If $$p=\sqrt{72-\sqrt{72-\sqrt{72-\sqrt{72-......\infty}}}}$$, then find the value of $$2p^2+1$$.
Expression : $$p=\sqrt{72-\sqrt{72-\sqrt{72-\sqrt{72-......\infty}}}}$$
=> $$p=\sqrt{72-p}$$
Squaring both sides, we get :
=> $$p^2=72-p$$
=> $$p^2+p-72=0$$
=> $$p^2+9p-8p-72=0$$
=> $$p(p+9)-8(p+9)=0$$
=> $$(p+9)(p-8)=0$$
=> $$p=-9,8$$
But $$p$$ cannot be negative, thus $$p=8$$
To find : $$2p^2+1$$
= $$2(8)^2+1=128+1=129$$
=> Ans - (C)
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