Question

20%5Csqrt%7B4x%20%2B%20%20%5Csqrt%7B4x%20%2B%20%20%5Csqrt%7B4x%20%2B%20%20%5Csqrt%7B...%7D%20%7D%20%7D%20%7D%20%20%5C%3A%20pls%20%5C%3A%20help%20%5C%3A%20me" id="TexFormula1" title=" \sqrt{x \sqrt{x \sqrt{x \sqrt{x....} } } } = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{...} } } } \: pls \: help \: me" alt=" \sqrt{x \sqrt{x \sqrt{x \sqrt{x....} } } } = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{...} } } } \: pls \: help \: me" align="absmiddle" class="latex-formula">
pls help me ......
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Answer 1

First observe that if $a+b>0$,

$(a + b)^2 = a^2 + 2ab + b^2 \\\\ \implies a + b = \sqrt{a^2 + 2ab + b^2} = \sqrt{a^2 + ab + b(a + b)} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b(a+b)}}} \\\\ \implies a + b = \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{a^2 + ab + b \sqrt{\cdots}}}}$

Let $a=0$ and $b=x$. It follows that

$a+b = x = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}}$

Now let $b=1$, so $a^2+a=4x$. Solving for $a$,

$a^2 + a - 4x = 0 \implies a = \dfrac{-1 + \sqrt{1+16x}}2$

which means

$a+b = \dfrac{1 + \sqrt{1+16x}}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + \sqrt{\cdots}}}}$

Now solve for $x$.

$x = \dfrac{1 + \sqrt{1 + 16x}}2 \\\\ 2x = 1 + \sqrt{1 + 16x} \\\\ 2x - 1 = \sqrt{1 + 16x} \\\\ (2x-1)^2 = \left(\sqrt{1 + 16x}\right)^2$

(note that we assume $2x-1\ge0$)

$4x^2 - 4x + 1 = 1 + 16x \\\\ 4x^2 - 20x = 0 \\\\ 4x (x - 5) = 0 \\\\ 4x = 0 \text{ or } x - 5 = 0 \\\\ \implies x = 0 \text{ or } \boxed{x = 5}$

(we omit $x=0$ since $2\cdot0-1=-1\ge0$ is not true)