Question

if one of the zeroes of the polynomial f(x)=(k square +8)x square +13x+6k is reciprocal of the other, then k is
(a) 4
(b) 2
(c) both (a)&(b)
(d) None of these

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Answer 1

Answer:

C

Step-by-step explanation: product of roots =1

That is c÷a=1

C=6k

a= k2+8

Answer 2

Answer:

 option (c)  correct  k=4  or  k=2

Step-by-step explanation:

Given,

$f\left ( x \right )=\left ( k^{2} +8\right )x^{2}+13x+6k=0$

$\alpha ,\beta$    are the roots of f(x)

Roots are reciprocal to each other

$\beta =1/\alpha$

sum of roots      $\alpha +\beta =-\left ( 13 \right )/\left ( k^{2}+8 \right )$

product of roots   $\alpha \times \beta =6k/\left ( k^{2}+8 \right )$

$\alpha \times 1/\alpha =6k/\left ( k^{2}+8 \right )$

$1=6k/(k^{2}+8)$

$k^{2}+8=6k$

$k^{2}-6k+8=0$

$k^{2}-4k-2k+8=0$

$k\left ( k-4 \right )-2\left ( k-4 \right )=0$

$\left ( k-4 \right )\left ( k-2 \right )=0$

If

$k-4=0$

$k=4$

if

$k-2=0$

$k=2$

k=4 or k=2