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CBSE Class 10 Maths Chapter 4 Quadratic Equations Model Questions - 2 Marks - Part 2

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SECTION B — Very Short Answer Type Questions [2 Marks Each]

• Q26. Solve: $4x^2 - 4ax + (a^2 - b^2) = 0$.
  Answer:
  1. Using the quadratic formula with $A = 4$, $B = -4a$, $C = a^2 - b^2$:
  2. Discriminant $D = (-4a)^2 - 4(4)(a^2 - b^2) = 16a^2 - 16a^2 + 16b^2 = 16b^2$.
  3. $x = \frac{-(-4a) \pm \sqrt{16b^2}}{2(4)} = \frac{4a \pm 4b}{8} = \frac{a \pm b}{2}$.
  
  Result: $x = \frac{a+b}{2}, \frac{a-b}{2}$
• Q27. Find the roots of the equation $3x^2 + 5\sqrt{5}x - 10 = 0$.
  Answer:
  1. Split middle term $5\sqrt{5}x$ into $6\sqrt{5}x - \sqrt{5}x$ (since $6\sqrt{5} \times -\sqrt{5} = -30$, matching $3 \times -10 = -30$):
  $3x^2 + 6\sqrt{5}x - \sqrt{5}x - 10 = 0$.
  2. Factor by grouping: $3x(x + 2\sqrt{5}) - \sqrt{5}(x + 2\sqrt{5}) = 0 \implies (3x - \sqrt{5})(x + 2\sqrt{5}) = 0$.
  3. Solve for $x$: $x = \frac{\sqrt{5}}{3}$ and $x = -2\sqrt{5}$.
  
  Result: $x = \frac{\sqrt{5}}{3}, -2\sqrt{5}$
• Q28. Solve by factorisation: $x^2 + 2\sqrt{2}x - 6 = 0$.
  Answer:
  1. Split middle term into $3\sqrt{2}x - \sqrt{2}x$: $x^2 + 3\sqrt{2}x - \sqrt{2}x - 6 = 0$.
  2. Factor by grouping: $x(x + 3\sqrt{2}) - \sqrt{2}(x + 3\sqrt{2}) = 0 \implies (x - \sqrt{2})(x + 3\sqrt{2}) = 0$.
  3. Solve for $x$: $x = \sqrt{2}$ or $x = -3\sqrt{2}$.
  
  Result: $x = \sqrt{2}, -3\sqrt{2}$
• Q29. Find the roots of $a^2x^2 - 3abx + 2b^2 = 0$.
  Answer:
  1. Split middle term into $-2abx - abx$: $a^2x^2 - 2abx - abx + 2b^2 = 0$.
  2. Factor by grouping: $ax(ax - 2b) - b(ax - 2b) = 0 \implies (ax - b)(ax - 2b) = 0$.
  3. Solve for $x$: $x = \frac{b}{a}$ and $x = \frac{2b}{a}$.
  
  Result: $x = \frac{b}{a}, \frac{2b}{a}$
• Q30. Solve for $x$: $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$ ($x \neq -4, 7$).
  Answer:
  1. Combine left fractions: $\frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} \implies \frac{-11}{x^2 - 3x - 28} = \frac{11}{30}$.
  2. Simplify to quadratic equation: $x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0$.
  
  Result: $x = 1, 2$
• Q31. Solve the quadratic equation $2x^2 - 7x + 3 = 0$ using the quadratic formula.
  Answer:
  1. Discriminant $D = (-7)^2 - 4(2)(3) = 49 - 24 = 25$.
  2. $x = \frac{-(-7) \pm \sqrt{25}}{2(2)} = \frac{7 \pm 5}{4}$.
  3. $x = \frac{12}{4} = 3$ and $x = \frac{2}{4} = \frac{1}{2}$.
  
  Result: $x = 3, \frac{1}{2}$
• Q32. Find the roots of $x^2 + 4x + 5 = 0$ using the quadratic formula, if they exist.
  Answer:
  1. Discriminant $D = 4^2 - 4(1)(5) = 16 - 20 = -4$.
  2. Since $D < 0$, real roots do not exist.
  
  Result: No real roots exist.
• Q33. Find the roots of $2x^2 + x - 4 = 0$ using the quadratic formula.
  Answer:
  1. Discriminant $D = 1^2 - 4(2)(-4) = 1 + 32 = 33$.
  2. $x = \frac{-1 \pm \sqrt{33}}{2(2)} = \frac{-1 \pm \sqrt{33}}{4}$.
  
  Result: $x = \frac{-1 \pm \sqrt{33}}{4}$
• Q34. Solve for $x$ using the quadratic formula: $p^2x^2 + (p^2-q^2)x - q^2 = 0$.
  Answer:
  1. Discriminant $D = (p^2 - q^2)^2 - 4(p^2)(-q^2) = p^4 - 2p^2q^2 + q^4 + 4p^2q^2 = p^4 + 2p^2q^2 + q^4 = (p^2 + q^2)^2$.
  2. $x = \frac{-(p^2 - q^2) \pm (p^2 + q^2)}{2p^2}$.
  3. Roots: $x = \frac{2q^2}{2p^2} = \frac{q^2}{p^2}$ and $x = \frac{-2p^2}{2p^2} = -1$.
  
  Result: $x = -1, \frac{q^2}{p^2}$
• Q35. Find the roots of the equation $x - \frac{1}{x} = 3$ ($x \neq 0$).
  Answer:
  1. Multiply by $x$: $x^2 - 1 = 3x \implies x^2 - 3x - 1 = 0$.
  2. Using quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}$.
  
  Result: $x = \frac{3 \pm \sqrt{13}}{2}$
• Q36. Solve for $x$: $\frac{1}{x} - \frac{1}{x-2} = 3$ ($x \neq 0, 2$).
  Answer:
  1. Combine fractions: $\frac{(x - 2) - x}{x(x - 2)} = 3 \implies \frac{-2}{x^2 - 2x} = 3$.
  2. Simplify: $-2 = 3x^2 - 6x \implies 3x^2 - 6x + 2 = 0$.
  3. Quadratic formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(2)}}{2(3)} = \frac{6 \pm \sqrt{12}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3}$.
  
  Result: $x = \frac{3 \pm \sqrt{3}}{3}$
• Q37. Use the quadratic formula to solve: $9x^2 - 9(a+b)x + (2a^2 + 5ab + 2b^2) = 0$.
  Answer:
  1. Factor constant term $2a^2 + 5ab + 2b^2 = (2a + b)(a + 2b)$.
  2. Discriminant $D = 81(a+b)^2 - 36(2a^2 + 5ab + 2b^2) = 9(a - b)^2$.
  3. $x = \frac{9(a+b) \pm 3(a-b)}{18}$.
  4. Roots: $x = \frac{12a + 6b}{18} = \frac{2a+b}{3}$ and $x = \frac{6a + 12b}{18} = \frac{a+2b}{3}$.
  
  Result: $x = \frac{2a+b}{3}, \frac{a+2b}{3}$
• Q38. Solve the equation $2x^2 + 5\sqrt{3}x + 6 = 0$ using the quadratic formula.
  Answer:
  1. Discriminant $D = (5\sqrt{3})^2 - 4(2)(6) = 75 - 48 = 27$.
  2. $x = \frac{-5\sqrt{3} \pm \sqrt{27}}{2(2)} = \frac{-5\sqrt{3} \pm 3\sqrt{3}}{4}$.
  3. Roots: $x = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2}$ and $x = \frac{-8\sqrt{3}}{4} = -2\sqrt{3}$.
  
  Result: $x = -\frac{\sqrt{3}}{2}, -2\sqrt{3}$
• Q39. Find the roots of $x^2 - 2ax + (a^2 - b^2) = 0$.
  Answer:
  1. Rewrite as $(x - a)^2 - b^2 = 0 \implies (x - a - b)(x - a + b) = 0$.
  2. Solve for $x$: $x = a + b$ and $x = a - b$.
  
  Result: $x = a+b, a-b$
• Q40. If $x = \frac{2}{3}$ is one root of the quadratic equation $6x^2 - x - k = 0$, find the value of $k$.
  Answer:
  1. Substitute $x = \frac{2}{3}$ into the equation: $6\left(\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right) - k = 0$.
  2. Simplify: $6\left(\frac{4}{9}\right) - \frac{2}{3} - k = 0 \implies \frac{8}{3} - \frac{2}{3} - k = 0 \implies \frac{6}{3} - k = 0 \implies 2 - k = 0 \implies k = 2$.
  
  Result: $k = 2$
• Q41. If $-5$ is a root of the quadratic equation $2x^2 + px - 15 = 0$ and the quadratic equation $p(x^2 + x) + k = 0$ has equal roots, find the value of $k$.
  Answer:
  1. Substitute $x = -5$ to find $p$: $2(-5)^2 + p(-5) - 15 = 0 \implies 35 - 5p = 0 \implies p = 7$.
  2. For equation $7x^2 + 7x + k = 0$ to have equal roots, $D = 0$: $7^2 - 4(7)(k) = 0 \implies 49 - 28k = 0 \implies k = \frac{7}{4}$.
  
  Result: $k = \frac{7}{4}$
• Q42. If $x = 2$ and $x = 3$ are roots of the equation $3x^2 - 2kx + 2m = 0$, find the values of $k$ and $m$.
  Answer:
  1. Substitute $x = 2$: $3(4) - 2k(2) + 2m = 0 \implies 12 - 4k + 2m = 0 \implies 2k - m = 6$ (Eq. 1).
  2. Substitute $x = 3$: $3(9) - 2k(3) + 2m = 0 \implies 27 - 6k + 2m = 0 \implies 3k - m = \frac{27}{2}$ (Eq. 2).
  3. Subtract Eq. 1 from Eq. 2: $k = \frac{15}{2}$, then $m = \frac{9}{2}$.
  
  Result: $k = \frac{15}{2}, m = \frac{9}{2}$
• Q43. Find the value of $k$ for which the given quadratic equation has real and equal roots: $(k-12)x^2 + 2(k-12)x + 2 = 0$.
  Answer:
  1. For equal roots, $D = 0$: $[2(k-12)]^2 - 4(k-12)(2) = 0$.
  2. Factor out $4(k-12)$: $4(k-12)[(k-12) - 2] = 0 \implies 4(k-12)(k-14) = 0$.
  3. Since $k \neq 12$ (otherwise $x^2$ coefficient becomes zero), $k = 14$.
  
  Result: $k = 14$
• Q44. If the roots of the quadratic equation $(a-b)x^2 + (b-c)x + (c-a) = 0$ are equal, prove that $2a = b + c$.
  Answer:
  1. For equal roots, $D = 0$: $(b-c)^2 - 4(a-b)(c-a) = 0$.
  2. Notice that sum of coefficients $(a-b) + (b-c) + (c-a) = 0$, meaning $x = 1$ is a root. Since roots are equal, both roots are $1$, implying product of roots $\frac{c-a}{a-b} = 1 \implies c-a = a-b \implies 2a = b+c$.
  
  Result: Proved.
• Q45. Find the values of $k$ for which the quadratic equation $kx^2 + 2x + 1 = 0$ has real and distinct roots.
  Answer:
  1. For real and distinct roots, $D > 0$: $2^2 - 4(k)(1) > 0 \implies 4 - 4k > 0 \implies k < 1$.
  2. Also, $k \neq 0$.
  
  Result: $k < 1$ and $k \neq 0$
• Q46. Find the range of values of $p$ for which the equation $x^2 + 5px + 16 = 0$ has no real roots.
  Answer:
  1. For no real roots, $D < 0$: $(5p)^2 - 4(1)(16) < 0 \implies 25p^2 - 64 < 0 \implies \left(5p - 8\right)\left(5p + 8\right) < 0$.
  
  Result: $-\frac{8}{5} < p < \frac{8}{5}$
• Q47. If $\sin\theta$ and $\cos\theta$ are the roots of the equation $ax^2 + bx + c = 0$, prove that $a^2 - b^2 + 2ac = 0$.
  Answer:
  1. Sum of roots: $\sin\theta + \cos\theta = -\frac{b}{a}$. Product of roots: $\sin\theta\cos\theta = \frac{c}{a}$.
  2. Square sum of roots: $(\sin\theta + \cos\theta)^2 = \left(-\frac{b}{a}\right)^2 \implies 1 + 2\sin\theta\cos\theta = \frac{b^2}{a^2}$.
  3. Substitute product: $1 + \frac{2c}{a} = \frac{b^2}{a^2} \implies \frac{a + 2c}{a} = \frac{b^2}{a^2} \implies a^2 + 2ac = b^2 \implies a^2 - b^2 + 2ac = 0$.
  
  Result: Proved.
• Q48. Find the quadratic equation whose roots are $2 + \sqrt{3}$ and $2 - \sqrt{3}$.
  Answer:
  1. Sum of roots = $(2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$.
  2. Product of roots = $(2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.
  3. Equation: $x^2 - (\text{Sum})x + \text{Product} = 0$.
  
  Result: $x^2 - 4x + 1 = 0$
• Q49. If the sum of the roots of the quadratic equation $ky^2 - 11y + (k-23) = 0$ is $\frac{13}{21}$ more than their product, find the value of $k$.
  Answer:
  1. Sum of roots = $\frac{11}{k}$, Product of roots = $\frac{k-23}{k}$.
  2. Given condition: $\frac{11}{k} = \frac{k-23}{k} + \frac{13}{21} \implies \frac{11 - k + 23}{k} = \frac{13}{21} \implies \frac{34 - k}{k} = \frac{13}{21}$.
  3. Cross-multiply: $21(34 - k) = 13k \implies 714 - 21k = 13k \implies 34k = 714 \implies k = 21$.
  
  Result: $k = 21$
• Q50. Solve using the quadratic formula: $x^2 - 5x - 6 = 0$.
  Answer:
  1. Discriminant $D = (-5)^2 - 4(1)(-6) = 25 + 24 = 49$.
  2. $x = \frac{-(-5) \pm \sqrt{49}}{2(1)} = \frac{5 \pm 7}{2}$.
  3. Roots: $x = \frac{12}{2} = 6$ and $x = \frac{-2}{2} = -1$.
  
  Result: $x = 6, -1$

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