#### Question

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

#### Solution

We have been given that,

*ax*^{2} + 7*x* + *b* = 0, x = 2/3, x = -3

We have to find *a *and *b*

Now, if x = 2/3 is a root of the equation, then it should satisfy the equation completely. Therefore we substitute x = 2/3 in the above equation. We get,

*a*(2/3)^{2} + 7(2/3) + *b* = 0

`(4a + 42+9b)/9=0`

`a=(-9b-42)/4` ......... (1)

Also, if x = -3 is a root of the equation, then it should satisfy the equation completely. Therefore we substitute x = -3 in the above equation. We get,

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

9a - 21 + b = 0 ......... (2)

Now, we multiply equation (2) by 9 and then subtract equation (1) from it. So we have,

81a + 9b - 189 - 4a - 9b - 42 = 0

77a - 231 = 0

`a = 231/77`

a = 3

Now, put this value of ‘*a*’ in equation (2) in order to get the value of ‘*b*’. So,

9(3) + b - 21 = 0

27 + b - 21 = 0

b = 21 - 27

b = -6

Therefore, we have a = 3 and b = -6.