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

Given:

Given equation is $ax^2+7x+b=0$.

To do:

We have to find the values of $a$ and $b$ if $x=\frac{2}{3}$ and $x=-3$ are the roots of the given equation.

Solution:

If $x=m$ is a root of $f(x)$ then $f(m)=0$.

For $x=\frac{2}{3}$

$ax^2+7x+b=0$

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

$\frac{4}{9}a+\frac{14}{3}+b=0$

$9(\frac{4}{9}a)+9(\frac{14}{3})+9(b)=9(0)$  (Multply by $9$ on both sides) 

 $4a+42+9b=0$

Let it be equation (1).

For $x=-3$

$ax^2+7x+b=0$

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

$9a-21+b=0$

Let it be equation (2).

Solving equations (1) and (2), we get,

$(4a+42+9b=0)-9(9a-21+b=0)$

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

$-77a+231=0$

$77a=231$

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

$a=3$

Substituting $a=3$ in equation (1), we get,

$4(3)+42+9b=0$

$12+42+9b=0$

$9b=-54$

$b=\frac{-54}{9}$

$b=-6$

The values of $a$ and $b$ are $3$ and $-6$ respectively.