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CBSE Class 10 Maths Chapter 4 Quadratic Equations Model Questions - 3 Marks - Part 1

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SECTION C — Short Answer Type Questions [3 Marks Each]

• Q1. Find the roots of the quadratic equation: $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ by factorization
  Answer:
  1. Multiply coefficient of $x^2$ and constant term: $\sqrt{2} \times 5\sqrt{2} = 10$.
  2. Split middle term $7x$ into $5x + 2x$: $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$.
  3. Factor by grouping: $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \implies (\sqrt{2}x + 5)(x + \sqrt{2}) = 0$.
  4. Set each factor to zero: $\sqrt{2}x + 5 = 0 \implies x = -\frac{5}{\sqrt{2}}$ (or $-\frac{5\sqrt{2}}{2}$) and $x + \sqrt{2} = 0 \implies x = -\sqrt{2}$.
  
  Result: $x = -\frac{5\sqrt{2}}{2}$ and $x = -\sqrt{2}$
• Q2. Solve for $x$: $4x^2 - 4ax + (a^2 - b^2) = 0$.
  Answer:
  1. Rewrite using quadratic formula or by grouping: Let's use the quadratic formula where $A = 4$, $B = -4a$, $C = a^2 - b^2$.
  2. Discriminant $D = B^2 - 4AC = (-4a)^2 - 4(4)(a^2 - b^2) = 16a^2 - 16(a^2 - b^2) = 16a^2 - 16a^2 + 16b^2 = 16b^2$.
  3. Solve for $x$: $x = \frac{-(-4a) \pm \sqrt{16b^2}}{2(4)} = \frac{4a \pm 4b}{8} = \frac{a \pm b}{2}$.
  
  Result: $x = \frac{a + b}{2}$ and $x = \frac{a - b}{2}$
• Q3. Find the value(s) of $k$ for which the quadratic equation $kx(x - 2) + 6 = 0$ has two real and equal roots.
  Answer:
  1. Expand the equation: $kx^2 - 2kx + 6 = 0$.
  2. Identify coefficients: $A = k$, $B = -2k$, $C = 6$.
  3. For real and equal roots, discriminant $D = B^2 - 4AC = 0$: $(-2k)^2 - 4(k)(6) = 0 \implies 4k^2 - 24k = 0$.
  4. Factor out $4k$: $4k(k - 6) = 0 \implies k = 0$ or $k = 6$.
  5. Since $k = 0$ makes the equation linear ($6 = 0$, invalid), $k = 6$.
  
  Result: $k = 6$
• Q4. Solve using the quadratic formula: $x^2 - 2ax + (a^2 - b^2) = 0$.
  Answer:
  1. Identify coefficients: $A = 1$, $B = -2a$, $C = a^2 - b^2$.
  2. Calculate discriminant $D = B^2 - 4AC = (-2a)^2 - 4(1)(a^2 - b^2) = 4a^2 - 4a^2 + 4b^2 = 4b^2$.
  3. Apply quadratic formula: $x = \frac{-(-2a) \pm \sqrt{4b^2}}{2(1)} = \frac{2a \pm 2b}{2} = a \pm b$.
  
  Result: $x = a + b$ and $x = a - b$
• Q5. Find the roots of the equation: $\frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}$, where $x \neq -4, 7$.
  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. Divide by $11$: $\frac{-1}{x^2 - 3x - 28} = \frac{1}{30} \implies x^2 - 3x - 28 = -30 \implies x^2 - 3x + 2 = 0$.
  3. Factorize: $(x - 1)(x - 2) = 0 \implies x = 1$ or $x = 2$.
  
  Result: $x = 1$ and $x = 2$
• Q6. Determine whether the equation $3x^2 - 5x + 9 = 0$ has real roots. If yes, find them.
  Answer:
  1. Find discriminant $D = B^2 - 4AC$: $A = 3$, $B = -5$, $C = 9$.
  2. $D = (-5)^2 - 4(3)(9) = 25 - 108 = -83$.
  3. Since $D < 0$, the equation has no real roots.
  
  Result: No real roots exist.
• Q7. Solve for $x$: $\frac{x+1}{x-1} + \frac{x-2}{x+2} = 3$, where $x \neq 1, -2$.
  Answer:
  1. Take LCM: $\frac{(x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)} = 3$.
  2. Expand numerator and denominator: $\frac{(x^2 + 3x + 2) + (x^2 - 3x + 2)}{x^2 + x - 2} = 3 \implies \frac{2x^2 + 4}{x^2 + x - 2} = 3$.
  3. Cross-multiply: $2x^2 + 4 = 3(x^2 + x - 2) \implies 2x^2 + 4 = 3x^2 + 3x - 6 \implies x^2 + 3x - 10 = 0$.
  4. Factorize: $(x + 5)(x - 2) = 0 \implies x = -5$ or $x = 2$.
  
  Result: $x = -5$ and $x = 2$
• Q8. Find the roots of the equation: $2x^2 - x + \frac{1}{8} = 0$.
  Answer:
  1. Multiply equation by $8$ to clear fractions: $16x^2 - 8x + 1 = 0$.
  2. Recognize as a complete square: $(4x - 1)^2 = 0$.
  3. Solve for $x$: $4x - 1 = 0 \implies x = \frac{1}{4}$ (repeated root).
  
  Result: $x = \frac{1}{4}, \frac{1}{4}$
• Q9. Solve the following equation for $x$: $x^2 - ({\sqrt{3} + 1})x + \sqrt{3} = 0$.
  Answer:
  1. The middle term is already split into $-\sqrt{3}x - 1x$: $x^2 - \sqrt{3}x - x + \sqrt{3} = 0$.
  2. Factor by grouping: $x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0 \implies (x - \sqrt{3})(x - 1) = 0$.
  3. Solve for $x$: $x = \sqrt{3}$ or $x = 1$.
  
  Result: $x = 1$ and $x = \sqrt{3}$
• Q10. Find the values of $k$ for which the quadratic equation $kx^2 + 2x + 1 = 0$ has real and distinct roots.
  Answer:
  1. Identify coefficients: $A = k$, $B = 2$, $C = 1$.
  2. For real and distinct roots, discriminant $D > 0$: $B^2 - 4AC > 0 \implies 2^2 - 4(k)(1) > 0 \implies 4 - 4k > 0$.
  3. Solve inequality: $4 > 4k \implies k < 1$. Also $k \neq 0$ for a quadratic equation.
  
  Result: $k < 1$ (and $k \neq 0$)
• Q11. Solve for $x$: $\frac{2x}{x-3} + \frac{1}{2x+3} + \frac{3x+9}{(x-3)(2x+3)} = 0$, $x \neq 3, -\frac{3}{2}$.
  Answer:
  1. Combine terms over common denominator $(x - 3)(2x + 3)$: $\frac{2x(2x + 3) + 1(x - 3) + (3x + 9)}{(x - 3)(2x + 3)} = 0$.
  2. Set numerator to zero: $4x^2 + 6x + x - 3 + 3x + 9 = 0 \implies 4x^2 + 10x + 6 = 0$.
  3. Divide by $2$: $2x^2 + 5x + 3 = 0 \implies 2x^2 + 2x + 3x + 3 = 0 \implies 2x(x + 1) + 3(x + 1) = 0 \implies (2x + 3)(x + 1) = 0$.
  4. Solve for $x$: $x = -1$ (since $x = -\frac{3}{2}$ is excluded).
  
  Result: $x = -1$
• Q12. Find the roots of: $x^2 - 3x - 10 = 0$ using the method of completing the square.
  Answer:
  1. Shift constant to the right: $x^2 - 3x = 10$.
  2. Add square of half the coefficient of $x$ ($\left(\frac{3}{2}\right)^2 = \frac{9}{4}$) to both sides: $x^2 - 3x + \frac{9}{4} = 10 + \frac{9}{4} \implies \left(x - \frac{3}{2}\right)^2 = \frac{49}{4}$.
  3. Take square root: $x - \frac{3}{2} = \pm \frac{7}{2}$.
  4. Solve for $x$: $x = \frac{3}{2} \pm \frac{7}{2} \implies x = \frac{10}{2} = 5$ or $x = -\frac{4}{2} = -2$.
  
  Result: $x = 5$ and $x = -2$
• Q13. 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 $x = -5$ into $2x^2 + px - 15 = 0$: $2(-5)^2 + p(-5) - 15 = 0 \implies 50 - 5p - 15 = 0 \implies 35 - 5p = 0 \implies p = 7$.
  2. Substitute $p = 7$ into $p(x^2 + x) + k = 0$: $7x^2 + 7x + k = 0$.
  3. For equal roots, discriminant $D = 0$: $7^2 - 4(7)(k) = 0 \implies 49 - 28k = 0 \implies k = \frac{49}{28} = \frac{7}{4}$.
  
  Result: $k = \frac{7}{4}$
• Q14. Solve for $x$: $a^2b^2x^2 + b^2x - a^2x - 1 = 0$.
  Answer:
  1. Factor by grouping: $b^2x(a^2x + 1) - 1(a^2x + 1) = 0$.
  2. Combine factors: $(b^2x - 1)(a^2x + 1) = 0$.
  3. Solve for $x$: $b^2x - 1 = 0 \implies x = \frac{1}{b^2}$ and $a^2x + 1 = 0 \implies x = -\frac{1}{a^2}$.
  
  Result: $x = \frac{1}{b^2}$ and $x = -\frac{1}{a^2}$
• Q15. Find the nature of roots of the quadratic equation $2x^2 - 4x + 3 = 0$. If the real roots exist, find them.
  Answer:
  1. Find discriminant $D = B^2 - 4AC$: $A = 2$, $B = -4$, $C = 3$.
  2. $D = (-4)^2 - 4(2)(3) = 16 - 24 = -8$.
  3. Since $D < 0$, the equation has no real roots.
  
  Result: No real roots exist.
• Q16. The sum of the reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is $\frac{1}{3}$. Find his present age.
  Answer:
  1. Let Rehman's present age be $x$ years.
  2. Age 3 years ago = $x - 3$, age 5 years from now = $x + 5$.
  3. Equation: $\frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \implies \frac{(x+5) + (x-3)}{(x-3)(x+5)} = \frac{1}{3} \implies \frac{2x+2}{x^2+2x-15} = \frac{1}{3}$.
  4. Cross-multiply: $3(2x+2) = x^2+2x-15 \implies 6x+6 = x^2+2x-15 \implies x^2 - 4x - 21 = 0$.
  5. Factorize: $(x - 7)(x + 3) = 0 \implies x = 7$ (rejecting negative age).
  
  Result: $7\text{ years}$
• Q17. In a class test, the sum of marks obtained by Shefali in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
  Answer:
  1. Let Maths marks be $x$, then English marks = $30 - x$.
  2. New Maths = $x + 2$, New English = $(30 - x) - 3 = 27 - x$.
  3. Product equation: $(x + 2)(27 - x) = 210 \implies 27x - x^2 + 54 - 2x = 210 \implies -x^2 + 25x + 54 = 210 \implies x^2 - 25x + 156 = 0$.
  4. Factorize: $(x - 12)(x - 13) = 0 \implies x = 12$ or $x = 13$.
  
  Result: Maths = $12$, English = $18$ (or Maths = $13$, English = $17$)
• Q18. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
  Answer:
  1. Let larger number be $x$ and smaller number be $y$.
  2. Equations: $x^2 - y^2 = 180$ and $y^2 = 8x$.
  3. Substitute $y^2$: $x^2 - 8x = 180 \implies x^2 - 8x - 180 = 0$.
  4. Factorize: $(x - 18)(x + 10) = 0 \implies x = 18$ (since larger number must be positive).
  5. Find $y$: $y^2 = 8(18) = 144 \implies y = \pm 12$.
  
  Result: Numbers are $18$ and $\pm 12$
• Q19. A train travels $360\text{ km}$ at a uniform speed. If the speed had been $5\text{ km/h}$ more, it would have taken 1 hour less for the same journey. Find the original speed of the train.
  Answer:
  1. Let original speed be $x\text{ km/h}$.
  2. Equation: $\frac{360}{x} - \frac{360}{x + 5} = 1$.
  3. Simplify: $\frac{360(x + 5) - 360x}{x(x + 5)} = 1 \implies \frac{1800}{x^2 + 5x} = 1 \implies x^2 + 5x - 1800 = 0$.
  4. Factorize: $(x + 45)(x - 40) = 0 \implies x = 40\text{ km/h}$.
  
  Result: $40\text{ km/h}$
• Q20. Two water taps together can fill a tank in $9\frac{3}{8}$ hours ($\frac{75}{8}\text{ hours}$). The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
  Answer:
  1. Let smaller tap take $x$ hours, larger take $x - 10$ hours.
  2. Equation: $\frac{1}{x} + \frac{1}{x - 10} = \frac{8}{75}$.
  3. Simplify: $\frac{2x - 10}{x^2 - 10x} = \frac{8}{75} \implies 75(2x - 10) = 8(x^2 - 10x) \implies 150x - 750 = 8x^2 - 80x \implies 8x^2 - 230x + 750 = 0 \implies 4x^2 - 115x + 375 = 0$.
  4. Factorize: $(4x - 25)(x - 15) = 0 \implies x = 25$ (since $x = 6.25$ is rejected because larger tap time $x-10 < 0$).
  5. Smaller tap = $25\text{ hours}$, Larger tap = $15\text{ hours}$.
  
  Result: Smaller tap = $25\text{ hours}$, Larger tap = $15\text{ hours}$
• Q21. Find two consecutive odd positive integers, sum of whose squares is 290.
  Answer:
  1. Let integers be $x$ and $x + 2$.
  2. Equation: $x^2 + (x + 2)^2 = 290 \implies x^2 + x^2 + 4x + 4 = 290 \implies 2x^2 + 4x - 286 = 0 \implies x^2 + 2x - 143 = 0$.
  3. Factorize: $(x + 13)(x - 11) = 0 \implies x = 11$ (positive integer).
  4. Second integer = $11 + 2 = 13$.
  
  Result: $11$ and $13$
• Q22. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
  Answer:
  1. Let son's age be $x$, father's age = $45 - x$.
  2. Five years ago: Son = $x - 5$, Father = $40 - x$.
  3. Equation: $(x - 5)(40 - x) = 124 \implies 40x - x^2 - 200 + 5x = 124 \implies -x^2 + 45x - 324 = 0 \implies x^2 - 45x + 324 = 0$.
  4. Factorize: $(x - 9)(x - 36) = 0 \implies x = 9$ (since son cannot be $36$ years old when father is $45$).
  5. Son = $9\text{ years}$, Father = $36\text{ years}$.
  
  Result: Son = $9\text{ years}$, Father = $36\text{ years}$
• Q23. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
  Answer:
  1. Let the natural number be $x$.
  2. Equation: $x + 12 = \frac{160}{x}$.
  3. Cross-multiply: $x(x + 12) = 160 \implies x^2 + 12x - 160 = 0$.
  4. Factorize: $(x + 20)(x - 8) = 0 \implies x = 8$ (natural number must be positive).
  
  Result: $8$
• Q24. The product of Shikha’s age (in years) 5 years ago and her age 8 years later is 30. Find her present age.
  Answer:
  1. Let Shikha's present age be $x$ years.
  2. Age 5 years ago = $x - 5$, age 8 years later = $x + 8$.
  3. Equation: $(x - 5)(x + 8) = 30 \implies x^2 + 3x - 40 = 30 \implies x^2 + 3x - 70 = 0$.
  4. Factorize: $(x + 10)(x - 7) = 0 \implies x = 7$ (rejecting negative age).
  
  Result: $7\text{ years}$
• Q25. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
  Answer:
  1. Let larger part be $x$, smaller part = $16 - x$.
  2. Equation: $2x^2 - (16 - x)^2 = 164$.
  3. Expand: $2x^2 - (256 - 32x + x^2) = 164 \implies x^2 + 32x - 256 = 164 \implies x^2 + 32x - 420 = 0$.
  4. Factorize: $(x + 42)(x - 10) = 0 \implies x = 10$ (positive part).
  5. Smaller part = $16 - 10 = 6$.
  
  Result: $10$ and $6$

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