Question

If one zero of the polynomial (k2+1) x2+16x +8k is reciprocal of the other then k is equal to

Answer 1

The polynomial is,

$\displaystyle\longrightarrow \sf{p(x)=(k^2+1)x^2+16x+8k}$

Here,

• $\displaystyle\sf {a=k^2+1}$

• $\displaystyle\sf {b=16}$

• $\displaystyle\sf {c=8k}$

One zero of this polynomial is reciprocal to the other.

Let the zeroes be $\displaystyle\sf {\alpha}$ and $\displaystyle\sf {\dfrac {1}{\alpha}.}$

Then, product of zeroes,

$\displaystyle\longrightarrow\sf{\alpha\times\dfrac {1}{\alpha}=\dfrac {c}{a}}$

$\displaystyle\longrightarrow\sf{1=\dfrac {8k}{k^2+1}}$

$\displaystyle\longrightarrow\sf{k^2+1=8k}$

$\displaystyle\longrightarrow\sf{k^2-8k+1=0}$

Solving this quadratic equation,

$\displaystyle\longrightarrow\sf{k=\dfrac {8\pm\sqrt{(-8)^2-4\times 1\times 1}}{2\times 1}}$

$\displaystyle\longrightarrow\sf{k=\dfrac {8\pm\sqrt{64-4}}{2}}$

$\displaystyle\longrightarrow\sf{k=\dfrac {8\pm\sqrt{60}}{2}}$

$\displaystyle\longrightarrow\sf{k=\dfrac {8\pm2\sqrt{15}}{2}}$

$\displaystyle\longrightarrow\underline {\underline {\sf{k=4\pm\sqrt{15}}}}$