87 Quadratic Equation. 100 Important Quadratic Equations Practice Questions for SSC Class 10

Part 1: Top 10 Solved Quadratic Equations

Q1. Solve by factorization: \(x^2 + 5x + 6 = 0\)

Step 1: Find two numbers whose sum is 5 and product is 6. The numbers are 2 and 3. Step 2: \(x^2 + 2x + 3x + 6 = 0\) Step 3: \(x(x + 2) + 3(x + 2) = 0\) Step 4: \((x + 2)(x + 3) = 0\) Step 5: \(x + 2 = 0\) or \(x + 3 = 0\) Final Answer: \(x = -2, x = -3\)

Q2. Solve by factorization: \(x^2 - 3x - 10 = 0\)

Sum = -3, Product = -10. Numbers are -5 and 2. \(x^2 - 5x + 2x - 10 = 0\) \(x(x - 5) + 2(x - 5) = 0\) \((x - 5)(x + 2) = 0\) Final Answer: \(x = 5, x = -2\)

Q3. Solve using Formula Method: \(x^2 + 6x + 5 = 0\)

Comparing with \(ax^2 + bx + c = 0\), \(a=1, b=6, c=5\) Discriminant \((\Delta) = b^2 - 4ac = 6^2 - 4(1)(5) = 36 - 20 = 16\) Using formula: \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\) \(x = \frac{-6 \pm \sqrt{16}}{2} = \frac{-6 \pm 4}{2}\) \(x = \frac{-2}{2} = -1\) or \(x = \frac{-10}{2} = -5\) Final Answer: \(x = -1, -5\)

Q4. Solve: \(2y^2 + 27y + 13 = 0\)

Sum = 27, Product = \(2 \times 13 = 26\). Numbers: 26, 1. \(2y^2 + 26y + 1y + 13 = 0\) \(2y(y + 13) + 1(y + 13) = 0\) \((y + 13)(2y + 1) = 0\) Final Answer: \(y = -13, y = -1/2\)

Q5. Find the value of Discriminant for \(x^2 + 7x - 1 = 0\)

\(a = 1, b = 7, c = -1\) \(\Delta = b^2 - 4ac\) \(\Delta = (7)^2 - 4(1)(-1) = 49 + 4 = 53\) Final Answer: \(\Delta = 53\)

Q6. Determine the nature of roots for \(x^2 - 4x + 4 = 0\)

\(\Delta = (-4)^2 - 4(1)(4) = 16 - 16 = 0\) Since \(\Delta = 0\), the roots are real and equal. Final Answer: Real and Equal roots.

Q7. Form a quadratic equation whose roots are 3 and 10.

Let \(\alpha = 3, \beta = 10\) \(\alpha + \beta = 13\), \(\alpha\beta = 30\) Equation: \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\) Final Answer: \(x^2 - 13x + 30 = 0\)

Q8. Find \(k\) if \(x = 3\) is a root of \(kx^2 - 10x + 3 = 0\)

Substitute \(x = 3\) in the equation: \(k(3)^2 - 10(3) + 3 = 0\) \(9k - 30 + 3 = 0 \implies 9k - 27 = 0\) \(9k = 27 \implies k = 3\) Final Answer: \(k = 3\)

Q9. Solve: \(x^2 - 15x + 54 = 0\)

Numbers: -9, -6 (Sum -15, Product 54) \((x - 9)(x - 6) = 0\) Final Answer: \(x = 9, 6\)

Q10. Solve by completing the square: \(x^2 + x - 20 = 0\)

\(x^2 + x = 20\) Add \((\frac{1}{2})^2 = \frac{1}{4}\) to both sides. \(x^2 + x + \frac{1}{4} = 20 + \frac{1}{4} \implies (x + \frac{1}{2})^2 = \frac{81}{4}\) \(x + \frac{1}{2} = \pm \frac{9}{2}\) \(x = \frac{8}{2} = 4\) or \(x = \frac{-10}{2} = -5\) Final Answer: \(x = 4, -5\)

Part 2: Practice Questions (Q11 - Q100)

Solve the following Quadratic Equations (Factorization/Formula):

11. \(x^2 - 4x - 5 = 0\)
12. \(x^2 + 8x + 15 = 0\)
13. \(x^2 - 7x + 12 = 0\)
14. \(2x^2 - 5x + 2 = 0\)
15. \(3x^2 - x - 10 = 0\)
16. \(x^2 - 11x + 24 = 0\)
17. \(x^2 + 2x - 48 = 0\)
18. \(5m^2 = 22m + 15\)
19. \(2x^2 - 2x + \frac{1}{2} = 0\)
20. \(6x - \frac{2}{x} = 1\)
21. \(x^2 - 25 = 0\)
22. \(3y^2 = 15y\)
23. \(x^2 + 4x + 1 = 0\)
24. \(m^2 - 5m - 3 = 0\)
25. \(x^2 + 5x + 5 = 0\)
26. \(y^2 + \frac{1}{3}y = 2\)
27. \(5x^2 + 13x + 8 = 0\)
28. \(x^2 + 10x + 24 = 0\)
29. \(x^2 - x - 72 = 0\)
30. \(x^2 - 16x + 63 = 0\)
31. \(2x^2 + 9x + 10 = 0\)
32. \(x^2 - 2x - 3 = 0\)
33. \(x^2 + 6x + 9 = 0\)
34. \(x^2 - 10x + 25 = 0\)
35. \(x^2 + 14x + 49 = 0\)
36. \(4x^2 - 4x + 1 = 0\)
37. \(x^2 - 1 = 0\)
38. \(2x^2 - 7x + 6 = 0\)
39. \(3x^2 + 8x + 5 = 0\)
40. \(x^2 - 12x + 32 = 0\)

Find the Discriminant and State Nature of Roots:

41. \(x^2 + x + 1 = 0\)
42. \(2x^2 - 5x - 3 = 0\)
43. \(x^2 - 6x + 9 = 0\)
44. \(3x^2 + 2x - 1 = 0\)
45. \(x^2 + 4x + 5 = 0\)
46. \(2x^2 - 7x + 3 = 0\)
47. \(x^2 - 2x + 1 = 0\)
48. \(x^2 + 5x + 6 = 0\)
49. \(4x^2 - 12x + 9 = 0\)
50. \(x^2 - 8x + 15 = 0\)
51. \(2x^2 + 5x + 5 = 0\)
52. \(x^2 - x - 1 = 0\)
53. \(x^2 + 10x + 25 = 0\)
54. \(3x^2 + 7x + 2 = 0\)
55. \(x^2 + 2x + 3 = 0\)
56. \(5x^2 - 4x + 1 = 0\)
57. \(x^2 - 4x + 3 = 0\)
58. \(2x^2 - 6x + 3 = 0\)
59. \(x^2 + 12x + 36 = 0\)
60. \(x^2 - 5x + 7 = 0\)

Form Quadratic Equations from Roots:

61. Roots: 2, 5
62. Roots: -3, -4
63. Roots: 0, 4
64. Roots: 1/2, 1/2
65. Roots: \(\sqrt{2}, -\sqrt{2}\)
66. Roots: 7, -7
67. Roots: 6, 1
68. Roots: -5, 2
69. Roots: 8, 3
70. Roots: -1, -1
71. Roots: 10, -2
72. Roots: 0, 0
73. Roots: 5, 5
74. Roots: -6, 6
75. Roots: 4, -3
76. Roots: 1, 9
77. Roots: -2, -8
78. Roots: 3, -3
79. Roots: 1/3, 3
80. Roots: -10, -10

Advanced and Word-Based Conditions:

81. Find \(k\) if roots of \(x^2 + kx + 12 = 0\) are real and equal.
82. Find \(k\) if one root of \(x^2 - kx + 18 = 0\) is 6.
83. Sum of roots is 10 and product is 21. Find equation.
84. One root is \(2 + \sqrt{3}\), find the other root.
85. Solve \(x^4 - 5x^2 + 4 = 0\) (Reducible to quadratic).
86. Solve \((x-3)(x+4) = 0\).
87. Solve \(x^2 = 49\).
88. Solve \(5x^2 = 20\).
89. Find \(k\) if \(\Delta = 0\) for \(kx(x-2) + 6 = 0\).
90. Solve \(x + 1/x = 2.5\).
91. Product of two consecutive natural numbers is 20.
92. Find \(x\) if \(x^2 - 9x + 20 = 0\).
93. Solve \(y^2 + 10y + 21 = 0\).
94. Solve \(x^2 - 11x + 30 = 0\).
95. Solve \(x^2 - 2x - 8 = 0\).
96. Roots are 1 and -1. Form equation.
97. Roots are 4 and 0. Form equation.
98. Find \(\Delta\) for \(x^2 + 5x + 5 = 0\).
99. Solve \(x^2 - 3x = 0\).
100. Solve \(2x^2 = 8\).

Answer Key (Q11 - Q100)

11. 5, -1

12. -3, -5

13. 4, 3

14. 2, 0.5

15. 2, -5/3

16. 8, 3

17. 6, -8

18. 5, -3/5

19. 0.5, 0.5

20. 2/3, -1/2

21. 5, -5

22. 0, 5

23. \(-2 \pm \sqrt{3}\)

24. \(\frac{5 \pm \sqrt{37}}{2}\)

25. \(\frac{-5 \pm \sqrt{5}}{2}\)

26. 4/3, -3/2

27. -1, -1.6

28. -4, -6

29. 9, -8

30. 9, 7

31. -2, -2.5

32. 3, -1

33. -3, -3

34. 5, 5

35. -7, -7

36. 0.5, 0.5

37. 1, -1

38. 2, 1.5

39. -1, -5/3

40. 8, 4

41. -3 (Not Real)

42. 49 (Real, Uniq)

43. 0 (Real, Equal)

44. 16 (Real, Uniq)

45. -4 (Not Real)

46. 25 (Real, Uniq)

47. 0 (Real, Equal)

48. 1 (Real, Uniq)

49. 0 (Real, Equal)

50. 4 (Real, Uniq)

51. -15 (Not Real)

52. 5 (Real, Uniq)

53. 0 (Real, Equal)

54. 25 (Real, Uniq)

55. -8 (Not Real)

56. -4 (Not Real)

57. 4 (Real, Uniq)

58. 12 (Real, Uniq)

59. 0 (Real, Equal)

60. -3 (Not Real)

61. \(x^2-7x+10=0\)

62. \(x^2+7x+12=0\)

63. \(x^2-4x=0\)

64. \(4x^2-4x+1=0\)

65. \(x^2-2=0\)

66. \(x^2-49=0\)

67. \(x^2-7x+6=0\)

68. \(x^2+3x-10=0\)

69. \(x^2-11x+24=0\)

70. \(x^2+2x+1=0\)

71. \(x^2-8x-20=0\)

72. \(x^2=0\)

73. \(x^2-10x+25=0\)

74. \(x^2-36=0\)

75. \(x^2-x-12=0\)

76. \(x^2-10x+9=0\)

77. \(x^2+10x+16=0\)

78. \(x^2-9=0\)

79. \(3x^2-10x+3=0\)

80. \(x^2+20x+100=0\)

81. \(k = \pm \sqrt{48}\)

82. \(k = 9\)

83. \(x^2-10x+21=0\)

84. \(2-\sqrt{3}\)

85. \(\pm 1, \pm 2\)

86. 3, -4

87. 7, -7

88. 2, -2

89. \(k = 6\)

90. 2, 0.5

91. 4, 5

92. 4, 5

93. -3, -7

94. 6, 5

95. 4, -2

96. \(x^2-1=0\)

97. \(x^2-4x=0\)

98. 5

99. 0, 3

100. 2, -2

10 Important Quadratic Equations Questions and Solutions for Class 10

10 Quadratic Equations Questions with Solution

Solutions

Question 1:

Solve the quadratic equation $x^2 - 5x + 6 = 0$ by factorization method.

Solution:

Given equation: $x^2 - 5x + 6 = 0$

To factorize, we look for two numbers whose sum is $-5$ and product is $6$. These numbers are $-2$ and $-3$.

$x^2 - 2x - 3x + 6 = 0$

$x(x - 2) - 3(x - 2) = 0$

$(x - 2)(x - 3) = 0$

Therefore, $x - 2 = 0$ or $x - 3 = 0$

$x = 2$ or $x = 3$

Question 2:

Solve the following equation using the Quadratic Formula: $2x^2 + 7x + 5 = 0$.

Solution:

Comparing $2x^2 + 7x + 5 = 0$ with $ax^2 + bx + c = 0$, we get:

$a = 2, b = 7, c = 5$

Quadratic Formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(5)}}{2(2)}$

$x = \frac{-7 \pm \sqrt{49 - 40}}{4}$

$x = \frac{-7 \pm \sqrt{9}}{4} = \frac{-7 \pm 3}{4}$

Case 1: $x = \frac{-7 + 3}{4} = \frac{-4}{4} = -1$

Case 2: $x = \frac{-7 - 3}{4} = \frac{-10}{4} = -2.5$

Roots: $x = -1, -2.5$

Question 3:

Determine the nature of the roots of the quadratic equation: $3x^2 - 4x + 1 = 0$.

Solution:

Here, $a = 3, b = -4, c = 1$

Discriminant ($D$) $= b^2 - 4ac$

$D = (-4)^2 - 4(3)(1)$

$D = 16 - 12 = 4$

Since $D > 0$ and $D$ is a perfect square, the roots are real, rational, and unequal.

Question 4:

Solve $x^2 + 6x + 9 = 0$.

Solution:

Given: $x^2 + 6x + 9 = 0$

This is in the form of $(a + b)^2 = a^2 + 2ab + b^2$.

$(x)^2 + 2(x)(3) + (3)^2 = 0$

$(x + 3)^2 = 0$

$x + 3 = 0$

$x = -3, -3$ (Equal roots)

Question 5:

Find the value of $k$ if one root of the quadratic equation $kx^2 - 14x + 8 = 0$ is $2$.

Solution:

Since $x = 2$ is a root, it must satisfy the equation.

$k(2)^2 - 14(2) + 8 = 0$

$4k - 28 + 8 = 0$

$4k - 20 = 0$

$4k = 20$

$k = 5$

Question 6:

The sum of two numbers is 15 and the sum of their reciprocals is $3/10$. Find the numbers.

Solution:

Let the numbers be $x$ and $15 - x$.

According to the condition: $\frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10}$

$\frac{15 - x + x}{x(15 - x)} = \frac{3}{10}$

$\frac{15}{15x - x^2} = \frac{3}{10}$

$150 = 3(15x - x^2)$

$50 = 15x - x^2$ (dividing by 3)

$x^2 - 15x + 50 = 0$

$(x - 10)(x - 5) = 0$

The numbers are 10 and 5.

Question 7:

Solve: $x^2 - 2x - 15 = 0$

Solution:

$x^2 - 5x + 3x - 15 = 0$

$x(x - 5) + 3(x - 5) = 0$

$(x - 5)(x + 3) = 0$

$x = 5$ or $x = -3$

Question 8:

Form a quadratic equation whose roots are $4$ and $-3$.

Solution:

Sum of roots ($\alpha + \beta$) $= 4 + (-3) = 1$

Product of roots ($\alpha\beta$) $= 4 \times (-3) = -12$

Equation: $x^2 - (\text{Sum})x + (\text{Product}) = 0$

$x^2 - (1)x + (-12) = 0$

$x^2 - x - 12 = 0$

Question 9:

Solve $4x^2 - 20x + 25 = 0$ using factorization.

Solution:

$4x^2 - 10x - 10x + 25 = 0$

$2x(2x - 5) - 5(2x - 5) = 0$

$(2x - 5)(2x - 5) = 0$

$2x - 5 = 0 \Rightarrow 2x = 5$

$x = 5/2$

Question 10:

Solve: $x + \frac{1}{x} = 2.5$

Solution:

Multiply the whole equation by $x$:

$x^2 + 1 = 2.5x$

$x^2 - 2.5x + 1 = 0$

Multiply by 2 to remove decimals: $2x^2 - 5x + 2 = 0$

$2x^2 - 4x - x + 2 = 0$

$2x(x - 2) - 1(x - 2) = 0$

$(x - 2)(2x - 1) = 0$

$x = 2$ or $x = 1/2$

10 Quadratic Equation Questions Important for Board Exam

Part 1: 10 Important Solved Questions

SSC Class 10 Algebra Quadratic Equations Important Solved Questions and Practice Set

Question 1: Factorization Method

Solve the quadratic equation by factorization: \(x^2 - 15x + 54 = 0\)

Solution:
Given: \(x^2 - 15x + 54 = 0\)
We need to find two numbers whose sum is \(-15\) and product is \(54\).
The numbers are \(-9\) and \(-6\).
\(x^2 - 9x - 6x + 54 = 0\)
\(x(x - 9) - 6(x - 9) = 0\)
\((x - 9)(x - 6) = 0\)
\(x - 9 = 0\) or \(x - 6 = 0\)
\(x = 9\) or \(x = 6\)
Roots: 9, 6

Question 2: Formula Method

Solve using formula: \(x^2 + 6x + 5 = 0\)

Solution:
Comparing with \(ax^2 + bx + c = 0\), we get:
\(a = 1, b = 6, c = 5\)
Discriminant \(\Delta = b^2 - 4ac = (6)^2 - 4(1)(5) = 36 - 20 = 16\)
Using formula \(x = \frac{-b \pm \sqrt{\Delta}}{2a}\):
\(x = \frac{-6 \pm \sqrt{16}}{2(1)} = \frac{-6 \pm 4}{2}\)
\(x = \frac{-6 + 4}{2} = \frac{-2}{2} = -1\) or \(x = \frac{-6 - 4}{2} = \frac{-10}{2} = -5\)
Roots: -1, -5

Question 3: Nature of Roots

Determine the nature of roots for \(2x^2 - 5x + 7 = 0\)

Solution:
\(a = 2, b = -5, c = 7\)
\(\Delta = b^2 - 4ac = (-5)^2 - 4(2)(7) = 25 - 56 = -31\)
Since \(\Delta < 0\), the roots are not real.

Question 4: Finding 'k'

Find \(k\) if the roots of \(kx(x - 2) + 6 = 0\) are real and equal.

Solution:
Equation: \(kx^2 - 2kx + 6 = 0\)
For real and equal roots, \(\Delta = 0\).
\(a = k, b = -2k, c = 6\)
\((-2k)^2 - 4(k)(6) = 0\)
\(4k^2 - 24k = 0\)
\(4k(k - 6) = 0\)
Since \(k \neq 0\) (as it is a quadratic equation), \(k - 6 = 0 \implies \mathbf{k = 6}\).

Question 5: Forming Equation

Form a quadratic equation whose roots are 3 and -10.

Solution:
Let \(\alpha = 3\) and \(\beta = -10\).
Sum of roots \(\alpha + \beta = 3 + (-10) = -7\)
Product of roots \(\alpha\beta = 3 \times (-10) = -30\)
Equation: \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
\(x^2 - (-7)x + (-30) = 0\)
Equation: \(x^2 + 7x - 30 = 0\)

Question 6: Sum and Product Relation

If \(\alpha\) and \(\beta\) are roots of \(x^2 + 5x - 1 = 0\), find \(\alpha^3 + \beta^3\).

Solution:
\(\alpha + \beta = -b/a = -5\), \(\alpha\beta = c/a = -1\)
\(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
\(= (-5)^3 - 3(-1)(-5)\)
\(= -125 - 15 = -140\)
Value: -140

Question 7: Word Problem (Numbers)

Product of Pragati's age 2 years ago and 3 years hence is 84. Find her present age.

Solution:
Let present age be \(x\).
\((x - 2)(x + 3) = 84\)
\(x^2 + 3x - 2x - 6 = 84\)
\(x^2 + x - 90 = 0\)
\((x + 10)(x - 9) = 0\)
\(x = -10\) (Discarded) or \(x = 9\).
Present age: 9 years

Question 8: Completing the Square

Solve \(x^2 + x - 20 = 0\) by completing the square method.

Solution:
\(x^2 + x = 20\)
Add \((1/2 \times \text{coeff of } x)^2 = (1/2)^2 = 1/4\) to both sides:
\(x^2 + x + 1/4 = 20 + 1/4\)
\((x + 1/2)^2 = 81/4\)
Taking square root: \(x + 1/2 = \pm 9/2\)
\(x = 9/2 - 1/2 = 4\) or \(x = -9/2 - 1/2 = -5\)
Roots: 4, -5

Question 9: Alpha/Beta Expression

If roots of \(x^2 - px + q = 0\) differ by 1, prove \(p^2 = 4q + 1\).

Solution:
Let roots be \(\alpha, \beta\). Given \(|\alpha - \beta| = 1\).
\(\alpha + \beta = p, \alpha\beta = q\).
We know \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
\(1^2 = p^2 - 4q\)
\(1 = p^2 - 4q \implies \mathbf{p^2 = 4q + 1}\). (Hence Proved)

Question 10: Speed/Distance

A train travels 360 km at uniform speed. If speed was 5 km/hr more, it would take 1 hr less. Find the speed.

Solution:
Let original speed be \(x\) km/hr. Time \(T_1 = 360/x\).
New speed \(x + 5\). Time \(T_2 = 360/(x + 5)\).
\(T_1 - T_2 = 1\)
\(360/x - 360/(x + 5) = 1\)
\(360(x + 5 - x) = x(x + 5)\)
\(1800 = x^2 + 5x \implies x^2 + 5x - 1800 = 0\)
\((x + 45)(x - 40) = 0\)
Speed cannot be negative, so \(x = 40\).
Speed: 40 km/hr

Part 2: 50 Practice Questions

1. \(x^2 - 4x + 3 = 0\)

2. \(x^2 + 7x + 10 = 0\)

3. \(x^2 - 5x + 6 = 0\)

4. \(x^2 - 9 = 0\)

5. \(2x^2 - 7x + 3 = 0\)

6. \(x^2 - 10x + 25 = 0\)

7. \(3x^2 - 5x + 2 = 0\)

8. \(x^2 + 2x - 8 = 0\)

9. \(x^2 - 1 = 0\)

10. \(x^2 - 6x + 8 = 0\)

11. \(x^2 + 5x + 4 = 0\)

12. \(2x^2 + x - 6 = 0\)

13. \(x^2 - 3x - 10 = 0\)

14. \(x^2 - 11x + 30 = 0\)

15. \(x^2 + 8x + 15 = 0\)

16. \(x^2 - 13x + 40 = 0\)

17. \(x^2 - 4 = 0\)

18. \(5x^2 - 18x - 8 = 0\)

19. \(x^2 + 4x + 4 = 0\)

20. \(x^2 - 12x + 32 = 0\)

21. \(2x^2 - 5x - 3 = 0\)

22. \(x^2 - 2x - 15 = 0\)

23. \(x^2 + 9x + 20 = 0\)

24. \(x^2 - 7x + 12 = 0\)

25. \(3x^2 - 10x + 3 = 0\)

26. \(x^2 - 14x + 49 = 0\)

27. \(x^2 - 16 = 0\)

28. \(2x^2 + 7x + 5 = 0\)

29. \(x^2 - x - 12 = 0\)

30. \(x^2 - 15x + 56 = 0\)

31. \(x^2 + 10x + 21 = 0\)

32. \(x^2 - 8x + 12 = 0\)

33. \(4x^2 - 4x + 1 = 0\)

34. \(x^2 - 25 = 0\)

35. \(x^2 + x - 6 = 0\)

36. \(x^2 - 5x - 14 = 0\)

37. \(x^2 - 11x + 28 = 0\)

38. \(x^2 + 6x + 9 = 0\)

39. \(2x^2 - 3x + 1 = 0\)

40. \(x^2 - 100 = 0\)

41. \(x^2 - 17x + 72 = 0\)

42. \(x^2 + 12x + 35 = 0\)

43. \(x^2 - 2x + 1 = 0\)

44. \(3x^2 + 4x + 1 = 0\)

45. \(x^2 - 9x + 20 = 0\)

46. \(x^2 - 36 = 0\)

47. \(x^2 + 2x - 15 = 0\)

48. \(x^2 - 13x + 42 = 0\)

49. \(x^2 - 64 = 0\)

50. \(2x^2 - x - 1 = 0\)

🔑 Answer Key

1: (3, 1)
2: (-2, -5)
3: (2, 3)
4: (3, -3)
5: (3, 0.5)
6: (5, 5)
7: (1, 2/3)
8: (2, -4)
9: (1, -1)
10: (4, 2)
11: (-1, -4)
12: (1.5, -2)
13: (5, -2)
14: (5, 6)
15: (-3, -5)
16: (5, 8)
17: (2, -2)
18: (4, -0.4)
19: (-2, -2)
20: (4, 8)
21: (3, -0.5)
22: (5, -3)
23: (-4, -5)
24: (3, 4)
25: (3, 1/3)
26: (7, 7)
27: (4, -4)
28: (-1, -2.5)
29: (4, -3)
30: (7, 8)
31: (-3, -7)
32: (2, 6)
33: (0.5, 0.5)
34: (5, -5)
35: (2, -3)
36: (7, -2)
37: (4, 7)
38: (-3, -3)
39: (1, 0.5)
40: (10, -10)
41: (8, 9)
42: (-5, -7)
43: (1, 1)
44: (-1/3, -1)
45: (4, 5)
46: (6, -6)
47: (3, -5)
48: (6, 7)
49: (8, -8)
50: (1, -0.5)

25 Solved Quadratic Equations: Step-by-Step Examples & Practice

Mastering Quadratic Equations

What you will learn

A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is \( ax^2 + bx + c = 0 \). Below are 25 fully solved examples to help you understand factorization and formula methods, followed by 10 practice problems to test your skills.

Note: Try to solve the equation yourself first, then click "Show Solution" to verify your method.

25 Solved Examples

01 Solve: \( x^2 + 5x + 6 = 0 \)

View Step-by-Step Solution

(i) Identify coefficients Here, \(a=1, b=5, c=6\). We need two numbers that multiply to 6 and add to 5.

(ii) Factorize The numbers are 2 and 3. \[ x^2 + 2x + 3x + 6 = 0 \] \[ x(x + 2) + 3(x + 2) = 0 \] \[ (x + 2)(x + 3) = 0 \]
(iii) Find Roots Either \(x+2=0\) or \(x+3=0\).
Answer: \( x = -2, x = -3 \)

02 Solve: \( x^2 - 5x + 6 = 0 \)

View Step-by-Step Solution

(i) Find factors We need numbers that multiply to \(+6\) and add to \(-5\). These are \(-2\) and \(-3\).

(ii) Split the middle term \[ x^2 - 2x - 3x + 6 = 0 \] \[ x(x - 2) - 3(x - 2) = 0 \] \[ (x - 2)(x - 3) = 0 \]
(iii) Solve Answer: \( x = 2, x = 3 \)

03 Solve: \( x^2 - 9 = 0 \)

View Step-by-Step Solution

(i) Use identity Recall \( a^2 - b^2 = (a+b)(a-b) \). Here \( 9 = 3^2 \).

(ii) Factorize \[ x^2 - 3^2 = 0 \] \[ (x + 3)(x - 3) = 0 \]
(iii) Solve Answer: \( x = -3, x = 3 \)

04 Solve: \( x^2 + 7x + 12 = 0 \)

View Step-by-Step Solution

(i) Find factors Product = 12, Sum = 7. Factors are 3 and 4.

(ii) Factorize \[ (x + 3)(x + 4) = 0 \]
(iii) Solve Answer: \( x = -3, x = -4 \)

05 Solve: \( 2x^2 + 3x + 1 = 0 \)

View Step-by-Step Solution

(i) AC Method Multiply \(a \times c = 2 \times 1 = 2\). We need sum = 3. Factors are 2 and 1.

(ii) Split middle term \[ 2x^2 + 2x + x + 1 = 0 \] \[ 2x(x + 1) + 1(x + 1) = 0 \] \[ (2x + 1)(x + 1) = 0 \]
(iii) Solve \( 2x = -1 \rightarrow x = -1/2 \) or \( x = -1 \).
Answer: \( x = -\frac{1}{2}, x = -1 \)

06 Solve: \( x^2 - 2x - 15 = 0 \)

View Step-by-Step Solution

(i) Factors Product = -15, Sum = -2. Factors: -5 and +3.

(ii) Factorize \[ (x - 5)(x + 3) = 0 \]
(iii) Solve Answer: \( x = 5, x = -3 \)

07 Solve: \( x^2 - 8x + 16 = 0 \)

View Step-by-Step Solution

(i) Analyze structure This is a perfect square trinomial because \( (-4)^2 = 16 \) and \( 2(-4) = -8 \).

(ii) Factorize \[ (x - 4)^2 = 0 \]
(iii) Solve Answer: \( x = 4 \) (Equal real roots)

08 Solve: \( 3x^2 - 5x + 2 = 0 \)

View Step-by-Step Solution

(i) AC Method \( a \times c = 3 \times 2 = 6 \). Sum = -5. Factors: -3 and -2.

(ii) Split middle term \[ 3x^2 - 3x - 2x + 2 = 0 \] \[ 3x(x - 1) - 2(x - 1) = 0 \] \[ (3x - 2)(x - 1) = 0 \]
(iii) Solve Answer: \( x = \frac{2}{3}, x = 1 \)

09 Solve: \( x^2 + 4x = 0 \)

View Step-by-Step Solution

(i) Take common factor There is no constant term \(c\). Take \(x\) common.

(ii) Factorize \[ x(x + 4) = 0 \]
(iii) Solve Answer: \( x = 0, x = -4 \)

10 Solve: \( 2x^2 - 7x + 3 = 0 \)

View Step-by-Step Solution

(i) AC Method Product = 6, Sum = -7. Factors: -6 and -1.

(ii) Split middle term \[ 2x^2 - 6x - x + 3 = 0 \] \[ 2x(x - 3) - 1(x - 3) = 0 \] \[ (2x - 1)(x - 3) = 0 \]
(iii) Solve Answer: \( x = \frac{1}{2}, x = 3 \)

11 Solve: \( x^2 - 4x - 21 = 0 \)

View Step-by-Step Solution

(i) Factors Product = -21, Sum = -4. Factors: -7 and +3.

(ii) Factorize \[ (x - 7)(x + 3) = 0 \]
(iii) Solve Answer: \( x = 7, x = -3 \)

12 Solve using Formula: \( x^2 + 4x + 2 = 0 \)

View Step-by-Step Solution

(i) Quadratic Formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \(a=1, b=4, c=2\).

(ii) Substitute \[ x = \frac{-4 \pm \sqrt{16 - 8}}{2} \] \[ x = \frac{-4 \pm \sqrt{8}}{2} \] \[ x = \frac{-4 \pm 2\sqrt{2}}{2} \]
(iii) Simplify Answer: \( x = -2 \pm \sqrt{2} \)

13 Solve: \( 6x^2 - x - 2 = 0 \)

View Step-by-Step Solution

(i) AC Method Product = -12, Sum = -1. Factors: -4 and +3.

(ii) Split middle term \[ 6x^2 - 4x + 3x - 2 = 0 \] \[ 2x(3x - 2) + 1(3x - 2) = 0 \] \[ (2x + 1)(3x - 2) = 0 \]
(iii) Solve Answer: \( x = -\frac{1}{2}, x = \frac{2}{3} \)

14 Solve: \( 4x^2 - 12x + 9 = 0 \)

View Step-by-Step Solution

(i) Identify square \( (2x)^2 - 2(2x)(3) + 3^2 = 0 \).

(ii) Factorize \[ (2x - 3)^2 = 0 \]
(iii) Solve Answer: \( x = \frac{3}{2} \) (Repeated root)

15 Solve: \( x^2 + x - 2 = 0 \)

View Step-by-Step Solution

(i) Factors Product = -2, Sum = +1. Factors: +2 and -1.

(ii) Factorize \[ (x + 2)(x - 1) = 0 \]
(iii) Solve Answer: \( x = -2, x = 1 \)

16 Solve: \( 5x^2 = 20 \)

View Step-by-Step Solution

(i) Isolate x squared Divide by 5: \( x^2 = 4 \).

(ii) Square root \( x = \pm\sqrt{4} \).
(iii) Solve Answer: \( x = 2, x = -2 \)

17 Solve: \( x^2 - 11x + 24 = 0 \)

View Step-by-Step Solution

(i) Factors Product = 24, Sum = -11. Factors: -8 and -3.

(ii) Factorize \[ (x - 8)(x - 3) = 0 \]
(iii) Solve Answer: \( x = 8, x = 3 \)

18 Solve: \( x^2 + 10x + 25 = 0 \)

View Step-by-Step Solution

(i) Perfect Square This fits \( (a+b)^2 \).

(ii) Factorize \[ (x + 5)^2 = 0 \]
(iii) Solve Answer: \( x = -5 \)

19 Solve: \( 2x^2 + 5x - 3 = 0 \)

View Step-by-Step Solution

(i) AC Method Product = -6, Sum = 5. Factors: +6 and -1.

(ii) Split middle term \[ 2x^2 + 6x - x - 3 = 0 \] \[ 2x(x + 3) - 1(x + 3) = 0 \] \[ (2x - 1)(x + 3) = 0 \]
(iii) Solve Answer: \( x = \frac{1}{2}, x = -3 \)

20 Solve using Formula: \( x^2 - 6x + 7 = 0 \)

View Step-by-Step Solution

(i) Apply Formula \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(7)}}{2(1)} \).

(ii) Calculate \[ x = \frac{6 \pm \sqrt{36 - 28}}{2} \] \[ x = \frac{6 \pm \sqrt{8}}{2} \] \[ x = \frac{6 \pm 2\sqrt{2}}{2} \]
(iii) Simplify Answer: \( x = 3 \pm \sqrt{2} \)

21 Solve: \( x^2 + 3x - 10 = 0 \)

View Step-by-Step Solution

(i) Factors Product = -10, Sum = 3. Factors: +5 and -2.

(ii) Factorize \[ (x + 5)(x - 2) = 0 \]
(iii) Solve Answer: \( x = -5, x = 2 \)

22 Solve: \( 3x^2 = 2x \)

View Step-by-Step Solution

(i) Rearrange \[ 3x^2 - 2x = 0 \]
(ii) Factorize common term \[ x(3x - 2) = 0 \]
(iii) Solve \( x = 0 \) or \( 3x - 2 = 0 \).
Answer: \( x = 0, x = \frac{2}{3} \)

23 Solve: \( x^2 - 13x + 42 = 0 \)

View Step-by-Step Solution

(i) Factors Product = 42, Sum = -13. Factors: -6 and -7.

(ii) Factorize \[ (x - 6)(x - 7) = 0 \]
(iii) Solve Answer: \( x = 6, x = 7 \)

24 Solve: \( 4x^2 - 1 = 0 \)

View Step-by-Step Solution

(i) Difference of Squares \( (2x)^2 - 1^2 = 0 \).

(ii) Factorize \[ (2x - 1)(2x + 1) = 0 \]
(iii) Solve Answer: \( x = \frac{1}{2}, x = -\frac{1}{2} \)

25 Solve: \( x^2 + 8x + 15 = 0 \)

View Step-by-Step Solution

(i) Factors Product = 15, Sum = 8. Factors: 3 and 5.

(ii) Factorize \[ (x + 3)(x + 5) = 0 \]
(iii) Solve Answer: \( x = -3, x = -5 \)

Practice Questions

Try solving these 10 questions on your own before checking the answer key below.

(i) \( x^2 + 7x + 10 = 0 \)

(ii) \( x^2 - 3x - 10 = 0 \)

(iii) \( 2x^2 + 5x + 3 = 0 \)

(iv) \( x^2 - 49 = 0 \)

(v) \( x^2 - 6x = 0 \)

(vi) \( x^2 + 12x + 36 = 0 \)

(vii) \( 3x^2 - x - 4 = 0 \)

(viii) \( x^2 - x - 30 = 0 \)

(ix) \( x^2 - 10x + 21 = 0 \)

(x) \( 2x^2 + 7x - 4 = 0 \)

Check Answer Key

(i) \( x = -2, x = -5 \)

(ii) \( x = 5, x = -2 \)

(iii) \( x = -1, x = -3/2 \)

(iv) \( x = 7, x = -7 \)

(v) \( x = 0, x = 6 \)

(vi) \( x = -6 \)

(vii) \( x = -1, x = 4/3 \)

(viii) \( x = 6, x = -5 \)

(ix) \( x = 3, x = 7 \)

(x) \( x = 1/2, x = -4 \)

6 Quadratic Equation

Part 1: Basic Concepts of Quadratic Equations

A Quadratic Equation is a polynomial equation of the second degree. The standard form is:

$$ax^2 + bx + c = 0$$

Where $a, b, c$ are real numbers and $a \neq 0$.

Key Methods to Solve

• Factorization: Splitting the middle term.
• Quadratic Formula: Used when factorization is difficult.

The Quadratic Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, the term $D = b^2 - 4ac$ is called the Discriminant.

• If $D > 0$, roots are real and distinct.
• If $D = 0$, roots are real and equal.
• If $D < 0$, roots are imaginary (no real solution).

Part 2: 6 Solved Problems

Problem 1: Solve by Factorization

Solve the equation for $x$: $$x^2 - 5x + 6 = 0$$

Solution:
We need two numbers that multiply to $+6$ and add up to $-5$. These numbers are $-2$ and $-3$.

Split the middle term: $$x^2 - 2x - 3x + 6 = 0$$

Group terms: $$x(x - 2) - 3(x - 2) = 0$$ $$(x - 2)(x - 3) = 0$$

Answer: $x = 2$ or $x = 3$

Problem 2: Using the Quadratic Formula

Find the roots of: $$2x^2 - 7x + 3 = 0$$

Solution:
Identify coefficients: $a = 2, b = -7, c = 3$.
Calculate the discriminant ($D$): $$D = b^2 - 4ac = (-7)^2 - 4(2)(3)$$ $$D = 49 - 24 = 25$$

Apply formula: $$x = \frac{-(-7) \pm \sqrt{25}}{2(2)}$$ $$x = \frac{7 \pm 5}{4}$$

Case 1: $x = \frac{7+5}{4} = \frac{12}{4} = 3$
Case 2: $x = \frac{7-5}{4} = \frac{2}{4} = \frac{1}{2}$

Answer: $x = 3, \frac{1}{2}$

Problem 3: Nature of Roots

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

Solution:
For equal roots, the Discriminant ($D$) must be zero.
$$D = b^2 - 4ac = 0$$

Substitute values ($a=2, b=k, c=3$): $$k^2 - 4(2)(3) = 0$$ $$k^2 - 24 = 0$$ $$k^2 = 24$$

Answer: $k = \pm \sqrt{24} = \pm 2\sqrt{6}$

Problem 4: Roots involving Irrational Numbers

Solve for $x$: $$x^2 + 4x - 4 = 0$$

Solution:
Here $a=1, b=4, c=-4$. Use the formula. $$D = 4^2 - 4(1)(-4) = 16 + 16 = 32$$

$$x = \frac{-4 \pm \sqrt{32}}{2}$$ Simplify $\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$. $$x = \frac{-4 \pm 4\sqrt{2}}{2}$$

Answer: $x = -2 \pm 2\sqrt{2}$

Problem 5: Geometry Word Problem

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:
Let the base be $x$ cm.
Then, altitude = $(x - 7)$ cm.
By Pythagoras theorem: $$(Base)^2 + (Altitude)^2 = (Hypotenuse)^2$$ $$x^2 + (x - 7)^2 = 13^2$$

Expand and simplify: $$x^2 + (x^2 - 14x + 49) = 169$$ $$2x^2 - 14x + 49 - 169 = 0$$ $$2x^2 - 14x - 120 = 0$$ Divide by 2: $$x^2 - 7x - 60 = 0$$

Factorize ($12$ and $-5$): $$(x - 12)(x + 5) = 0$$ $x = 12$ or $x = -5$. Since length cannot be negative, $x = 12$.

Answer: Base = 12 cm, Altitude = 5 cm

Problem 6: Speed and Distance Word Problem

A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.

Solution:
Let the usual speed be $x$ km/h.
Time taken usually = $\frac{480}{x}$ hours.
New speed = $(x - 8)$ km/h.
New time = $\frac{480}{x - 8}$ hours.

According to the question, the difference in time is 3 hours: $$\frac{480}{x - 8} - \frac{480}{x} = 3$$

Solve for $x$: $$480 \left[ \frac{x - (x - 8)}{x(x - 8)} \right] = 3$$ $$480 \left[ \frac{8}{x^2 - 8x} \right] = 3$$ $$3840 = 3(x^2 - 8x)$$ $$1280 = x^2 - 8x$$ $$x^2 - 8x - 1280 = 0$$

Using factorization ($x^2 - 40x + 32x - 1280 = 0$): $$(x - 40)(x + 32) = 0$$ $x = 40$ or $x = -32$. Speed cannot be negative.

Answer: Speed of the train = 40 km/h

Complete Solution: Quadratic Equation 2x² – x – 4 = 0 (Formula Method)

Mathematics Solutions

(vi) \( 2x^2 – x – 4 = 0 \)

Sol. \( 2x^2 – x – 4 = 0 \)

Comparing with \( ax^2 + bx + c = 0 \) we have \( a = 2, b = – 1, c = – 4 \)

$$ \begin{align} b^2 – 4ac &= (– 1)^2 – 4 (2) (– 4) \\ &= 1 + 32 \\ &= 33 \end{align} $$

By Formula method,

$$ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} $$

$$ x = \frac{-(-1) \pm \sqrt{33}}{2(2)} $$

$$ \therefore x = \frac{1 \pm \sqrt{33}}{4} $$

$$ \therefore x = \frac{1 + \sqrt{33}}{4} \quad \text{or} \quad \frac{1 - \sqrt{33}}{4} $$

Solution for Quadratic Equation 5x² – 22x – 15 = 0

Problem Statement

Solve the following quadratic equation:

\[ (xi) \quad 5x^2 – 22x – 15 = 0 \]

Detailed Solution

Sol.

\[ 5x^2 – 22x – 15 = 0 \]
\[ \therefore 5x^2 – 25x + 3x – 15 = 0 \]
\[ \therefore 5x(x – 5) + 3(x – 5) = 0 \]
\[ \therefore (x – 5) (5x + 3) = 0 \]
\[ \therefore x – 5 = 0 \quad \text{or} \quad 5x + 3 = 0 \]
\[ \therefore x = 5 \quad \text{or} \quad 5x = –3 \]

\[ \therefore x = 5 \quad \text{or} \quad x = \frac{-3}{5} \]

QUADRATIC EQUATIONS

EXERCISE - 2.1

1. Which of the following are quadratic equations ?

(i) 11 = – 4x2 – x3          [Ans.]

(ii) -¾ y2 = 2y + 7          [Ans.]

(iii) (y – 2) (y + 2) = 0      [Ans.]

(iv) 3/y – 4 = y      [Ans.]

(v) m3 + m + 2 = 4m      [Ans.]

(vi) n – 3 = 4n      [Ans.]

(vii) y2 – 4 = 11y      [Ans.]

(viii) z – 7/z = 4z + 5    [Ans.]

(ix) 3y2 – 7 = √3 y      [Ans.]

(x) (q2 – 4)/q2 = - 3  [Ans.]

2. Write the following quadratic equations in standard form ax2 + bx + c = 0

(i) 7 – 4x –x2 = 0  [Ans]

(ii) 3y2 = 10y + 7     [Ans]

(iii) (m + 4) (m – 10) = 0    [Ans]

(iv) p(p – 6) = 0    [Ans]

(v) (x2/25)  – 4 = 0     [Ans]

(vi) n – (7/n) = 4     [Ans]

(vii) y2 – 9 = 13y     [Ans]

(viii) 2z – (5/z) = z – 6     [Ans]

(ix) x2 = –7 – √10 x     [Ans]

(x) (m2 +5)/m2  = –3     [Ans]

EXERCISE - 2.2

1. In each of the examples given below determine whether the values given against each of the quadratic equation are the roots of the equation or not.

(i) x2 + 3x – 4 = 0,  x = 1, –2, – 3 [Ans]

(ii) 4m2 – 9 = 0, m = 2, 2/3, 3/2   [Ans]

(iii) x2 + 5x – 14 = 0, x = √2 , –7, 3    [Ans]

(iv) 2p2 + 5p – 3 = 0, p = 1, ½,  –3    [Ans]

(v) n2 + 4n = 0, n = 0, – 2, – 4   [Ans]

2. If one root of the quadratic equation x2 – 7x + k = 0 is 4, then find the value of k.     [Ans]

3. If one root of the quadratic equation 3y2 – ky + 8 = 0 is 2/3,  then find the value of k.      [Ans]

4. State whether k is the root of the given equation y2 – (k – 4)y – 4k = 0.    [Ans]

5. If one root of the quadratic equation kx2 – 7x + 12 = 0 is 3, then find the value of k.     [Ans]

EXERCISE - 2.3

Solve the following quadratic equations by

factorization method..

(i) x2 – 5x + 6 = 0 [Ans.]

(ii) x2 + 10x + 24 = 0 [Ans.]

(iii) x2 – 13x – 30 = 0 [Ans.]

(iv) x2 – 17x + 60 = 0 [Ans.]

(v). m2 – 84 = 0 [Ans.]

(vi) x + 20/x – 12 = 0 [Ans.]

(vii) x2 = 2(11x – 48) [Ans.]

(viii) 21x = 196 – x2  [Ans.]

(ix) 2x - 10/x = 1 [Ans.]

(x) x2 – x – 132 = 0 [Ans.]

(xi) 5x2 – 22x – 15 = 0 [Ans.]

(xii) 3x2 – x – 10 = 0 [Ans.]

(xiii) 2x2 – 5x – 3 = 0 [Ans]

(xiv) x (2x + 3) = 35 [Ans.]

(xv) 7x2 + 4x – 20 = 0 [Ans]

(xvi) 10x2 + 3x – 4 = 0   [Ans.]

(xvii) 6x2 – 7x – 13 = 0  [Ans.]

(xviii) 3x2 + 34x + 11 = 0 [Ans.]

(xix) 3x2 – 11x + 6 = 0 [Ans.]

(xx) 3x2 – 10x + 8 = 0  [Ans.]

(xxi) 2m2 + 19m + 30 = 0 [Ans.]

(xxii) 7m2 – 84 = 0 [Ans.]

(xxiii)  x2 – 3√3 x + 6 = 0 [Ans.]

EXERCISE - 2.4

Solve the following quadratic equations by completing square.

(i) x2 + 8x + 9 = 0 [Ans]

(ii) z2 + 6z – 8 = 0 [Ans]

(iii) m2 – 3m – 1 = 0 [Ans]

(iv) y2 = 3 + 4y [Ans]

(v) p2 – 12p + 32 = 0 [Ans]

(vi) x (x – 1) = 1 [Ans]

(vii) 3y2 + 7y + 1 = 0 [Ans]

(viii) 4p2 + 7 = 12p [Ans]

(ix) 6m2 + m = 2  [Ans]

EXERCISE - 2.5

1. Solve the following quadratic equations by using formula.

(i) m2 – 3m – 10 = 0 [Ans]

(ii) x2 + 3x – 2 = 0 [Ans]

(iii) x2 + (x – 1)/3 = 0 [Ans]

(iv) 5m2 – 2m = 2 [Ans.]

(v) 7x + 1 = 6x2 [Ans.]

(vi) 2x2 – x – 4 = 0 [Ans.]

(vii) 3y2 + 7y + 4 = 0 [Ans.]

(viii) 2n2 + 5n + 2 = 0 [Ans.]

(ix) 7p2 – 5p – 2 = 0 [Ans.]

(x) 9s2 – 4 = – 6s [Ans.]

(xi) 3q2 = 2q + 8 [Ans.]

(xii) 4x2 + 7x + 2 = 0 [Ans.]

EXERCISE - 2.6

1. Find the value of discriminant of each of the following equations :

(i) x2 + 4x + 1 = 0   [Ans]

(ii) 3x2 + 2x – 1 = 0  [Ans]

(iii) x2 + x + 1 = 0  [Ans]

(iv) √3 x2 + 2√2 x – 2√3 = 0 [Ans]

(v) 4x2 + kx + 2 = 0 [Ans]

(vi) x2 + 4x + k = 0  [Ans]

2. Determine the nature of the roots of the following equations from their  discriminants :

(i) y2 – 4y – 1 = 0 [Ans.]

(ii) y2 + 6y – 2 = 0  [Ans.]

(iii) y2 + 8y + 4 = 0  [Ans.]

(iv) 2y2 + 5y – 3 = 0  [Ans.]

(v) 3y2 + 9y + 4 = 0  [Ans.]

(vi) 2x2 + 5√3 x + 16 = 0   [Ans.]

3. Find the value of k for which given equation has real and equal roots :

(i) (k – 12)x2 + 2 (k – 12)x + 2 = 0  [Ans.]

(ii) k2x2 – 2 (k – 1)x + 4 = 0 [Ans.]

EXERCISE - 2.7

1. If one root of the quadratic equation kx2 – 5x + 2 = 0 is 4 times the other, find k.   [Ans.]

2. Find k, if the roots of the quadratic equation x2 + kx + 40 = 0 are in the ratio 2 : 5.   [Ans.]

3. Find k, if one of the roots of the quadratic equation kx2 – 7x + 12 = 0 is 3.   [Ans.]

4. If the roots of the equation x2 + px + q = 0 differ by 1, prove that p2= 1 + 4q.  [Ans.]

5. Find k, if the sum of the roots of the quadratic equation 4x2 + 8kx + k + 9 = 0 is equal to their product. [Ans.]

6. If  α  and β  are the roots of the equation x2 – 5x + 6 = 0, find [Ans.]

(i) α2+β2

(ii) α/β +β/α

7. If one root of the quadratic equation kx2 – 20x + 34 = 0 is 5 – 2√2 , find k. [Ans.]

EXERCISE - 2.8

1. Form the quadratic equation if its roots are

(i) 5 and – 7     [Ans.]

(ii) ½  and – ¾     [Ans.]

(iii) - 3 and –11    [Ans.]

(iv) -2 and  11/2    [Ans.]

(v) ½  and – ½      [Ans.]

(vi) 0 and – 4       [Ans.]

2. Form the quadratic equation if one of the root is

(i) 3 – 2√ 5 [Ans.]

(ii) 4 – 3√ 2  [Ans.]

(iii) √ 2 + √3 [Ans.]

(iv) 2√3 – 4 [Ans.]

(v) 2+√5  [Ans.]

(vi) √ 5 - √3 [Ans.]

3. If the sum of the roots of the quadratic is 3 and sum of their cubes is 63, find the quadratic equation. [Ans.]

4. If the difference of the roots of the quadratic equation is 5 and the difference  of their cubes is 215, find the quadratic equation. [Ans.]

EXERCISE - 2.9

Solve the following equations.

(i) x4 – 3x2 + 2 = 0 [Ans.]

(ii) (x2 + 2x) (x2 + 2x – 11) + 24 = 0 [Ans.]

(iii) 2(x2 + 1/x2 ) – 9(x+1/x) + 14 = 0 [Ans.]

(iv) 35y2 + 12/y2 = 44 [Ans.]

(v) x2 + 12/x2 = 7 [Ans.]

(vi) (x2 + x) (x2 + x – 7) + 10 = 0 [Ans.]

(vii) 3x4 – 13x2 + 10 = 0 [Ans.]

(viii) 2y2 + 15/y2 = 12 [Ans.]

EXERCISE - 2.10

1. The sum of the squares of two consecutive natural numbers is 113. Find the numbers. [Ans]

2. Tinu is younger than Pinky by three years. The product of their ages is 180. Find their ages.[Ans]

3. The length of the rectangle is greater than its breadth by 2 cm. The area of the rectangle is 24 sq.cm, find its length and breadth.[Ans]

4. The sum of the squares of two consecutive even natural numbers is 100. Find the numbers. [Ans]

5. A natural number is greater than twice its square root by 3. Find the number. [Ans]

6. The sum of a natural number and its reciprocal is 10/3 . Find the number. [Ans]

7. The sum of the ages of father and his son is 42 years. The product of their ages is 185, find their ages. [Ans]

8. Three times the square of a natural numbers is 363. Find the numbers. [Ans]

9. The length of one diagonal of a rhombus is less than the second diagonal by 4 cm. The area of the rhombus is 30 sq.cm. Find the length of the diagonals. [Ans]

10. A natural number is greater than the other  by 5. The sum of their squares is 73. Find those numbers. [Ans]

11. The sum ‘S’ of the first ‘n’ natural numbers is given by S = n (n + 1)/2 . Find ‘n’, if the sum (S) is 276. [Ans]

12. A rectangular playground is 420 sq.m. If its length is increases by 7 m and breadth is decreased by 5 metres, the area remains the same. Find the length and breadth of the playground ? [Ans]

13. The cost of bananas is increased by Re. 1 per dozen, one can get 2 dozen less for Rs. 840. Find the original cost of one dozen of banana.  [Ans]