Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.4 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

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

Question 1.

Compare the given quadratic equations to the general form and write values of a, b, c.

i. x^{2} – 7x + 5 = 0

ii. 2m^{2} = 5m – 5

iii. y^{2} = 7y

Solution:

i. x^{2} – 7x + 5 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 1, b = -7, c = 5

ii. 2m^{2} = 5m – 5

∴ 2m^{2} – 5m + 5 = 0

Comparing the above equation with

am^{2} + bm + c = 0, we get

a = 2, b = -5, c = 5

iii. y^{2} = 7y

∴ y^{2} – 7y + 0 = 0

Comparing the above equation with

ay^{2} + by + c = 0, we get

a = 1, b = -7, c = 0

Question 2.

Solve using formula.

i. x^{2} + 6x + 5 = 0

ii. x^{2} – 3x – 2 = 0

iii. 3m^{2} + 2m – 7 = 0

iv. 5m^{2} – 4m – 2 = 0

v. y^{2} + \(\frac { 1 }{ 3 } \) y = 2

vi. 5x^{2} + 13x + 8 = 0

Solution:

i. x^{2} + 6x + 5 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 1, b = 6, c = 5

∴ b^{2} – 4ac = (6)2 – 4 × 1 × 5

= 36 – 20 = 16

∴ x = -3 + 2 or x = -3 -2

∴ x = -1 or x = -5

∴ The roots of the given quadratic equation are -1 and -5.

ii. x^{2} – 3x – 2 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 1, b = -3, c = -2

∴ b^{2} – 4ac = (-3)2 – 4 × 1 × (-2)

= 9 + 8 = 17

iii. 3m^{2} + 2m – 7 = 0

Comparing the above equation with

am^{2} + bm + c = 0, we get

a = 3, b = 2, c = -7

∴ b^{2} – 4ac = (2)^{2} – 4 × 3 × ( -7)

= 4 + 84 = 88

iv. 5m^{2} – 4m – 2 = 0

Comparing the above equation with

am^{2} + bm + c = 0, we get

a = 5, b = -4, c = -2

∴ b^{2} – 4ac = (-4)^{2} – 4 × 5 × (-2)

= 16 + 40 = 56

v. y^{2} + \(\frac { 1 }{ 3 } \)y = 2

∴ 3y^{2} + y = 6 …(Multiplying both sides by 3]

∴ 3y^{2} + y – 6 = 0

Comparing the above equation with

ay^{2} + by + c = 0, we get

a = 3, b = 1, c = -6

∴ b^{2} – 4ac = (1)^{2} – 4 × 3 × (-6)

= 1 + 72 = 73

vi. 5x^{2} + 13x + 8 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 5, b = 13, c = 8

∴ b^{2} – 4ac = (13)2 – 4 × 5 × 8

= 169 – 160 = 9

The roots of the given quadratic equation are -1 and \(\frac { -8 }{ 5 } \).

Question 3.

With the help of the flow chart given below solve the equation x^{2} + 2√3 x + 3 = 0 using the formula.

Solution:

i. x^{2} + 2√3 x + 3 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 1, b = 2√3 ,c = 3

ii. b^{2} – 4ac = (2√3)2 -4 × 1 × 3

= 12 – 12

= 0

Question 1.

Solve the equation 2x^{2} + 13x + 15 = 0 by factorisation method, by completing the square method and by using the formula. Verify that you will get the same roots every time. (Textbook pg. no. 43)

Solution:

By using the property, if the product of two numbers is zero, then at least zero, we get

∴ x + 5 = 0 or 2x + 3 = 0

∴ x + -5 = 0 or 2x = -3 = 0

∴ x + -5 = or x = \(\frac { -3 }{ 2 } \)

∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

ii. Completing the square method:

2x² + 13x + 15 = 0

∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

iii. Formula method:

2x^{2} + 13x + 15 = 0

Comparing the above equation with

ax^{2} + bx + c = 0, we get

a = 2, b = 13, c = 15

∴ b^{2} – 4ac = (13)2 – 4 × 2 × 15

= 169 – 120 = 49

∴ The roots of the given quadratic equation are \(\frac { -3 }{ 2 } \) and -5.

∴ By all the above three methods, we get the same roots of the given quadratic equation.