -b and rs = c = 3. Step 2: We know r² + s² = (r + s)² - 2rs = 13. Step 3: Substituting: (-b)² - 2(3) = 13 → b² - 6 = 13 → b² = 19. Step 4: The discriminant is b² - 4c = 19 - 4(3) = 19 - 12 = 7.

Question 15

If α and β are roots of x² - 5x + k = 0, and α/β + β/α = 13/6, then the value of k is:

Accepted Answer

Step 1: From Vieta's formulas, α + β = 5 and αβ = k. Step 2: Simplify α/β + β/α = (α² + β²)/(αβ) = [(α+β)² - 2αβ]/(αβ). Step 3: Substitute: [(5)² - 2k]/k = 13/6 → (25 - 2k)/k = 13/6. Step 4: Cross-multiply: 6(25 - 2k) = 13k → 150 - 12k = 13k → 150 = 25k → k = 6.

Question 16

A quadratic equation with integer coefficients has roots (3 + √5) and (3 - √5). If the equation is written as x² + px + q = 0, what is the value of p + q?

Accepted Answer

Step 1: Sum of roots = (3 + √5) + (3 - √5) = 6. By Vieta's formulas, p = -6. Step 2: Product of roots = (3 + √5)(3 - √5) = 9 - 5 = 4. By Vieta's formulas, q = 4. Step 3: Therefore, p + q = -6 + 4 = -2.

SBI PO Quadratic Equations

SBI PO Quadratic Equations — Past Exam Questions

Quadratic Equations— Rules & Concept

Key Points to Remember

Exam-Specific Tips

60-Second Revision — Quadratic Equations