If x=\sqrt{15\sqrt{15\sqrt{15\sqrt{15\sqrt{15....\infty}}}}} and x>0, then find the value of x^2+4

Question 3

If $$x=\sqrt{15\sqrt{15\sqrt{15\sqrt{15\sqrt{15....\infty}}}}}$$ and x>0, then find the value of $$x^2+4$$

Solution

Expression : $$x=\sqrt{15\sqrt{15\sqrt{15\sqrt{15\sqrt{15....\infty}}}}}$$

=> $$x=\sqrt{15x}$$

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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