# R.D. Sharma Solutions Class 10th: Ch 8 Quadratic Equations Exercise 8.1

## Chapter 8 Quadratic Equations R.D. Sharma Solutions for Class 10th Math Exercise 8.1

##
1. Which of the following are quadratic equations ?

Solution

2. In each of the following , determine whether the given values are solutions of the given equation or not :

Solution

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

3. In each of the following , find the value of k for which the given value is a solution of the given equation :

Solution

(i)

(ii)

(iii)

(iv)

4. Determine if, 3 is a root of the equation given below:

Solution

5. If x = 2/3 and x = -3 are the roots of the equation ax^{2} + 7x + b = 0, find the values of a and b .

Solution

We know that x = -2/3 and x = -3 are the roots of the equation ax^{2} + 7x + b. When we the value of x in this equation, we will get the result as 0.
Putting x = -2/3
a(2/3)^{2} + 7(2/3) + b = 0
⇒ 4/9 a +14/3 + b =0
⇒ 4/9 a + b = -14/3 --- (i)

Putting x = -3
a(-3)^{2} + 7(-3) + b = 0

⇒ 9a - 21 + b =0

⇒ 9a + b = 21 --- (ii)

Subtracting (i) from (ii), we get

9a + b - (4/9 a + b) = 21 + 14/3

⇒ 9a + b - 4/9 a -b = 77/3

⇒ 77/9 a = 49/3
⇒ a = 77/3 × 9/77
⇒ a = 3

Now, putting a =3 in equation (ii),
9×3 + b = 21
⇒ 27 + b = 21
⇒ b = 21 - 27
⇒ b = -6

Thus, a = 3 and b = -6

4. Determine if, 3 is a root of the equation given below:

5. If x = 2/3 and x = -3 are the roots of the equation ax

^{2}+ 7x + b = 0, find the values of a and b .

We know that x = -2/3 and x = -3 are the roots of the equation ax

^{2}+ 7x + b. When we the value of x in this equation, we will get the result as 0.
Putting x = -2/3

a(2/3)

^{2}+ 7(2/3) + b = 0
⇒ 4/9 a +14/3 + b =0

⇒ 4/9 a + b = -14/3 --- (i)

Putting x = -3

a(-3)

⇒ 9a - 21 + b =0

⇒ 9a + b = 21 --- (ii)

Subtracting (i) from (ii), we get

9a + b - (4/9 a + b) = 21 + 14/3

⇒ 9a + b - 4/9 a -b = 77/3

^{2}+ 7(-3) + b = 0⇒ 9a - 21 + b =0

⇒ 9a + b = 21 --- (ii)

Subtracting (i) from (ii), we get

9a + b - (4/9 a + b) = 21 + 14/3

⇒ 9a + b - 4/9 a -b = 77/3

⇒ 77/9 a = 49/3

⇒ a = 77/3 × 9/77

⇒ a = 3

Now, putting a =3 in equation (ii),

9×3 + b = 21

⇒ 27 + b = 21

⇒ b = 21 - 27

⇒ b = -6

Thus, a = 3 and b = -6

Thus, a = 3 and b = -6