**Chapter 2 Quadratic Equations Set 2.1**

## Chapter 2 Quadratic Equations Set 2.1

**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 quadratici. x**

^{2}– 7y + 2 = 0ii. y

^{2}= 5y – 10

v. (m + 2) (m – 5) = 03

vi. m

Solution:

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} += 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 + = -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 + 3iii. x^{2} + 5x = – (3 – x)iv. 3m^{2} = 2m^{2} – 9v. P (3 + 6p) = – 5vi. x^{2} – 9 = 13Solution:**

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.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 = 52 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 = 279

∴ k = 3

**Question 6.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 = 0ii. m

^{3}– 5m

^{2}+ 4 = 0iii. (l + 2)(l – 5) = 0Solution:

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:**