Chapter 2 Quadratic Equations Practice Set 2.4 Solutions

Practice Set 2.4 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

ii. 2m²= 5m – 5
∴ 2m²– 5m + 5 = 0
Comparing the above equation with
am²+ bm + c = 0, we get
a = 2, b = -5, c = 5

iii. y²= 7y
∴ y²– 7y + 0 = 0
Comparing the above equation with
ay²+ by + c = 0, we get
a = 1, b = -7, c = 0

ii. x²– 3x – 2 = 0
Comparing the above equation with
ax²+ bx + c = 0, we get
a = 1, b = -3, c = -2
∴ b²– 4ac = (-3)2 – 4 × 1 × (-2)
= 9 + 8 = 17

iii. 3m²+ 2m – 7 = 0
Comparing the above equation with
am²+ bm + c = 0, we get
a = 3, b = 2, c = -7
∴ b²– 4ac = (2)²– 4 × 3 × ( -7)
= 4 + 84 = 88

iv. 5m²– 4m – 2 = 0
Comparing the above equation with
am²+ bm + c = 0, we get
a = 5, b = -4, c = -2
∴ b²– 4ac = (-4)²– 4 × 5 × (-2)
= 16 + 40 = 56

v. y²+ y = 2
∴ 3y²+ y = 6 …(Multiplying both sides by 3]
∴ 3y²+ y – 6 = 0
Comparing the above equation with
ay²+ by + c = 0, we get
a = 3, b = 1, c = -6
∴ b²– 4ac = (1)²– 4 × 3 × (-6)
= 1 + 72 = 73

vi. 5x²+ 13x + 8 = 0
Comparing the above equation with
ax²+ bx + c = 0, we get
a = 5, b = 13, c = 8
∴ b²– 4ac = (13)2 – 4 × 5 × 8
= 169 – 160 = 9


The roots of the given quadratic equation are -1 and .

ii. b²– 4ac = (2√3)2 -4 × 1 × 3
= 12 – 12
= 0

ii. Completing the square method:
2x² + 13x + 15 = 0


∴ The roots of the given quadratic equation are and -5.

iii. Formula method:
2x²+ 13x + 15 = 0
Comparing the above equation with
ax²+ bx + c = 0, we get
a = 2, b = 13, c = 15
∴ b²– 4ac = (13)2 – 4 × 2 × 15
= 169 – 120 = 49

∴ The roots of the given quadratic equation are and -5.
∴ By all the above three methods, we get the same roots of the given quadratic equation.