CBSE Class 10 Quadratic Equations Sure Shot Questions Set 04

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This Class 10 Maths practice set on Quadratic Equations, Chapter 4, covers the nature of roots. Students find the value of k or p for which a quadratic equation has real and equal roots, real roots, or distinct roots, and prove conditions on the coefficients using the discriminant D = b² - 4ac.

PRACTICE QUESTIONS
CLASS X : CHAPTER 4
QUADRATIC EQUATIONS
NATURE OF ROOTS

1. Find the value of k for which the quadratic equation 2x² + kx + 3 = 0 has two real equal roots.

Answer:
1. Comparing with ax² + bx + c = 0, a = 2, b = k, c = 3.
2. For equal roots, D = b² - 4ac = 0, so k² - 4(2)(3) = 0.
3. k² - 24 = 0, so k² = 24.
4. k = ±2√6.

Teacher's Note:
a) Equal roots need D = 0, not D greater than 0.
b) Remember both signs: k = 2√6 and k = -2√6.

2. Find the value of k for which the quadratic equation kx(x - 3) + 9 = 0 has two real equal roots.

Answer:
1. Expanding, kx² - 3kx + 9 = 0, so a = k, b = -3k, c = 9.
2. D = (-3k)² - 4(k)(9) = 9k² - 36k = 0.
3. 9k(k - 4) = 0, so k = 0 or k = 4.
4. k = 0 makes the equation 9 = 0, which is not a quadratic, so k = 4.

Teacher's Note:
a) Expand the bracket before comparing coefficients.
b) Reject k = 0 because the coefficient of x² must not be zero.

3. Find the value of k for which the quadratic equation 4x² - 3kx + 1 = 0 has two real equal roots.

Answer:
1. a = 4, b = -3k, c = 1.
2. D = (-3k)² - 4(4)(1) = 9k² - 16 = 0.
3. 9k² = 16, so k² = 16/9.
4. k = ±4/3.

Teacher's Note:
a) The square of -3k is 9k², not -9k².
b) Both k = 4/3 and k = -4/3 give equal roots.

4. If -4 is a root of the equation x² + px - 4 = 0 and the equation x² + px + q = 0 has equal roots, find the value of p and q.

Answer:
1. Substituting x = -4 in x² + px - 4 = 0 gives 16 - 4p - 4 = 0.
2. 12 - 4p = 0, so p = 3.
3. The second equation x² + 3x + q = 0 has equal roots, so D = 9 - 4q = 0.
4. 4q = 9, so q = \( \frac{9}{4} \).
5. Hence, p = 3 and q = \( \frac{9}{4} \).

Teacher's Note:
a) Substituting the given root is the first step in every such problem.
b) Check the answer: x² + 3x - 4 = 0 has root -4, since 16 - 12 - 4 = 0.
c) Equal roots of x² + 3x + q = 0 mean 3² = 4q.

5. If -5 is a root of the equation 2x² + px - 15 = 0 and the equation p(x² + x) + k = 0 has equal roots, find the value of k.

Answer:
1. Substituting x = -5 in 2x² + px - 15 = 0 gives 2(25) - 5p - 15 = 0.
2. 50 - 5p - 15 = 0, so 35 = 5p and p = 7.
3. The second equation becomes 7x² + 7x + k = 0.
4. For equal roots, D = 7² - 4(7)(k) = 49 - 28k = 0.
5. 28k = 49, so k = \( \frac{7}{4} \).

Teacher's Note:
a) Find p first, then substitute it into the second equation.
b) Do not forget that p(x² + x) means 7 multiplied by both x² and x.

6. Find the value of k for which the quadratic equation (k - 12)x² + 2(k - 12)x + 2 = 0 has two real equal roots..

Answer:
1. a = k - 12, b = 2(k - 12), c = 2.
2. D = 4(k - 12)² - 8(k - 12) = 0.
3. 4(k - 12)(k - 12 - 2) = 0, so (k - 12)(k - 14) = 0.
4. k = 12 or k = 14.
5. For k = 12, the equation becomes 2 = 0, which is not a quadratic equation, so k = 14.

Teacher's Note:
a) Factorising the discriminant as 4(k - 12)(k - 14) gives both values quickly.
b) Always check that the coefficient of x² is not zero before accepting a value of k.

7. Find the value of k for which the quadratic equation k²x² - 2(k - 1)x + 4 = 0 has two real equal roots..

Answer:
1. a = k², b = -2(k - 1), c = 4.
2. D = 4(k - 1)² - 4(k²)(4) = 0.
3. 4(k - 1)² = 16k², so (k - 1)² = 4k².
4. k - 1 = 2k gives k = -1; k - 1 = -2k gives k = \( \frac{1}{3} \).
5. Hence, k = -1 or k = \( \frac{1}{3} \).

Teacher's Note:
a) Taking the square root of both sides gives two cases, and both must be solved.
b) Both values are valid because neither makes k² zero.

8. If the roots of the equation (a - b)x² + (b - c)x + (c - a) = 0 are equal, prove that b + c = 2a.

Answer:
1. Sum of the coefficients = (a - b) + (b - c) + (c - a) = 0.
2. So x = 1 is a root of the equation.
3. The roots are equal, so both roots equal 1, and their sum is 2.
4. Sum of roots = \( -\frac{(b - c)}{(a - b)} \) = 2.
5. -(b - c) = 2(a - b), so -b + c = 2a - 2b.
6. Hence, b + c = 2a.

Teacher's Note:
a) When the coefficients add up to zero, x = 1 is always a root.
b) The proof needs a - b not equal to 0, otherwise the equation is not quadratic.
c) Using the discriminant gives the same result but takes more algebra.

9. Prove that both the roots of the equation (x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0 are real but they are equal only when a = b = c.

Answer:
1. Expanding, x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca = 0.
2. So 3x² - 2(a + b + c)x + (ab + bc + ca) = 0.
3. D = 4(a + b + c)² - 12(ab + bc + ca).
4. D = 4(a² + b² + c² - ab - bc - ca) = 2[(a - b)² + (b - c)² + (c - a)²].
5. D is never negative, so both roots are real.
6. D = 0 only when a - b = 0, b - c = 0 and c - a = 0, that is a = b = c.

Teacher's Note:
a) Expand each product before adding the three terms.
b) The identity a² + b² + c² - ab - bc - ca = \( \frac{1}{2} \)[(a - b)² + (b - c)² + (c - a)²] is the key step.
c) Common mistake: forgetting that a square of a real number cannot be negative.

10. Find the positive value of k for which the equation x² + kx + 64 = 0 and x² - 8x + k = 0 will have real roots.

Answer:
1. For x² + kx + 64 = 0 to have real roots, k² - 256 ≥ 0, so k ≥ 16 or k ≤ -16.
2. For x² - 8x + k = 0 to have real roots, 64 - 4k ≥ 0, so k ≤ 16.
3. Both conditions hold together only for k = 16 or k ≤ -16 and k ≤ 16, that is k = 16 or k ≤ -16.
4. The positive value of k is 16.

Teacher's Note:
a) Each equation needs its own discriminant condition.
b) Common mistake: taking only k ≥ 16 and forgetting the second equation.
c) Check k = 16: the first equation gives 256 - 256 = 0 and the second gives 64 - 64 = 0.

11. Find the value of k for which the quadratic equation kx² - 6x - 2 = 0 has two real roots.

Answer:
1. a = k, b = -6, c = -2.
2. D = 36 - 4(k)(-2) = 36 + 8k.
3. For real roots, D ≥ 0, so 36 + 8k ≥ 0, that is k ≥ \( -\frac{9}{2} \).
4. The equation is quadratic only if k ≠ 0.
5. Hence, k ≥ \( -\frac{9}{2} \) and k ≠ 0.

Teacher's Note:
a) Real roots need D greater than or equal to 0, while equal roots need D = 0.
b) Remember the condition k ≠ 0, because k is the coefficient of x².

12. Find the value of k for which the quadratic equation 3x² + 2x + k = 0 has two real roots.

Answer:
1. a = 3, b = 2, c = k.
2. D = 2² - 4(3)(k) = 4 - 12k.
3. For real roots, 4 - 12k ≥ 0, so 12k ≤ 4.
4. k ≤ \( \frac{1}{3} \).

Teacher's Note:
a) The sign of 12k decides the result, so divide carefully.
b) Real roots include equal roots, so the value k = \( \frac{1}{3} \) is also accepted.

13. Find the value of k for which the quadratic equation 2x² + kx + 2 = 0 has two real roots.

Answer:
1. a = 2, b = k, c = 2.
2. D = k² - 4(2)(2) = k² - 16.
3. For real roots, k² - 16 ≥ 0, so k² ≥ 16.
4. k ≥ 4 or k ≤ -4.

Teacher's Note:
a) A square is at least 16 only when the number is 4 or more, or -4 or less.
b) Writing only k ≥ 4 misses the negative values.

14. Show that the equation 3x² + 7x + 8 = 0 is not true for any real value of x.

Answer:
1. a = 3, b = 7, c = 8.
2. D = b² - 4ac = 49 - 4(3)(8) = 49 - 96 = -47.
3. Since D is less than 0, the equation has no real roots.
4. Hence, the equation is not satisfied by any real value of x.

Teacher's Note:
a) A negative discriminant means no real roots.
b) Watch the sign when multiplying 4 by 3 and 8.

15. Show that the equation 2(a² + b²)x² + 2(a + b)x + 1 = 0 has no real roots, when a ≠ b.

Answer:
1. a = 2(a² + b²), b = 2(a + b), c = 1.
2. D = 4(a + b)² - 4(2)(a² + b²)(1) = 4(a + b)² - 8(a² + b²).
3. D = 4[a² + 2ab + b² - 2a² - 2b²] = 4[-a² + 2ab - b²].
4. D = -4(a - b)².
5. Since a ≠ b, (a - b)² is positive, so D is less than 0.
6. Hence, the equation has no real roots.

Teacher's Note:
a) The expression a² - 2ab + b² is (a - b)², which gives the key result.
b) If a = b, then D = 0 and the roots are real and equal, which is why a ≠ b is given.

16. Find the value of k for which the quadratic equation kx² + 2x + 1 = 0 has two real and distinct roots.

Answer:
1. a = k, b = 2, c = 1.
2. D = 2² - 4(k)(1) = 4 - 4k.
3. For distinct real roots, D greater than 0, so 4 - 4k greater than 0.
4. k less than 1.
5. The equation is quadratic only if k ≠ 0, so k less than 1 and k ≠ 0.

Teacher's Note:
a) Distinct roots need D strictly greater than 0.
b) Common mistake: forgetting to exclude k = 0.

17. Find the value of p for which the quadratic equation 2x² + px + 8 = 0 has two real and distinct roots.

Answer:
1. a = 2, b = p, c = 8.
2. D = p² - 4(2)(8) = p² - 64.
3. For distinct real roots, p² - 64 greater than 0, so p² greater than 64.
4. p greater than 8 or p less than -8.

Teacher's Note:
a) Both the positive and negative values of p must be included.
b) A useful check: p = 8 gives D = 0, which means equal roots, not distinct roots.

18. If the equation (1 + m²)x² + 2mcx + (c² - a²) = 0 has equal roots, prove that c² = a²(1 + m²).

Answer:
1. a = 1 + m², b = 2mc, c-term = c² - a².
2. D = (2mc)² - 4(1 + m²)(c² - a²) = 0.
3. 4m²c² - 4(c² - a² + m²c² - m²a²) = 0.
4. m²c² - c² + a² - m²c² + m²a² = 0.
5. -c² + a² + m²a² = 0, so c² = a²(1 + m²).

Teacher's Note:
a) The m²c² terms cancel, which makes the proof short.
b) Expand 4(1 + m²)(c² - a²) carefully, since the sign of each term matters.

19. If the roots of the equation (c² - ab)x² - 2(a² - bc)x + (b² - ac) = 0 are real and equal, show that either a = 0 or (a³ + b³ + c³) = 3abc.

Answer:
1. a = c² - ab, b = -2(a² - bc), c-term = b² - ac.
2. D = 4(a² - bc)² - 4(c² - ab)(b² - ac) = 0.
3. (a² - bc)² = (c² - ab)(b² - ac).
4. Left side: a⁴ - 2a²bc + b²c².
5. Right side: b²c² - ac³ - ab³ + a²bc.
6. a⁴ - 3a²bc + ac³ + ab³ = 0.
7. a(a³ + b³ + c³ - 3abc) = 0.
8. So a = 0 or a³ + b³ + c³ = 3abc.

Teacher's Note:
a) Equal roots give D = 0, which is the starting point of the proof.
b) Cancelling the common term a² bc on both sides is a common error; expand fully instead.
c) The identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) is useful here.
d) The factor a comes out of the whole expression, which explains the case a = 0.

20. Find the value of k for which the quadratic equation 9x² + 8kx + 16 = 0 has two real equal roots.

Answer:
1. a = 9, b = 8k, c = 16.
2. D = (8k)² - 4(9)(16) = 64k² - 576 = 0.
3. k² = 9, so k = ±3.

Teacher's Note:
a) Divide by 64 first to make the arithmetic easier.
b) Both k = 3 and k = -3 are correct.

21. Find the value of k for which the quadratic equation (k + 4)x² + (k + 1)x + 1 = 0 has two real equal roots.

Answer:
1. a = k + 4, b = k + 1, c = 1.
2. D = (k + 1)² - 4(k + 4)(1) = 0.
3. k² + 2k + 1 - 4k - 16 = 0, so k² - 2k - 15 = 0.
4. (k - 5)(k + 3) = 0, so k = 5 or k = -3.

Teacher's Note:
a) Expanding (k + 1)² correctly gives the 2k term.
b) Neither value makes k + 4 zero, so both values give quadratic equations.

22. Prove that the equation x²(a² + b²) + 2x(ac + bd) + (c² + d²) = 0 has no real root, if ad ≠ bc.

Answer:
1. a = a² + b², b = 2(ac + bd), c = c² + d².
2. D = 4(ac + bd)² - 4(a² + b²)(c² + d²).
3. D = 4[a²c² + 2abcd + b²d² - a²c² - a²d² - b²c² - b²d²].
4. D = 4[2abcd - a²d² - b²c²] = -4(ad - bc)².
5. Since ad ≠ bc, (ad - bc)² is positive, so D is less than 0.
6. Hence, the equation has no real roots.

Teacher's Note:
a) The result follows from the identity (ad - bc)² = a²d² - 2abcd + b²c².
b) The condition a² + b² ≠ 0 is needed for the equation to be quadratic.

23. If the roots of the equation x² + 2cx + ab = 0 are real unequal, prove that the equation x² - 2(a + b)x + a² + b² + 2c² = 0 has no real roots.

Answer:
1. For x² + 2cx + ab = 0, D = 4c² - 4ab, and real unequal roots mean D greater than 0, so c² greater than ab.
2. For x² - 2(a + b)x + a² + b² + 2c² = 0, D = 4(a + b)² - 4(a² + b² + 2c²).
3. D = 4(a² + 2ab + b² - a² - b² - 2c²) = 8(ab - c²).
4. Since c² is greater than ab, ab - c² is less than 0, so D is less than 0.
5. Hence, the second equation has no real roots.

Teacher's Note:
a) The printed second equation has no x after 2(a + b); the correct equation is x² - 2(a + b)x + a² + b² + 2c² = 0.
b) Real unequal roots give a strict inequality, which is used in the last step.

24. Find the positive values of k for which the equation x² + kx + 64 = 0 and x² - 8x + k = 0 will both have real roots.

Answer:
1. The first equation has real roots when k² - 256 ≥ 0, so k ≥ 16 or k ≤ -16.
2. The second equation has real roots when 64 - 4k ≥ 0, so k ≤ 16.
3. Both conditions are met only when k = 16 or k ≤ -16.
4. The positive value is k = 16.

Teacher's Note:
a) This is the same pair of equations as Question 10, and the answer is the same.
b) Only one positive value exists, so the plural "values" gives just k = 16.

25. Find the value of k for which the quadratic equation (k + 4)x² + (k + 1)x + 1 = 0 has equal roots.

Answer:
1. This is the same equation as Question 21.
2. D = (k + 1)² - 4(k + 4) = 0, so k² - 2k - 15 = 0.
3. (k - 5)(k + 3) = 0, so k = 5 or k = -3.

Teacher's Note:
a) Both values are accepted, and the working is identical to Question 21.
b) Check k = -3: the equation becomes x² - 2x + 1 = 0, which is (x - 1)² = 0.

26. Find the value of k for which the quadratic equation x² - 2(k + 1)x + k² = 0 has real and equal roots.

Answer:
1. a = 1, b = -2(k + 1), c = k².
2. D = 4(k + 1)² - 4k² = 0.
3. 4(k² + 2k + 1 - k²) = 0, so 8k + 4 = 0.
4. k = \( -\frac{1}{2} \).

Teacher's Note:
a) The k² terms cancel, leaving a linear equation in k.
b) Watch the sign: the middle term is -2(k + 1), so its square is 4(k + 1)².

27. Find the value of k for which the quadratic equation k²x² - 2(2k - 1)x + 4 = 0 has real and equal roots.

Answer:
1. a = k², b = -2(2k - 1), c = 4.
2. D = 4(2k - 1)² - 16k² = 0.
3. (2k - 1)² = 4k², so 4k² - 4k + 1 = 4k².
4. -4k + 1 = 0, so k = \( \frac{1}{4} \).

Teacher's Note:
a) The 4k² terms cancel, which leaves a simple linear equation.
b) k = 0 is not a solution here, since the value of k must be nonzero for the equation to be quadratic.

28. Find the value of k for which the quadratic equation (k + 1)x² - 2(k - 1)x + 1 = 0 has real and equal roots.

Answer:
1. a = k + 1, b = -2(k - 1), c = 1.
2. D = 4(k - 1)² - 4(k + 1) = 0.
3. (k - 1)² = k + 1, so k² - 3k = 0.
4. k(k - 3) = 0, so k = 0 or k = 3.

Teacher's Note:
a) Both k = 0 and k = 3 are valid, since k + 1 is not zero in either case.
b) The value k = -1 is excluded because it would make the equation linear.

29. Find the value of k for which the quadratic equation (4 - k)x² + (2k + 4)x + (8k + 1) = 0 has real and equal roots.

Answer:
1. a = 4 - k, b = 2k + 4, c = 8k + 1.
2. D = (2k + 4)² - 4(4 - k)(8k + 1) = 0.
3. 4k² + 16k + 16 - 4(-8k² + 31k + 4) = 0.
4. 36k² - 108k = 0, so 36k(k - 3) = 0.
5. k = 0 or k = 3.

Teacher's Note:
a) Expand (4 - k)(8k + 1) completely before subtracting.
b) Check k = 0: the equation becomes 4x² + 4x + 1 = 0, which is (2x + 1)² = 0.

30. Find the value of k for which the quadratic equation (2k + 1)x² + 2(k + 3)x + (k + 5) = 0 has real and equal roots.

Answer:
1. a = 2k + 1, b = 2(k + 3), c = k + 5.
2. D = 4(k + 3)² - 4(2k + 1)(k + 5) = 0.
3. (k + 3)² = (2k + 1)(k + 5).
4. k² + 6k + 9 = 2k² + 11k + 5.
5. k² + 5k - 4 = 0.
6. k = \( \frac{-5 \pm \sqrt{41}}{2} \).

Teacher's Note:
a) Bring every term to one side before using the quadratic formula.
b) Both roots, about 0.70 and -5.70, are valid because 2k + 1 is not zero for either.

Question. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number. 
Answer: 8

Question. By increasing the list price of a book by ₹ 10, a person can buy 10 books less for ₹ 1200. Find the original list price of the book. 
Answer: ₹ 30

Question. The hypotenuse of a right-angled triangle is \( 1\text{ cm} \) more than twice the shortest side. If the third side is \( 2\text{ cm} \) less than the hypotenuse, find the sides of the triangle. 
Answer: \( 8\text{ cm} \), \( 15\text{ cm} \), \( 17\text{ cm} \)

Question. A passenger train takes 2 hours less for a journey of \( 300\text{ km} \), if its speed is increased by \( 5\text{ km/hr} \) from its usual speed. Find its usual speed. 
Answer: \( 25\text{ km/hr} \)

Question. The numerator of a fraction is one less than its denominator. If three is added to each of the numerator and denominator, the fraction is increased by \( \frac{3}{28} \). Find the fraction.
Answer: \( \frac{3}{4} \)

Question. The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers. 
Answer: 9 and 6

Question. Solve: \( x^2 + 5\sqrt{5}x - 70 = 0 \). 
Answer: \( 2\sqrt{5} \); \( -7\sqrt{5} \)

Question. A train travels a distance of \( 300\text{ km} \) at a uniform speed. If the speed of the train is increased by \( 5\text{ km} \) an hour, the journey would have taken two hours less. Find the original speed of the train. 
Answer: \( 25\text{ km/hr} \)

Question. The speed of a boat in still water is \( 11\text{ km/hr} \). It can go \( 12\text{ km} \) upstream and returns downstream to the original point in 2 hours 45 minutes. Find the speed of the stream. 
Answer: \( 5\text{ km/hr} \)

Question. Using quadratic formula, solve the following equation for ‘\( x \)’: \( abx^2 + (b^2 - ac)x - bc = 0 \) 
Answer: \( x = \frac{c}{b} \), \( x = -\frac{b}{a} \)

Question. The sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \( \frac{1}{2} \). Find the fraction. 
Answer: \( \frac{5}{7} \)

Question. Rewrite the following as a quadratic equation in \( x \) and then solve for \( x \): \( \frac{4}{x} - 3 = \frac{5}{2x + 3}; x \neq 0, -\frac{3}{2} \) 
Answer: \( -2, 1 \)

Question. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number. 
Answer: 92

Question. A train covers a distance of \( 90\text{ km} \) at a uniform speed. Had the speed been \( 15\text{ km/hr} \) more, it would have taken 30 minutes less for the journey. Find the original speed of the train.  
Answer: \( 45\text{ km/hr} \)

Question. Solve for \( x \): \( \frac{x+1}{x-1} + \frac{x-2}{x+2} = 3; (x \neq 1, -2) \) 
Answer: \( x = -5, 2 \)

Question. Using quadratic formula, solve the following for \( x \): \( 9x^2 - 3(a^2 + b^2)x + a^2b^2 = 0 \) 
Answer: \( \frac{a^2}{3}, \frac{b^2}{3} \)

Question. Find the equation whose roots are reciprocal of the roots of \( 3x^2 - 5x + 7 = 0 \)
Answer: \( 7x^2 - 5x + 3 = 0 \)

Question. A number consists of two digits whose product is 18. When 27 is subtracted from the number, the digits change their places. Find the number. 
Answer: 63

Question. A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number. 
Answer: 63

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers.
Answer: 13 and 15

Question. An aeroplane left 30 minutes later than its scheduled time and in order to reach its destination \( 1500\text{ km} \) away in time, it has to increase its speed by \( 250\text{ km/hr} \) from its usual speed. Determine its usual speed.
Answer: \( 750\text{ km/h} \)

Question. Using quadratic formula, solve for \( x \): \( 9x^2 - 3(a + b)x + ab = 0 \) 
Answer: \( \frac{a}{3}, \frac{b}{3} \)

Question. Find the number which exceeds its positive square root by 20.
Answer: \( x = 25 \) (Rejecting \( x = 16 \))

Question. The sum of two numbers is 15 and the sum of their reciprocals is \( \frac{1}{3} \). Find the numbers. 
Answer: 5, 10

Question. A two digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number. 
Answer: 27

Question. A two digit number is such that the product of its digits is 20. If 9 is added to the number, the digits interchange their places. Find the number. 
Answer: 45

Question. A two digit number is such that the product of its digits is 15. If 8 is added to the number, the digits interchange their places. Find the number. 
Answer: 35

Question. Solve for \( x \): \( \frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3}; (x \neq 2, 4) \) 
Answer: \( \frac{5}{2}, 5 \)

Question. Solve for \( x \): \( abx^2 + (b^2 - ac)x - bc = 0 \) 
Answer: \( x = -\frac{b}{a}, \frac{c}{b} \)

Question. The sum of two numbers is 18. The sum of their reciprocals is \( \frac{1}{4} \). Find the numbers. 
Answer: 6, 12

Question. The sum of two numbers is 16, and the sum of their reciprocals is \( \frac{1}{3} \). Find the numbers. 
Answer: 12, 4

Question. The sum of two numbers ‘\( a \)’ and ‘\( b \)’ is 15, and sum of their reciprocals \( \frac{1}{a} \) and \( \frac{1}{b} \) is \( \frac{3}{10} \). Find the numbers ‘\( a \)’ and ‘\( b \)’. 
Answer: \( a = 5 \) or 10, \( b = 10 \) or 5

Question. Solve for \( x \): \( \frac{1}{a+b+x} = \frac{1}{a} + \frac{1}{b} + \frac{1}{x}; a \neq 0, b \neq 0, x \neq 0 \) 
Answer: \( x = -a, -b \)

Question. Find the roots of the following quadratic equation: \( \frac{2}{5}x^2 - x - \frac{3}{5} = 0 \) 
Answer: \( -\frac{1}{2} \) and \( x = 3 \)

Question. Find the roots of the equation: \( \frac{1}{2x-3} + \frac{1}{x-5} = 1; x \neq \frac{3}{2}, 5 \) 
Answer: \( 4, \frac{3\sqrt{2}}{2} \)

Question. A natural number when subtracted from 28, becomes equal to 160 times its reciprocal. Find the number. 
Answer: 8

Question. Find two consecutive odd positive integers, sum of whose squares is 290.
Answer: 11 and 13

Question. Solve for \( x \): \( \frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, -1. \) 
Answer: \( x = \pm 4 \)

Question. Solve for \( x \): \( \frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}; x \neq 3, 5 \) 
Answer: \( x = \frac{7}{2} \)