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

##
1.Â 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)Â

(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)Â

(iii)Â

(iv)Â

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

Solution

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

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}Â + 7(2/3) + b = 0
â‡’ 4/9 aÂ +14/3 + b =0
â‡’ 4/9 aÂ + b = -14/3 --- (i)

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

â‡’ 9aÂ - 21 + b =0

â‡’ 9aÂ + b = 21 --- (ii)

Subtracting (i)Â from (ii), we get

9aÂ + b - (4/9 aÂ + b) = 21 + 14/3

â‡’Â 9aÂ + b -Â 4/9 aÂ -b = 77/3

â‡’ 77/9 aÂ = 49/3
â‡’ Â aÂ = 77/3 Ã— 9/77
â‡’Â Â aÂ =Â 3

Now, putting a =3 in equation (ii),
9Ã—3 + b = 21
â‡’Â 27 + bÂ =Â 21
â‡’Â Â bÂ =Â 21 - 27
â‡’Â Â bÂ =Â -6

Thus, a = 3 and b = -6

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

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

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

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}Â + 7(2/3) + b = 0
â‡’ 4/9 aÂ +14/3 + b =0

â‡’ 4/9 aÂ + b = -14/3 --- (i)

Putting x = -3

a(-3)

â‡’ 9aÂ - 21 + b =0

â‡’ 9aÂ + b = 21 --- (ii)

Subtracting (i)Â from (ii), we get

9aÂ + b - (4/9 aÂ + b) = 21 + 14/3

â‡’Â 9aÂ + b -Â 4/9 aÂ -b = 77/3

^{2}Â + 7(-3) + b = 0â‡’ 9aÂ - 21 + b =0

â‡’ 9aÂ + b = 21 --- (ii)

Subtracting (i)Â from (ii), we get

9aÂ + b - (4/9 aÂ + b) = 21 + 14/3

â‡’Â 9aÂ + b -Â 4/9 aÂ -b = 77/3

â‡’ 77/9 aÂ = 49/3

â‡’ Â aÂ = 77/3 Ã— 9/77

â‡’Â Â aÂ =Â 3

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

9Ã—3 + b = 21

â‡’Â 27 + bÂ =Â 21

â‡’Â Â bÂ =Â 21 - 27

â‡’Â Â bÂ =Â -6

Thus, a = 3 and b = -6

Thus, a = 3 and b = -6