Question

if one zero of a polynomial (a square + 9) X square + 13 x + 6a is reciprocal of the Other find the value of a​

Answer 1

$\huge\frak{\underline{\underline{\frak{\red{Question}}}}}$

If one zero of a polynomial (a² + 9) x²+ 13 x + 6a is reciprocal of the other, find the value of a.

$\huge\frak{\underline{\underline{\frak{\red{To\:find}}}}}$

The value of a

$\huge\frak{\underline{\underline{\frak{\red{Solution}}}}}$

(a² + 9) x²+ 13 x + 6a

Let one zero of the polynomial be a

then the other zero is 1/a

Multiply both zeros

so, Product of zeros = $\sf a×\large\frac{1}{a}=1$

And

Product of zeros :-

$\sf \large\frac{constant\:term}{coefficient\:of\:x^2}$ = $\sf \frac{6a}{(a^2+9)}$

$\sf \large\frac{6a}{(a^2+9)}=1$

$\implies\sf a^2+9=6a$

$\implies\sf a^2+9-6a=0$

Using this identity (a-b)²=a²+b²-2ab

$\implies\sf (a-3)^2=0$

$\implies\sf a-3=0$

$\implies\sf a=3$

$\large{\boxed{\bf{a=3}}}$