1. If $$x=\sqrt{15\sqrt{15\sqrt{15\sqrt{15\sqrt{15....\infty}}}}}$$ and x>0, then find the value of $$x^2+4$$
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By: anil on 05 May 2019 03.00 pm
Expression : $$x=sqrt{15sqrt{15sqrt{15sqrt{15sqrt{15....infty}}}}}$$ => $$x=sqrt{15x}$$ Squaring both sides, we get : => $$x^2=15x$$ => $$x=15$$ To find : $$x^2+4$$ = $$(15)^2+4=225+4=229$$ => Ans - (D)
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