If $x=2 / 3$ and $x=-3$ are the roots of the equation $a x^{2}+7 x+b=0$, find the values of $a$ and $b$.
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 $b=-6$.
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