Question

If $\alpha$ and $\beta$ are the zeros of polynomial $x^2-2x-8$ then form a quadratic polynomial whose zeros are $3\alpha$ and 3$\beta$.

Answer 1

Given $\alpha$ and $\beta$ be the zeroes of polynomials, so the quadratic polynomial is $x^2-(\alpha +\beta )x+\alpha \beta$ , comparing the above equation to the given equation, $x^2-2x-8$, so we have,

$\alpha +\beta =2$ and $\alpha \beta =-8$

So, $3\alpha +3\beta =3(\alpha +\beta )=3\times 2=6$ and $3\alpha \times 3\beta =9\alpha \beta =9\times (-8)=-72$

Thus, the required quadratic polynomial is $x^2-3(\alpha +\beta )x+9\alpha \beta$

Putting the values of $\alpha +\beta$ and $\alpha \beta$ in above equation, $x^2-6x-72$