Balbharti Maharashtra State Board Class 10 Maths Solutions covers the Practice Set 2.1 Algebra 10th Class Maths Part 1 Answers Solutions Chapter 2 Quadratic Equations.

## Practice Set 2.1 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1.

Write any two quadratic equations.

Solution:

i. y^{2} – 7y + 12 = 0

ii. x^{2} – 8 = 0

Question 2.

Decide which of the following are quadratic

i. x^{2} – 7y + 2 = 0

ii. y^{2} = 5y – 10

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

iv. x + \(\frac { 1 }{ x } \) = -2

v. (m + 2) (m – 5) = 03

vi. m^{3} + 3m^{2} – 2 = 3m^{3}

Solution:

i. The given equation is x^{2} + 5x – 2 = 0

Here, x is the only variable and maximum index of the variable is 2.

a = 1, b = 5, c = -2 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

ii. The given equation is

y^{2} = 5y – 10

∴ y^{2} – 5y + 10 = 0

Here, y is the only variable and maximum index of the variable is 2.

a = 1, b = -5, c = 10 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

iii. The given equation is

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

∴ y^{3} + 1 = 2y …[Multiplying both sides by y]

∴ y^{3} – 2y + 1 = 0

Here, y is the only variable and maximum index of the variable is not 2.

∴ The given equation is not a quadratic equation.

iv. The given equation is

x + \(\frac { 1 }{ x } \) = -2

∴ x^{2} + 1 = -2x …[Multiplying both sides by x]

∴ x^{2} + 2x+ 1 = 0

Here, x is the only variable and maximum index of the variable is 2.

a = 1, b = 2, c = 1 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

v. The given equation is

(m + 2) (m – 5) = 0

∴ m(m – 5) + 2(m – 5) = 0

∴ m^{2} – 5m + 2m – 10 = 0

∴ m^{2} – 3m – 10 = 0

Here, m is the only variable and maximum index of the variable is 2.

a = 1, b = -3, c = -10 are real numbers and a ≠ 0.

∴ The given equation is a quadratic equation.

vi. The given equation is

m^{3} + 3m^{2} – 2 = 3m^{3}

∴ 3m^{3} – m^{3} – 3m^{2} + 2 = 0

∴ 2m^{3} – 3m^{2} + 2 = 0

Here, m is the only variable and maximum

index of the variable is not 2.

∴ The given equation is not a quadratic equation.

Question 3.

Write the following equations in the form ax^{2} + bx + c = 0, then write the values of a, b, c for each equation.

i. 2y = 10 – y^{2}

ii. (x – 1)^{2} = 2x + 3

iii. x^{2} + 5x = – (3 – x)

iv. 3m^{2} = 2m^{2} – 9

v. P (3 + 6p) = – 5

vi. x^{2} – 9 = 13

Solution:

i. 2y – 10 – y^{2}

∴ y^{2} + 2y – 10 = 0

Comparing the above equation with

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

a = 1, b = 2, c = -10

ii. (x – 1)^{2} = 2x + 3

∴ x^{2} – 2x + 12x + 3

x^{2 }– 2x + 1 – 2x – 30

∴ x^{2} – 4x – 2 = 0

Comparing the above equation with

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

a = 1, b = -4, c = -2

iii. x^{2} + 5x = – (3 – x)

∴ x^{2} + 5x = -3 + x

∴ x^{2} + 5x – x + 3 = 0

∴ x^{2} + 4x + 3 = 0

Comparing the above equation with

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

a = 1, b = 4, c = 3

iv. 3m^{2} = 2m^{2} – 9

∴ 3m^{2} – 2m^{2} + 9 = 0

∴ m^{2} + 9 = 0

∴ m^{2} + 0m + 9 = 0

Comparing the above equation with

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

a = 1, b = 0, c = 9

v. p (3 + 6p) = – 5

∴ 3p + 6p^{2} = -5

∴ 6p^{2} + 3p + 5 = 0

Comparing the above equation with

ap^{2} + bp + c = 0, we get

a = 6, b = 3, c = 5

vi. x^{2} – 9 = 13

∴ x^{2} – 9 – 13 = 0

∴ x^{2} – 22 = 0

∴ x^{2} + 0x – 22 = 0

Comparing the above equation with

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

a = 1, b = 0, c = -22

Question 4.

Determine whether the values given against each of the quadratic equation are the roots of the equation.

i. x^{2} + 4x – 5 = 0; x = 1,-1

ii. 2m^{2} – 5m = 0; m = 2, \(\frac { 5 }{ 2 } \)

Solution:

i. The given equation is

x^{2} + 4x – 5 = 0 …(i)

Putting x = 1 in L.H.S. of equation (i), we get

L.H.S. = (1)^{2} + 4(1) – 5 = 1 + 4 – 5 = 0

∴ L.H.S. = R.H.S.

∴ x = 1 is the root of the given quadratic equation.

Putting x = -1 in L.H.S. of equation (i), we get

L.H.S. = (-1)^{2} + 4(-1) – 5 = 1 – 4 – 5 = -8

∴ LH.S. ≠ R.H.S.

∴ x = -1 ¡s not the root of the given quadratic equation.

ii. The given equation is

2m^{2} – 5m = 0 …(i)

Putting m = 2 in L.H.S. of equation (i), we get

L.H.S. = 2(2)^{2} – 5(2) = 2(4) -10 = 8 – 10 = -2

∴ L.H.S. ≠ R.H.S.

∴ m = 2 is not the root of the given quadratic equation.

Putting m = \(\frac { 5 }{ 2 } \) in L.H.S. of equation (i), we get

Question 5.

Find k if x = 3 is a root of equation kx^{2} – 10x + 3 = 0.

Solution:

x = 3 is the root of the equation kx^{2} – 10x + 3 = 0.

Putting x = 3 in the given equation, we get

k(3)^{2} – 10(3) + 3 = 0

∴ 9k – 30 +3 = 0

∴ 9k – 27 = 0

∴ 9k = 27

∴ k = \(\frac { 27 }{ 9 } \)

∴ k = 3

Question 6.

One of the roots of equation 5m^{2} + 2m + k = 0 is \(\frac { -7 }{ 5 } \) Complete the following activity to find the value of ‘k’.

Solution:

Question 1.

x^{2} + 3x – 5, 3x^{2} – 5x, 5x^{2}; Write the polynomials In the index form. Observe the coefficients and fill in the boxes. (Textbook p. no. 31)

Answer:

Index form of the given polynomials:

x^{2} + 3x – 5, 3x^{2} – 5x + 0, 5x^{2} + 0x + 0

i. Coefficients of x2 are [1], [3] and [5] respectively, and these coefficients are non zero.

ii. Coefficients of x are 3, [-5] and [0] respectively.

iii. Constant terms are [-5], [0] and [0] respectively.

Here, constant terms of second and third polynomial is zero.

Question 2.

Complete the following table (Textbook p. no. 31)

Answer:

Question 3.

Decide which of the following are quadratic equations? (Textbook pg. no. 31)

i. 9y^{2} + 5 = 0

ii. m^{3} – 5m^{2} + 4 = 0

iii. (l + 2)(l – 5) = 0

Solution:

i. In the equation 9y^{2} + 5 = 0, [y] is the only variable and maximum index of the variable is [2].

∴ It [is] a quadratic equation.

ii. In the equation m^{3} – 5m^{2} + 4 = 0, [m] is the only variable and maximum index of the variable is not 2.

∴ It [is not] a quadratic equation.

iii. (l + 2)(l – 5) = 0

∴ l(l – 5) + 2(l – 5) = 0

∴ l^{2} – 5l + 2l – 10 = 0

∴ l^{2} – 3l – 10 = 0.

In this equation [l] is the only variable and maximum index of the variable is [2]

∴ it [is] a quadratic equation.

Question 4.

If x = 5 is a root of equation kx^{2} – 14x – 5 = 0, then find the value of k by completing the following activity. (Textbook pg, no. 33)

Solution: