Question 1

If one root of the equation 2x² − 3x + k = 0 is twice the other root, then the value of k is:

Accepted Answer

Setup: Let the roots be α and 2α. We use Vieta's formulas and the given relationship to find k.

Step 1: Let the roots be α and 2α.

Step 2: By Vieta's formulas for 2x² − 3x + k = 0:
  Sum of roots: α + 2α = 3/2
  Product of roots: α · 2α = k/2

Step 3: From the sum equation:
  3α = 3/2
  α = 1/2

Step 4: From the product equation:
  2α² = k/2
  2(1/2)² = k/2
  2(1/4) = k/2
  1/2 = k/2
  k = 1

Verification: The equation becomes 2x² − 3x + 1 = 0, which factors as (2x − 1)(x − 1) = 0, giving roots x = 1/2 and x = 1. Indeed, 1 = 2 · (1/2). ✓

Question 2

The polynomial P(x) = x³ − 6x² + 11x − 6 has roots in arithmetic progression. If the roots are (a − d), a, and (a + d), then the value of a is:

Accepted Answer

Setup: For a cubic polynomial with roots in arithmetic progression, we use Vieta's formulas. Let the roots be (a − d), a, and (a + d).

Step 1: By Vieta's formula for the sum of roots of x³ − 6x² + 11x − 6 = 0:
  (a − d) + a + (a + d) = 6
  3a = 6
  a = 2

Step 2: Verify using the product of roots:
  (a − d) · a · (a + d) = 6
  a(a² − d²) = 6
  2(4 − d²) = 6
  4 − d² = 3
  d² = 1
  d = ±1

Step 3: The roots are either (2 − 1), 2, (2 + 1) = 1, 2, 3 or (2 + 1), 2, (2 − 1) = 3, 2, 1 (same set).

Verification: P(1) = 1 − 6 + 11 − 6 = 0 ✓, P(2) = 8 − 24 + 22 − 6 = 0 ✓, P(3) = 27 − 54 + 33 − 6 = 0 ✓

Question 3

If the roots of x² − (p + q)x + pq = 0 are p and q, then which of the following is true?

Accepted Answer

Setup: We are given that p and q are the roots of the equation x² − (p + q)x + pq = 0. We use Vieta's formulas to verify consistency.

Step 1: By Vieta's formulas, for the equation x² − (p + q)x + pq = 0:
  Sum of roots = p + q
  Product of roots = pq

Step 2: If the roots are p and q, then by Vieta's formulas:
  p + q = p + q ✓ (This is always true)
  p · q = pq ✓ (This is always true)

Step 3: The equation is consistent for any values of p and q. Both p and q satisfy the equation:
  For root p: p² − (p + q)p + pq = p² − p² − pq + pq = 0 ✓
  For root q: q² − (p + q)q + pq = q² − pq − q² + pq = 0 ✓

Conclusion: The statement is true for all real values of p and q. The equation is constructed such that p and q are always its roots.

Question 4

If one root of the quadratic equation 2x² + 5x + m = 0 is the reciprocal of the other, then m equals:

Accepted Answer

Setup: Using Vieta's formulas with the reciprocal root condition.

Step 1: Let the roots be α and 1/α (reciprocals of each other).

Step 2: Apply Vieta's formula for the product of roots in 2x² + 5x + m = 0:
Product of roots = c/a = m/2

Step 3: Since the roots are reciprocals:
α · (1/α) = 1

Step 4: Equate the two expressions for the product:
m/2 = 1
m = 2

Verification: For 2x² + 5x + 2 = 0, factoring gives (2x + 1)(x + 2) = 0, so roots are −1/2 and −2.
Check reciprocals: −2 = 1/(−1/2) ✓
Product: (−1/2) × (−2) = 1 ✓
Product by Vieta: m/2 = 2/2 = 1 ✓

Question 5

If α and β are the roots of the quadratic equation x² − 5x + 6 = 0, then the value of α³ + β³ is:

Accepted Answer

Setup: Using Vieta's formulas and the algebraic identity for sum of cubes.

Step 1: Find sum and product of roots using Vieta's formulas.
For equation x² − 5x + 6 = 0:
- α + β = 5 (sum of roots = −b/a)
- αβ = 6 (product of roots = c/a)

Step 2: Use the identity α³ + β³ = (α + β)³ − 3αβ(α + β)
α³ + β³ = (5)³ − 3(6)(5)
α³ + β³ = 125 − 90
α³ + β³ = 35

Verification: Solving x² − 5x + 6 = 0 gives (x − 2)(x − 3) = 0, so α = 2, β = 3.
Check: 2³ + 3³ = 8 + 27 = 35 ✓

Question 6

The quadratic equation x² + px + q = 0 has roots that are reciprocals of each other. Which of the following must be true?

Accepted Answer

Setup: If roots are reciprocals, let them be r and 1/r. Apply Vieta's formulas to establish a constraint.

Step 1: Let the roots be α = r and β = 1/r.

Step 2: By Vieta's formulas for x² + px + q = 0:
  • Sum of roots: α + β = −p
  • Product of roots: αβ = q

Step 3: Calculate the product:
  αβ = r · (1/r) = 1

Step 4: Therefore:
  q = 1

This is the necessary and sufficient condition for roots to be reciprocals.

Verification: If q = 1, then x² + px + 1 = 0 has roots satisfying αβ = 1, meaning β = 1/α. ✓

Question 7

If the roots of x² − 6x + k = 0 are real and distinct, then the range of k is:

Accepted Answer

Setup: For a quadratic ax² + bx + c = 0 to have real and distinct roots, the discriminant Δ = b² − 4ac must be strictly positive.

Step 1: For x² − 6x + k = 0, identify coefficients:
  • a = 1
  • b = −6
  • c = k

Step 2: Calculate discriminant:
  Δ = b² − 4ac = (−6)² − 4(1)(k) = 36 − 4k

Step 3: For real and distinct roots, require Δ > 0:
  36 − 4k > 0
  36 > 4k
  9 > k
  k < 9

Step 4: Since k can be any real number less than 9, the range is:
  k ∈ (−∞, 9)

Verification: If k = 8, then Δ = 36 − 32 = 4 > 0 ✓ (distinct roots)
              If k = 9, then Δ = 36 − 36 = 0 (equal roots, not distinct) ✗

Question 8

The polynomial P(x) = x³ − 7x² + 14x − 8 has a root at x = 1. What is the sum of the other two roots?

Accepted Answer

Setup: Use Vieta's formulas for cubic polynomials. If one root is known, we can find the sum of the remaining roots.

Step 1: Verify that x = 1 is a root:
  P(1) = 1³ − 7(1)² + 14(1) − 8 = 1 − 7 + 14 − 8 = 0 ✓

Step 2: For a cubic x³ + ax² + bx + c = 0 with roots α, β, γ:
  By Vieta's formula: α + β + γ = −a

Step 3: For P(x) = x³ − 7x² + 14x − 8, the coefficient of x² is −7, so:
  Sum of all three roots = −(−7) = 7

Step 4: If one root is 1, and the sum of all three is 7:
  1 + (sum of other two roots) = 7
  Sum of other two roots = 6

Verification: Factor P(x) = (x − 1)(x² − 6x + 8) = (x − 1)(x − 2)(x − 4).
  Roots are 1, 2, 4. Sum of other two: 2 + 4 = 6 ✓

Question 9

If one root of the equation 2x² − 5x + c = 0 is twice the other, then the value of c is:

Accepted Answer

Setup: Use the relationship between roots and Vieta's formulas. If one root is twice the other, express both roots in terms of a single variable.

Step 1: Let the roots be α and 2α (one is twice the other).

Step 2: Apply Vieta's formulas for 2x² − 5x + c = 0:
  • Sum of roots: α + 2α = 5/2
  • Product of roots: α · 2α = c/2

Step 3: From the sum:
  3α = 5/2
  α = 5/6

Step 4: Calculate the product:
  α · 2α = 2α² = 2(5/6)² = 2 · 25/36 = 50/36 = 25/18

Step 5: By Vieta's formula:
  c/2 = 25/18
  c = 50/18 = 25/9

Verification: Substitute back into 2x² − 5x + 25/9 = 0.
  Multiply by 9: 18x² − 45x + 25 = 0
  Using quadratic formula: x = [45 ± √(2025 − 1800)]/36 = [45 ± √225]/36 = [45 ± 15]/36
  Roots: x = 60/36 = 5/3 and x = 30/36 = 5/6
  Check: 5/3 = 2 · (5/6) ✓

Question 10

The quadratic equation x² + px + q = 0 has roots that are reciprocals of each other. Which of the following is necessarily true?

Accepted Answer

Setup: Using Vieta's formulas to relate coefficients to root properties.

Step 1: Let the roots be α and 1/α (reciprocals of each other).

Step 2: Apply Vieta's formulas to x² + px + q = 0:
- Sum of roots: α + 1/α = −p
- Product of roots: α · (1/α) = q

Step 3: Simplify the product:
α · (1/α) = 1 = q

Therefore, q = 1 is necessarily true.

Verification: Consider x² − 5x + 1 = 0. Roots are (5 ± √21)/2. Product = 1 ✓. The condition p can be any real number, but q must equal 1.

Question 11

If the polynomial P(x) = x³ − 6x² + 11x − 6 has roots α, β, and γ, then α + β + γ equals:

Accepted Answer

Setup: Using Vieta's formulas for cubic polynomials.

For a cubic equation x³ + ax² + bx + c = 0 with roots α, β, γ:
- Sum of roots = −a
- Sum of products of pairs = b
- Product of all roots = −c

Step 1: Identify coefficients in P(x) = x³ − 6x² + 11x − 6:
- Coefficient of x²: −6
- Coefficient of x: 11
- Constant term: −6

Step 2: Apply Vieta's formula for sum of roots:
α + β + γ = −(coefficient of x²)/(coefficient of x³)
α + β + γ = −(−6)/1 = 6

Verification: Factoring P(x) = (x − 1)(x − 2)(x − 3), roots are 1, 2, 3.
Sum = 1 + 2 + 3 = 6 ✓

Question 12

The quadratic equation x² − 2kx + (k² − 1) = 0 has equal roots. The value of k is:

Accepted Answer

Setup: Condition for equal roots using the discriminant.

For a quadratic ax² + bx + c = 0 to have equal roots, the discriminant Δ = b² − 4ac must equal 0.

Step 1: Identify coefficients in x² − 2kx + (k² − 1) = 0:
- a = 1
- b = −2k
- c = k² − 1

Step 2: Set discriminant equal to zero:
Δ = b² − 4ac = 0
(−2k)² − 4(1)(k² − 1) = 0
4k² − 4(k² − 1) = 0
4k² − 4k² + 4 = 0
4 = 0

This is impossible, so the equation never has equal roots for any real k.

Note: If the original equation were x² − 2kx + k² = 0, then Δ = 4k² − 4k² = 0 for all k, giving equal roots. The constant term k² − 1 prevents this.

Question 13

If the roots of the equation x² + px + q = 0 are in the ratio 2:3, then which of the following is correct?

Accepted Answer

Setup: Using Vieta's formulas with roots in a given ratio to establish a relationship between coefficients p and q.

Step 1: Let the roots be 2k and 3k (where k ≠ 0), since they are in ratio 2:3.

Step 2: By Vieta's formulas for x² + px + q = 0:
- Sum of roots: 2k + 3k = -p
- Therefore: 5k = -p, so k = -p/5

Step 3: Product of roots: (2k)(3k) = q
- Therefore: 6k² = q

Step 4: Substitute k = -p/5 into 6k² = q:
6(-p/5)² = q
6(p²/25) = q
6p² = 25q

Step 5: Rearranging: 6p² - 25q = 0 or equivalently 6p² = 25q

Question 14

The polynomial P(x) = x³ - 6x² + 11x - 6 has roots α, β, and γ. If one root is 1, then the sum of the other two roots is:

Accepted Answer

Setup: Using Vieta's formulas for cubic polynomials and the given information that one root is known.

Step 1: Verify that x = 1 is a root:
P(1) = 1³ - 6(1)² + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 ✓

Step 2: For the cubic x³ - 6x² + 11x - 6 = 0, by Vieta's formulas:
- Sum of all three roots: α + β + γ = 6 (negative of coefficient of x² divided by coefficient of x³)

Step 3: Given that one root is 1, let α = 1.
- Therefore: 1 + β + γ = 6
- So: β + γ = 5

Step 4: Verification by factoring:
P(x) = (x - 1)(x² - 5x + 6) = (x - 1)(x - 2)(x - 3)
Roots are 1, 2, 3.
Sum of other two roots: 2 + 3 = 5 ✓

Question 15

If the roots of x² + px + q = 0 are in the ratio 2:3, and their sum is 10, then the value of q is:

Accepted Answer

Setup: We use the relationship between roots and coefficients (Vieta's formulas) combined with the given ratio constraint.

Solution:
Step 1: Let the roots be 2k and 3k (in ratio 2:3).

Step 2: Sum of roots = 2k + 3k = 5k = 10
  Therefore: k = 2

Step 3: The roots are:
  α = 2k = 2(2) = 4
  β = 3k = 3(2) = 6

Step 4: By Vieta's formulas for x² + px + q = 0:
  - Sum of roots: α + β = -p
  - Product of roots: αβ = q

Step 5: Calculate the product:
  q = αβ = 4 × 6 = 24

Verification: The equation is x² - 10x + 24 = 0
  Discriminant: Δ = 100 - 96 = 4 > 0 ✓
  Roots: x = (10 ± 2)/2 = 6 or 4 ✓
  Ratio: 4:6 = 2:3 ✓

Question 16

The quadratic equation x² - (k+1)x + (k-1) = 0 has real roots for all real values of k if and only if:

Accepted Answer

Setup: For a quadratic ax² + bx + c = 0 to have real roots for all real values of a parameter, the discriminant must be non-negative for all values of that parameter.

Solution:
Step 1: For the equation x² - (k+1)x + (k-1) = 0:
  a = 1, b = -(k+1), c = (k-1)

Step 2: The discriminant is:
  Δ = b² - 4ac
  Δ = [-(k+1)]² - 4(1)(k-1)
  Δ = (k+1)² - 4(k-1)

Step 3: Expand:
  Δ = k² + 2k + 1 - 4k + 4
  Δ = k² - 2k + 5

Step 4: For real roots for ALL real k, we need Δ ≥ 0 for all k:
  k² - 2k + 5 ≥ 0 for all real k

Step 5: Check if k² - 2k + 5 can be negative. Complete the square:
  k² - 2k + 5 = (k - 1)² - 1 + 5 = (k - 1)² + 4

Step 6: Since (k - 1)² ≥ 0 for all real k:
  (k - 1)² + 4 ≥ 4 > 0 for all real k

Conclusion: The discriminant is always positive (≥ 4), so the equation has real roots for all real values of k.

Question 17

If the quadratic equation x² - (k + 1)x + (k² - 3k + 2) = 0 has equal roots, then the value of k is:

Accepted Answer

Setup: A quadratic equation has equal roots if and only if its discriminant equals zero.

Step 1: For the equation x² - (k + 1)x + (k² - 3k + 2) = 0, identify coefficients:
- a = 1
- b = -(k + 1)
- c = k² - 3k + 2

Step 2: The discriminant is Δ = b² - 4ac.
For equal roots, Δ = 0.

Step 3: Calculate the discriminant:
Δ = [-(k + 1)]² - 4(1)(k² - 3k + 2)
Δ = (k + 1)² - 4(k² - 3k + 2)
Δ = k² + 2k + 1 - 4k² + 12k - 8
Δ = -3k² + 14k - 7

Step 4: Set Δ = 0:
-3k² + 14k - 7 = 0
3k² - 14k + 7 = 0

Step 5: Using the quadratic formula:
k = [14 ± √(196 - 84)] / 6
k = [14 ± √112] / 6
k = [14 ± 4√7] / 6
k = [7 ± 2√7] / 3

Step 6: Simplify:
k = (7 + 2√7)/3  or  k = (7 - 2√7)/3

Note: Both values are valid. If the question expects a single answer, verify the problem context. The sum of both roots is 14/3.

Question 18

If α and β are the roots of x² + bx + c = 0, and γ and δ are the roots of x² + px + q = 0, and if α + γ = β + δ = -1, then which of the following is correct?

Accepted Answer

Setup: Using Vieta's formulas for two quadratic equations with a constraint relating roots from both equations.

Step 1: For x² + bx + c = 0 with roots α and β:
- α + β = -b
- αβ = c

Step 2: For x² + px + q = 0 with roots γ and δ:
- γ + δ = -p
- γδ = q

Step 3: Given constraint: α + γ = -1 and β + δ = -1.

Step 4: From α + γ = -1, we get γ = -1 - α.
From β + δ = -1, we get δ = -1 - β.

Step 5: Find γ + δ:
γ + δ = (-1 - α) + (-1 - β) = -2 - (α + β) = -2 - (-b) = b - 2
Therefore: -p = b - 2, which gives p = 2 - b.

Step 6: Find γδ:
γδ = (-1 - α)(-1 - β) = 1 + α + β + αβ = 1 + (-b) + c = 1 - b + c
Therefore: q = 1 - b + c.

Step 7: Relationship: p + b = 2 and q = 1 - b + c = 1 + c - b.
Alternatively: p = 2 - b and q - c = 1 - b, so q = c + 1 - b.
Simplifying: b + p = 2 and q - c = 1 - b, which gives b + p = 2 and c + q = 1 + 2b - b = 1 + b.

Question 19

If α and β are the roots of the quadratic equation x² - 5x + 6 = 0, then the equation whose roots are (α + β) and αβ is:

Accepted Answer

Setup: We use Vieta's formulas to find the sum and product of roots, then construct a new quadratic equation.

Step 1: For the equation x² - 5x + 6 = 0, by Vieta's formulas:
- Sum of roots: α + β = 5
- Product of roots: αβ = 6

Step 2: The new roots are (α + β) = 5 and αβ = 6.

Step 3: For a quadratic with roots r₁ and r₂, the equation is:
x² - (r₁ + r₂)x + r₁r₂ = 0

Step 4: Sum of new roots = 5 + 6 = 11
Product of new roots = 5 × 6 = 30

Step 5: The required equation is:
x² - 11x + 30 = 0

Verification: Factoring gives (x - 5)(x - 6) = 0, so roots are 5 and 6. ✓

Question 20

Let P(x) = x⁴ - 4x³ + 6x² - 4x + 1. Which of the following statements is correct?

Accepted Answer

Setup: We recognize the structure of the polynomial and use binomial expansion to identify it as a perfect power.

Step 1: Observe the coefficients: 1, -4, 6, -4, 1
These match the binomial expansion coefficients: C(4,0), C(4,1), C(4,2), C(4,3), C(4,4)

Step 2: Recall the binomial expansion of (a - b)⁴:
(a - b)⁴ = a⁴ - 4a³b + 6a²b² - 4ab³ + b⁴

Step 3: Compare with P(x) = x⁴ - 4x³ + 6x² - 4x + 1:
If a = x and b = 1:
(x - 1)⁴ = x⁴ - 4x³(1) + 6x²(1)² - 4x(1)³ + (1)⁴
(x - 1)⁴ = x⁴ - 4x³ + 6x² - 4x + 1 ✓

Step 4: Therefore P(x) = (x - 1)⁴

Step 5: The roots of P(x) = 0 are found from (x - 1)⁴ = 0
This gives x = 1 as a root with multiplicity 4.

Step 6: P'(x) = 4(x - 1)³
At x = 1: P'(1) = 0, confirming x = 1 is a repeated root.

Step 7: The polynomial has only one distinct root (x = 1) with multiplicity 4.

CDS Quadratic Equations & Polynomials

Quadratic Equations & Polynomials— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Quadratic Equations & Polynomials — Practice Questions

60-Second Revision — Quadratic Equations & Polynomials