Problem

Solve for a:
\[
\sqrt{4+\sqrt{16+16 a}}+\sqrt{1+\sqrt{1+a}}=6 .
\]
$\left(7 b^{3}\right)^{2} \cdot\left(4 b^{2}\right)^{-3}$, given that bis non-zero.

Answer

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Hide Steps
Answer

Solve for a: \(\boxed{a = 8}\)

Steps

Step 1 :Square both sides of the equation: \(\left(\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}\right)^2 = 6^2\)

Step 2 :Solve for a: \(\boxed{a = 8}\)

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