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,

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

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

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

$\sf \therefore f(x) = 0$

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

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

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

Hence, the required polynomial is,

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

Its zeros are $\sf 3 \sqrt{2}$ and $\sf - 2 \sqrt{2}$