Question

If 2/3 and -3 are the zeroes of the polynomial ax^2+7x+b,then find the values of a and b

Answer 1

$\mathfrak{\huge{Answer:}}$

Given that the zeroes of p(x) = $\tt{ax^{2} + 7x + b}$ are $\tt{\frac{2}{3} \:and\:(-3)}$

We already are aware that :-

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

Where, $\sf{\alpha\:and\:\beta}$ are the zeroes of the polynomial.

$\tt{ax^{2} + 7x + b}$, here the values of the variables will be :
a = a
b = 7
c = b

Now, firstly find the value of $\sf{\alpha + \beta }$

=》 $\sf{\alpha + \beta = \frac{2}{3} + (-3)}$

=》 $\sf{\alpha + \beta = \frac{-7}{3}}$

We even have another value of $\sf{\alpha + \beta}$ which is :

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

Bring the values in and then solve :

=》 $\sf{\frac{-7}{3} = \frac{-7}{a}}$

=》 $\boxed{\tt{a = 3}}$

We even know another thing about the $\sf{\alpha\:and\:\beta}$, which is:

=》 $\sf{\alpha\beta = \frac{c}{a}}$

Bring in the values and solve :

=》 $\sf{\frac{2}{3}\times 3 = \frac{b}{3}}$

Solve this formed equation further

=》 $\sf{2 = \frac{c}{3}}$

Last one step:

=》 $\boxed{\tt{c = b = 6}}$

Answer 2

Answer

Refer to the attachment