Question

Find the quadratic polynomial, the sum of whose zeros is √2 and their product is -12. Hence find the zeros of the polynomial​

Answer 1

Answer

Let α and β be the zeros of the required polynomial f(x). Then,

$\sf \dashrightarrow (\alpha + \beta) = \sqrt{2} \: \: and \: \: \alpha \beta = -12$

$\sf \dashrightarrow f (x) = {x}^{2} - (\alpha + \beta)x + \alpha \beta$

$\sf \dashrightarrow {x}^{2} - \sqrt{2} x - 12$

So, the required polynomial is,

$\sf \dashrightarrow f (x) = {x}^{2} - \sqrt{2} x - 12$

Now,

$\sf \dashrightarrow f (x) = {x}^{2} - \sqrt{2} x - 12$

$\sf \dashrightarrow {x}^{2} - 3 \sqrt{2} x + 2 \sqrt{2} x - 12$

$\sf \dashrightarrow x (x - 3 \sqrt{2}) + 2 \sqrt{2} (x - 3 \sqrt{2})$

$\sf \dashrightarrow (x - 3 \sqrt{2}) (x + 2 \sqrt{2})$

So, $\sf f (x) = 0$

$\sf \dashrightarrow (x - 3 \sqrt{2}) (x + 2 \sqrt{2}) = 0$

$\sf \dashrightarrow x - 3 \sqrt{2} = 0 \: \: or \: \: x + 2 \sqrt{2} = 0$

$\sf \dashrightarrow x = 3 \sqrt{2} \: \: or \: \: x = -2 \sqrt{2}$

Thus, the required polynomial is $\sf f (x) = {x}^{2} - \sqrt{2} x - 12$ whose zeros are $\sf 3 \sqrt{2}$ and $\sf -2 \sqrt{2}$.