If x = 2/3 and x = −3 are the roots of the equation

Solution:

We have been given that,

$a x^{2}+7 x+b=0, x=\frac{2}{3}, x=-3$

We have to find a and b

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

$a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$

$\frac{4 a+42+9 b}{9}=0$

$a=\frac{-9 b-42}{4} \ldots \ldots$(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,

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

$9 a-21+b=0 \ldots \ldots(2)$

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

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

$77 a-231=0$

$a=\frac{231}{77}$

$a=3$

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

$9(3)+b-21=0$

$b=-6$

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