Question

if p(x) =3x² - 9, find the value of p(a+ 1/a) where a ≠ 0​

Answer 1

Answer:

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Answer 2

Answer:

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

Step-by-step explanation:

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

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

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

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