Question

plzzzzz solve following 14th question ??????

Answer 1

According to the Question;
the zeroes of the polynomial be:-
$\alpha \: and \: 1 \div \alpha$
P (x) =
$({k}^{2} + 4) {x}^{2} + 23x + 4k$
Therefore
$a = {k}^{2} + 4$
and
c = 4k.

product of zeroes = c÷a
$\alpha \times 1 \div \alpha = 4k \div ({k}^{2} + 4)$

$1 = 4k \div ({k}^{2} + 4)$

${k}^{2} - 4k + 4 = 0$
To solve this equation,

we get the value of k = 2 or 2

Therefore, the answer of this question is 2.

☆☆☆☆HOPE THIS WILL HELP YOU....