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

Given $\frac{2}{3}$ and ‒3 are the roots of the quadratic polynomial equation. 

The relation between the zeros, say $\alpha$and $\beta$, and the coefficients of a quadratic polynomial $a{x}^2+bx+c$ is given as follows, 

$\alpha +\beta =\frac{-b}{a}$

$\alpha \beta =\frac{c}{a}$

So, comparing the equation $a{x}^2+7x+b=0$ with the general form, we get the coefficients as a = a, b = 7 and c = b. Now, as per the relation between the zeros and the coefficients of a quadratic polynomial, we have

$\alpha +\beta =\frac{-7}{a}$

$\alpha \beta =\frac{b}{a}$

Next, we have the roots of the given equation.

So,  $\alpha =\frac{2}{3}$ and $\beta =-3$.

Hence, we can rewrite the above relations as follows,
$\frac{2}{3}+(-3)=\frac{-7}{a}$ -----(i)

$\frac{2}{3}\times (-3)=\frac{b}{a}$ -----(ii)

Now solving equation (i), we get,

$\frac{2}{3}+(-3)=\frac{-7}{a}$

$\Rightarrow \frac{2-9}{3}=\frac{-7}{a}$

$\Rightarrow \frac{-7}{3}=\frac{-7}{a}$

$\Rightarrow a=3$

Now, substituting this value of a in the equation (ii), we get,
$\frac{2}{3}\times (-3)=\frac{b}{3}$

$\Rightarrow -2=\frac{b}{3}$

$\Rightarrow -2\times 3=b$

$\Rightarrow b=-6$

So, the values of a and b are 3 and –6 respectively.