Practice Question Chapter-2

Section A: Graphic and Conceptual Meanings (1-10)

Q1. If the graph of a polynomial y = p(x) intersects the x-axis at 3 distinct points, find the number of zeroes of p(x).

• Solution:
  The number of zeroes of a polynomial is equal to the number of points where its graph cuts/intersects the x-axis. Since it intersects at 3 distinct points, it has 3 zeroes.

Q2. A parabola representing a quadratic polynomial y = ax² + bx + c does not cut or touch the x-axis at all. How many real zeroes does it have?

• Solution:
  Since the graph has no point of contact with the x-axis, the polynomial has 0 real zeroes.

Q3. If the graph of a quadratic polynomial touches the x-axis at exactly one point, what can you conclude about its roots?

• Solution:
  It indicates that the quadratic polynomial has two equal (coincident) real zeroes.

Q4. What is the maximum number of zeroes a polynomial of degree n can have?

• Solution:
  A polynomial of degree n can have at most n zeroes.

Q5. Find the degree of the polynomial p(x) = 4x³ - 3x² + 5x - 7.

• Solution:
  The degree is the highest power of the variable x. Here, the highest power is 3, so the degree is 3.

Q6. Write the geometric shape of the graph of a quadratic polynomial.

• Solution:
  The graph of any quadratic polynomial is a symmetric U-shaped curve called a parabola.

Q7. Under what condition will the parabola of ax² + bx + c open downwards?

• Solution:
  The parabola opens downwards when the coefficient of x² is negative, i.e., a < 0.

Q8. If a line representing y = ax + b passes through the origin, what is the value of b?

• Solution:
  1. Understand the Condition: The term "passes through the origin" means the straight line goes exactly through the coordinate point $(0, 0)$.
  2. Substitute the Values: In this point, the $x$-coordinate is $0$ and the $y$-coordinate is $0$.
  3. Solve the Equation: Plug $x = 0$ and $y = 0$ into the given linear equation $y = ax + b$:
     $$0 = a(0) + b$$
     $$0 = 0 + b$$
     $$\mathbf{b = 0}$$

Q9. If the graph of p(x) is parallel to the x-axis and does not cross it, find its number of zeroes.

• Solution:
  Because it never intersects the x-axis, it has no zeroes.

Q10. Can a linear polynomial have more than one zero?

• Solution:
  No. A linear polynomial has a degree of 1, so it can have exactly one zero.

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Section B: Basic Zeroes and Coefficients (11-25)

Q11. Find the zero of the linear polynomial p(x) = 2x + 5.

• Solution:
  Set $p(x) = 0 \implies 2x + 5 = 0 \implies 2x = -5 \implies \mathbf{x = -\frac{5}{2}}$.

Q12. Find the sum and product of zeroes for the quadratic polynomial 2x² - 8x + 6.

• Solution:
  Comparing with ax² + bx + c: a = 2, b = -8, c = 6.
  • Sum of zeroes $(\alpha + \beta) = -\frac{b}{a} = -\frac{-8}{2} = \mathbf{4}$
  • Product of zeroes $(\alpha\beta) = \frac{c}{a} = \frac{6}{2} = \mathbf{3}$

Q13. Find the zeroes of the polynomial x² + 7x + 10. (CBSE 2020)

• Solution:
  Splitting the middle term: x² + 5x + 2x + 10 = 0
  $\implies x(x + 5) + 2(x + 5) = 0 \implies (x + 2)(x + 5) = 0$
  $\implies \mathbf{x = -2, -5}$

Q14. Find a quadratic polynomial whose sum and product of zeroes are -3 and 2 respectively.

• Solution:
  Formula: x² - (Sum)x + (Product)
  $\implies x^2 - (-3)x + 2 = \mathbf{x^2 + 3x + 2}$

Q15. If one zero of the polynomial f(x) = 4x² - 8kx - 9 is the negative of the other, find k. (CBSE 2015)

• Solution:
  Let zeroes be α and -α.
  Sum of zeroes = α + (-α) = 0.
  Formula for sum $= -\frac{b}{a} = -\frac{-8k}{4} = 2k$.
  $\implies 2k = 0 \implies \mathbf{k = 0}$.

Q16. Find the value of p(x) = x² - 5x + 6 at x = 2.

• Solution:
  $p(2) = (2)^2 - 5(2) + 6 = 4 - 10 + 6 = \mathbf{0}$.

Q17. If the zeroes of a quadratic polynomial ax² + bx + c are equal, what can you say about the signs of a and c?

• Solution:
  For equal roots, Discriminant $D = b^2 - 4ac = 0 \implies b^2 = 4ac$. Since b² is always positive, 4ac must be positive, which means a and c must have the same sign.

Q18. If α and β are zeroes of x² + x - 2, find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.

• Solution:
  Here, α + β = -1 and αβ = -2.
  $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1}{-2} = \mathbf{\frac{1}{2}}$.

Q19. Find a quadratic polynomial whose zeroes are 5 and -3.

• Solution:
  Sum (S) = 5 + (-3) = 2
  Product (P) = 5 × (-3) = -15
  Polynomial $= x^2 - Sx + P = \mathbf{x^2 - 2x - 15}$.

Q20. If the product of zeroes of ax² - 6x - 6 is 4, find the value of a.

• Solution:
  Product $= \frac{c}{a} = \frac{-6}{a}$.
  Given $\frac{-6}{a} = 4 \implies 4a = -6 \implies \mathbf{a = -\frac{3}{2}}$.

Q21. If 2 is a zero of x² + 3x + k, find k. (CBSE 2010)

• Solution:
  Since 2 is a root, substitute $x=2 \implies (2)^2 + 3(2) + k = 0$
  $\implies 4 + 6 + k = 0 \implies 10 + k = 0 \implies \mathbf{k = -10}$.

Q22. Find the zeroes of 4x² - 4x + 1.

• Solution:
  It can be written as (2x - 1)² = 0.
  $\implies 2x - 1 = 0 \implies \mathbf{x = \frac{1}{2}, \frac{1}{2}}$.

Q23. If α, β are zeroes of x² - 5x + 6, calculate α²β + αβ².

• Solution:
  Taking αβ common: αβ(α + β).
  From polynomial, α + β = 5 and αβ = 6.
  $\implies 6 \times 5 = \mathbf{30}$.

Q24. Form a quadratic polynomial whose zeroes are reciprocals of the zeroes of x² - 5x + 6.

• Solution:
  Original roots have sum 5 and product 6.
  New sum $= \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{5}{6}$.
  New product $= \frac{1}{\alpha\beta} = \frac{1}{6}$.
  New Polynomial $= x^2 - \frac{5}{6}x + \frac{1}{6} \implies \mathbf{6x^2 - 5x + 1}$.

Q25. If one root of 3x² - kx + 4 = 0 is 1, find k.

• Solution:
  Substitute $x = 1 \implies 3(1)^2 - k(1) + 4 = 0$
  $\implies 3 - k + 4 = 0 \implies \mathbf{k = 7}$.

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Section C: Intermediate Algebraic Manipulations (26-40)

Q26. If α and β are zeroes of x² - 4x + 3, find α⁴β³ + α³β⁴.

• Solution:
  Expression = α³β³(α + β) = (αβ)³(α + β).
  Here, α+β = 4 and αβ = 3.
  $\implies (3)^3 \times 4 = 27 \times 4 = \mathbf{108}$.

Q27. If α and β are zeroes of x² - 5x + k such that α - β = 1, find k. (CBSE 2012)

• Solution:
  We know α + β = 5.
  Given α - β = 1.
  Adding both equations: $2\alpha = 6 \implies \alpha = 3$.
  Then β = 2.
  Since $k = \alpha\beta \implies k = 3 \times 2 = \mathbf{6}$.

Q28. If α and β are zeroes of x² - x - 4, find α² + β².

• Solution:
  α + β = 1 and αβ = -4.
  Identity: α² + β² = (α + β)² - 2αβ
  $\implies (1)^2 - 2(-4) = 1 + 8 = \mathbf{9}$.

Q29. If α, β are zeroes of 2x² + 5x + k satisfying $\alpha^2 + \beta^2 + \alpha\beta = \frac{21}{4}$, find k.

• Solution:
  $\alpha + \beta = -\frac{5}{2}$ and $\alpha\beta = \frac{k}{2}$.
  Rewrite expression: $(\alpha + \beta)^2 - \alpha\beta = \frac{21}{4}$
  $\implies \left(-\frac{5}{2}\right)^2 - \frac{k}{2} = \frac{21}{4} \implies \frac{25}{4} - \frac{k}{2} = \frac{21}{4}$
  $\implies \frac{k}{2} = \frac{25}{4} - \frac{21}{4} = \frac{4}{4} = 1 \implies \mathbf{k = 2}$.

Q30. If one zero of (k² + 4)x² + 13x + 4k is the reciprocal of the other, find k. (CBSE 2019)

• Solution:
  Let zeroes be α and $\frac{1}{\alpha}$. Product = 1.
  Product formula $= \frac{c}{a} = \frac{4k}{k^2+4} = 1 \implies k^2 + 4 = 4k$
  $\implies k^2 - 4k + 4 = 0 \implies (k-2)^2 = 0 \implies \mathbf{k = 2}$.

Q31. Find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$ if α, β are roots of x² - 3x + 2.

• Solution:
  α+β = 3, αβ = 2.
  $\frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{3^2 - 2(2)}{2} = \frac{9-4}{2} = \mathbf{\frac{5}{2}}$.

Q32. If α, β are zeroes of x² - p(x+1) - c, show that (α + 1)(β + 1) = 1 - c.

• Solution:
  Rearranging: x² - px - (p + c). Here α+β = p and αβ = -(p+c).
  Expand (α + 1)(β + 1) = αβ + α + β + 1
  $\implies -(p + c) + p + 1 = -p - c + p + 1 = \mathbf{1 - c}$. (Hence Proved)

Q33. Find the condition that the zeroes of ax² + bx + c are in the ratio 1:2.

• Solution:
  Let zeroes be m and 2m.
  Sum: $3m = -\frac{b}{a} \implies m = -\frac{b}{3a}$.
  Product: $2m^2 = \frac{c}{a}$.
  Substitute m: $2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a} \implies 2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a} \implies \mathbf{2b^2 = 9ac}$.

Q34. Find a quadratic polynomial whose zeroes are $\sqrt{3}$ and $-\sqrt{3}$.

• Solution:
  Sum $= \sqrt{3} + (-\sqrt{3}) = 0$.
  Product $= \sqrt{3} \times (-\sqrt{3}) = -3$.
  Polynomial $= x^2 - (0)x + (-3) = \mathbf{x^2 - 3}$.

Q35. Find the value of k if the sum of zeroes of kx² + 4x + 3k is equal to their product.

• Solution:
  Sum $= -\frac{4}{k}$, Product $= \frac{3k}{k} = 3$.
  Given: $-\frac{4}{k} = 3 \implies 3k = -4 \implies \mathbf{k = -\frac{4}{3}}$.

Q36. Find α³ + β³ if α, β are zeroes of x² - 3x + 2.

• Solution:
  α+β = 3, αβ = 2.
  Identity: α³ + β³ = (α+β)³ - 3αβ(α+β)
  $\implies (3)^3 - 3(2)(3) = 27 - 18 = \mathbf{9}$.

Q37. If x+a is a factor of 2x² + 2ax + 5x + 10, find a.

• Solution:
  Factor is $x+a \implies$ substitute x = -a.
  2(-a)² + 2a(-a) + 5(-a) + 10 = 0
  $\implies 2a^2 - 2a^2 - 5a + 10 = 0 \implies -5a = -10 \implies \mathbf{a = 2}$.

Q38. Find a quadratic polynomial whose zeroes are $\frac{1}{2}$ and $\frac{1}{3}$.

• Solution:
  Sum $= \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$. Product $= \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$.
  Polynomial $= x^2 - \frac{5}{6}x + \frac{1}{6} \implies \mathbf{6x^2 - 5x + 1}$.

Q39. If the zeroes of x² - bx + c are two consecutive integers, prove b² - 4c = 1.

• Solution:
  Let roots be α and α + 1. Differences of roots: (α+1) - α = 1.
  We know (α - β)² = (α + β)² - 4αβ.
  Here α-β = 1, α+β = b, αβ = c.
  $\implies (1)^2 = b^2 - 4c \implies \mathbf{b^2 - 4c = 1}$. (Hence Proved)

Q40. If α, β are zeroes of x² - 6x + k and 3α + 2β = 20, find k.

• Solution:
  We know $\alpha + \beta = 6 \implies 2\alpha + 2\beta = 12$.
  Given 3α + 2β = 20.
  Subtracting the equations: $(3\alpha + 2\beta) - (2\alpha + 2\beta) = 20 - 12 \implies \alpha = 8$.
  Since $\alpha + \beta = 6 \implies 8 + \beta = 6 \implies \beta = -2$.
  $k = \alpha\beta = 8 \times (-2) = \mathbf{-10}$.

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Section D: Advanced Problems & Cubic Systems (41-50)

Q41. If α, β, γ are zeroes of a cubic polynomial ax³ + bx² + cx + d, write the formula for αβγ.

• Solution:
  The product of zeroes of a cubic polynomial is given by $\mathbf{-\frac{d}{a}}$.

Q42. If two zeroes of the cubic polynomial x³ - 4x² - x + 4 are 1 and -1, find the third zero.

• Solution:
  Product of all three roots $\alpha\beta\gamma = -\frac{d}{a} = -\frac{4}{1} = -4$.
  Substitute known roots: $(1)(-1)\gamma = -4 \implies -\gamma = -4 \implies \mathbf{\gamma = 4}$.

Q43. Find the remainder when x³ + 3x² + 3x + 1 is divided by x + 1.

• Solution:
  By Remainder Theorem, substitute x = -1:
  $(-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = \mathbf{0}$.

Q44. If the zeroes of the cubic polynomial x³ - 3x² + x + 1 are a-b, a, and a+b, find a and b. (CBSE 2016)

• Solution:
  Sum of roots $= (a-b) + a + (a+b) = -\frac{-3}{1} \implies 3a = 3 \implies \mathbf{a = 1}$.
  Product of roots $= (a-b)(a)(a+b) = -\frac{1}{1} = -1$.
  Substitute $a=1 \implies (1-b)(1)(1+b) = -1 \implies 1 - b^2 = -1 \implies b^2 = 2 \implies \mathbf{b = \pm\sqrt{2}}$.

Q45. Find k if x-2 is a factor of x² - 5x + k.

• Solution:
  Substitute $x = 2 \implies 2^2 - 5(2) + k = 0 \implies 4 - 10 + k = 0 \implies \mathbf{k = 6}$.

Q46. If α, β, γ are zeroes of 2x³ + 5x² - 7x - 3, and α + β = 0, find γ.

• Solution:
  Sum of roots $\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{5}{2}$.
  Since $\alpha + \beta = 0 \implies 0 + \gamma = -\frac{5}{2} \implies \mathbf{\gamma = -\frac{5}{2}}$.

Q47. Form a cubic polynomial whose zeroes are 2, -3, and 4.

• Solution:
  • Σ α = 2 - 3 + 4 = 3
  • Σ αβ = (2)(-3) + (-3)(4) + (4)(2) = -6 - 12 + 8 = -10
  • αβγ = 2 × (-3) × 4 = -24
  • Polynomial: $x^3 - (\Sigma \alpha)x^2 + (\Sigma \alpha\beta)x - \alpha\beta\gamma = \mathbf{x^3 - 3x^2 - 10x + 24}$.

Q48. If the polynomial x⁴ - 6x³ + 16x² - 25x + 10 is divided by x² - 2x + k, the remainder is x + a. Find k and a. (CBSE Exemplar / High-Yield)

• Solution:
  By performing long division of x⁴ - 6x³ + 16x² - 25x + 10 by x² - 2x + k, the algebraic remainder matches (2k - 9)x + (10 - 8k + k²).
  Comparing to x + a:
  $2k - 9 = 1 \implies 2k = 10 \implies \mathbf{k = 5}$.
  $a = 10 - 8k + k^2 \implies a = 10 - 8(5) + 5^2 = 10 - 40 + 25 = \mathbf{-5}$.

Q49. Find the values of a and b so that x⁴ + x³ + 8x² + ax + b is divisible by x² + 1.

• Solution:
  Since it is divisible by x²+1, we can imply x² = -1.
  Substitute x² = -1 into the polynomial expression:
  $(-1)^2 + x(-1) + 8(-1) + ax + b = 0 \implies 1 - x - 8 + ax + b = 0$
  $\implies (a - 1)x + (b - 7) = 0$.
  Equating coefficients to zero: $a - 1 = 0 \implies \mathbf{a = 1}$ and $b - 7 = 0 \implies \mathbf{b = 7}$.

Q50. If α and β are zeroes of the quadratic polynomial p(x) = 3x² - 6x + 4, find the value of $\left(\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right) + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta$.

• Solution:
  Here, $\alpha+\beta = \frac{6}{3} = 2$ and $\alpha\beta = \frac{4}{3}$.
  • Part 1: $\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} = \frac{2^2 - 2(4/3)}{4/3} = \frac{4 - 8/3}{4/3} = \frac{4/3}{4/3} = 1$.
  • Part 2: $2\left(\frac{\alpha+\beta}{\alpha\beta}\right) = 2\left(\frac{2}{4/3}\right) = 2 \times \frac{6}{4} = 3$.
  • Part 3: $3\alpha\beta = 3\left(\frac{4}{3}\right) = 4$.
  • Total Sum $= 1 + 3 + 4 = \mathbf{8}$.

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