Question

If p(x)=3x^2-9, find the value of p(a+1/a) where a is not equal to 0

Answer 1

Given:  $p(x)=3x^{2} -9$

To find : $p(a+\frac{1}{a} )$

Step by step solution :

Substitute $x=a+\frac{1}{a}$  in  $p(x)=3x^{2} -9$

$p(a+\frac{1}{a} )$ = $3(a+\frac{1}{a}) ^{2}-9$

$p(a+\frac{1}{a} )$ = $3(a^{2} +\frac{1}{a^{2} }+2) -9$

$p(a+\frac{1}{a} )$ = $3a^{2} +\frac{3}{a^{2} }+6 -9$

$p(a+\frac{1}{a} )$ = $\frac{3a^{4}+3+6a^{2}-9a^{2} }{a^{2} }$

$p(a+\frac{1}{a} )$ = $\frac{3a^{4}-3a^{2} +3 }{a^{2} }$

Hence the value of $p(\frac{a+1}{a} )$   is $\frac{3a^{4}-3a^{2}+3 }{a^{2} }$

Answer 2

Answer:

Here is your answer.

Step-by-step explanation:

I hope it helps.