CBSE Class 10 Mathematics Quadratic Equations Assignment Set 23

Very Short Answer Type Questions (1 Mark Each)

Question 1. Check whether the following are quadratic equations: (i) \( x^2 + 2x + 1 = (4 - x)^2 + 3 \) (ii) \( x^3 - x^2 = (x - 1)^3 \) (iii) \( (k + 1)x^2 + \tfrac{3}{2}x = 7 \), where k = −1 (iv) \( 2(x - 1)^2 = 4x^2 - 2x + 1 \) (v) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^2 \).
Answer:
(i) Expanding, \( x^2 + 2x + 1 = x^2 - 8x + 19 \), so \( 10x - 18 = 0 \): linear, not quadratic.
(ii) \( x^3 - x^2 = x^3 - 3x^2 + 3x - 1 \) gives \( 2x^2 - 3x + 1 = 0 \): quadratic.
(iii) For k = −1 the coefficient of \( x^2 \) is 0, leaving \( \tfrac{3}{2}x = 7 \): not quadratic.
(iv) \( 2x^2 - 4x + 2 = 4x^2 - 2x + 1 \) gives \( 2x^2 + 2x - 1 = 0 \): quadratic.
(v) \( x^4 + 4x^3 + 4x^2 = x^4 + 4x^2 + 3 \) gives \( 4x^3 - 3 = 0 \): cubic, not quadratic.
In simple words: Simplify to the form ax² + bx + c = 0 and look at the highest power.

Teacher's Note:
1. A quadratic needs a non-zero x² term and degree exactly 2.
2. Always cancel terms before judging.

Question 2. Check whether 2 is a root of (i) \( x^2 - 4x + 5 = 0 \) (ii) \( 2x^2 - 7x + 6 = 0 \) (iii) \( 3x^2 - 6x - 2 = 0 \).
Answer:
(i) 4 − 8 + 5 = 1 ≠ 0: not a root.
(ii) 8 − 14 + 6 = 0: 2 is a root.
(iii) 12 − 12 − 2 = −2 ≠ 0: not a root.
In simple words: Substitute x = 2.

Teacher's Note:
1. A root makes the left side zero.

Question 3. If \( \tfrac{1}{2} \) is a root of \( x^2 + kx - \tfrac{5}{4} = 0 \), find k.
Answer:
\( \tfrac{1}{4} + \tfrac{k}{2} - \tfrac{5}{4} = 0 \), so \( \tfrac{k}{2} = 1 \) and k = 2.
In simple words: Put x = 1/2 in the equation.

Teacher's Note:
1. Check: x² + 2x − 5/4 at 1/2 gives 1/4 + 1 − 5/4 = 0.

Question 4. For which values of k does \( 2x^2 - kx + k = 0 \) have equal roots?
Answer:
Equal roots need D = 0: \( k^2 - 8k = 0 \), so k(k − 8) = 0. Hence k = 0 or k = 8.
In simple words: Set b² − 4ac = 0.

Teacher's Note:
1. k = 0 gives 2x² = 0 (both roots 0), a legitimate equal-roots case.

Question 5. Which constant must be added and subtracted to solve \( 9x^2 + \tfrac{3}{4}x - \sqrt{2} = 0 \) by completing the square?
Answer:
\( 9x^2 + \tfrac{3}{4}x = (3x)^2 + 2(3x)\left(\tfrac{1}{8}\right) \). Adding \( \left(\tfrac{1}{8}\right)^2 \) completes \( \left(3x + \tfrac{1}{8}\right)^2 \).
The constant is 1/64.
In simple words: Write the first two terms as (3x)² + 2·(3x)·(something).

Teacher's Note:
1. Half of the middle coefficient, squared, after taking out the leading term.

Question 6. Find the nature of the roots of (i) \( 2x^2 - \sqrt{5}x + 1 = 0 \) (ii) \( 2x^2 - 3\sqrt{2}x + \tfrac{9}{4} = 0 \) (iii) \( x^2 + 3x + 2\sqrt{2} = 0 \) (iv) \( x^2 + x - 5 = 0 \).
Answer:
(i) D = 5 − 8 = −3 < 0: no real roots.
(ii) D = 18 − 18 = 0: equal real roots.
(iii) D = 9 − 8√2 ≈ 9 − 11.31 < 0: no real roots.
(iv) D = 1 + 20 = 21 > 0: two distinct real roots.
In simple words: Compute D = b² − 4ac in each case.

Teacher's Note:
1. D > 0 distinct, D = 0 equal, D < 0 none.

Short Answer Type Questions (2 Marks Each)

Question 7. Find the roots using the quadratic formula: (i) \( 2x^2 - 3x - 5 = 0 \) (ii) \( -3x^2 + 5x + 12 = 0 \) (iii) \( \tfrac{1}{2}x^2 - \sqrt{11}x + 1 = 0 \).
Answer:
(i) D = 9 + 40 = 49, \( x = \dfrac{3 \pm 7}{4} \): x = 5/2, −1.
(ii) Multiply by −1: \( 3x^2 - 5x - 12 = 0 \), D = 25 + 144 = 169, \( x = \dfrac{5 \pm 13}{6} \): x = 3, −4/3.
(iii) D = 11 − 2 = 9, \( x = \dfrac{\sqrt{11} \pm 3}{1} \): x = √11 + 3, √11 − 3.
In simple words: x = (−b ± √D) / 2a.

Teacher's Note:
1. Clear a negative leading coefficient first.
2. Verify by substituting.

Question 8. Find the roots by factorisation: (i) \( 2x^2 + \tfrac{5}{3}x - 2 = 0 \) (ii) \( 3\sqrt{2}x^2 - 5x - \sqrt{2} = 0 \).
Answer:
(i) Multiply by 3: \( 6x^2 + 5x - 6 = 0 \), i.e. \( (3x - 2)(2x + 3) = 0 \): x = 2/3, −3/2.
(ii) \( 3\sqrt{2}x^2 - 6x + x - \sqrt{2} = 3\sqrt{2}\,x\left(x - \sqrt{2}\right) + 1\left(x - \sqrt{2}\right) = 0 \) (using \( 3\sqrt{2}\cdot\sqrt{2} = 6 \)), so \( (x - \sqrt{2})(3\sqrt{2}x + 1) = 0 \): x = √2, −1/(3√2).
In simple words: Split the middle term so that the product matches a × c.

Teacher's Note:
1. (ii) Check: −6 + 1 = −5 and (−6)(1) = 3√2 × (−√2).

Question 9. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.
Answer:
Let the number be n. \( n^2 - 84 = 3(n + 8) \), so \( n^2 - 3n - 108 = 0 \), \( (n - 12)(n + 9) = 0 \). Reject −9: n = 12.
In simple words: Form n² − 84 = 3(n + 8).

Teacher's Note:
1. Check: 144 − 84 = 60 = 3 × 20.

Question 10. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number.
Answer:
\( n + 12 = \dfrac{160}{n} \), so \( n^2 + 12n - 160 = 0 \), \( (n + 20)(n - 8) = 0 \). Reject −20: n = 8.
In simple words: Reciprocal of n is 1/n.

Teacher's Note:
1. Check: 8 + 12 = 20 = 160/8.

Question 11. A train, travelling at a uniform speed for 360 km, would have taken 48 minutes less to travel the same distance if its speed were 5 km/h more. Find the original speed.
Answer:
Let the speed be x km/h. \( \dfrac{360}{x} - \dfrac{360}{x + 5} = \dfrac{4}{5} \), so \( x(x + 5) = 2250 \), \( x^2 + 5x - 2250 = 0 \), \( (x + 50)(x - 45) = 0 \).
Reject −50: the original speed is 45 km/h.
In simple words: 48 minutes = 4/5 hour.

Teacher's Note:
1. Check: 8 h − 7.2 h = 0.8 h.

Question 12. If Zeba were younger by 5 years than what she really is, the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?
Answer:
Let her age be x. \( (x - 5)^2 = 5x + 11 \), so \( x^2 - 15x + 14 = 0 \), \( (x - 1)(x - 14) = 0 \). Age 1 is impossible since x − 5 would be negative: x = 14 years.
In simple words: Her younger age is x − 5.

Teacher's Note:
1. Check: 81 = 70 + 11.

Long Answer Type Questions (4 Marks Each)

Question 13. At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both.
Answer:
Let Nisha's age be n; then Asha's age is \( n^2 + 2 \). Nisha reaches Asha's present age after \( n^2 + 2 - n \) years, when Asha is \( n^2 + 2 + n^2 + 2 - n = 2n^2 - n + 4 \).
\( 2n^2 - n + 4 = 10n - 1 \), so \( 2n^2 - 11n + 5 = 0 \), \( (2n - 1)(n - 5) = 0 \). Taking the sensible age n = 5, Asha's age is 27.
Nisha is 5 years and Asha is 27 years (n = 1/2 is rejected as unrealistic).
In simple words: Both ages increase by the same number of years.

Teacher's Note:
1. Check: 27 + 22 = 49 = 10(5) − 1.

Question 14. Check whether \( 6x^2 - 7x + 2 = 0 \) has real roots, and if it has, find them by completing the square.
Answer:
D = 49 − 48 = 1 > 0, so the roots are real.
\( x^2 - \tfrac{7}{6}x = -\tfrac{1}{3} \), so \( \left(x - \tfrac{7}{12}\right)^2 = \tfrac{49}{144} - \tfrac{48}{144} = \tfrac{1}{144} \). Hence \( x - \tfrac{7}{12} = \pm\tfrac{1}{12} \): x = 2/3, 1/2.
In simple words: Divide by 6 first.

Teacher's Note:
1. Half of 7/6 is 7/12.

Teacher's Note:
1. Equal borders mean each side shrinks by 2x.
2. The figure shows grass on all four sides.

Question 16. At t minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than t²/4 minutes. Find t.
Answer:
Time left to 3 pm = 60 − t. \( 60 - t = \dfrac{t^2}{4} - 3 \), so \( t^2 + 4t - 252 = 0 \), \( (t + 18)(t - 14) = 0 \). Reject −18: t = 14.
In simple words: Time to 3 pm is (60 − t) minutes.

Teacher's Note:
1. Check: 46 = 49 − 3.