Question 59

If $$p=3+\frac{1}{p}$$, the value of $$p^4+\frac{1}{p^4}$$ is

Solution

Given : $$p=3+\frac{1}{p}$$

=> $$p-\frac{1}{p}=3$$

Squaring both sides,

=> $$(p-\frac{1}{p})^2=(3)^2$$

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

=> $$p^2+\frac{1}{p^2}-2=9$$

=> $$p^2+\frac{1}{p^2}=9+2=11$$

Again squaring both sides, we get :

=> $$(p^2+\frac{1}{p^2})^2=(11)^2$$

=> $$p^4+\frac{1}{p^4}+2(p^2)(\frac{1}{p^2})=121$$

=> $$p^4+\frac{1}{p^4}=121-2=119$$

=> Ans - (D)

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