Question

`\sqrt{x \sqrt{x \sqrt{x √(x....) } } } = \sqrt{4x + \sqrt{4x + \sqrt{4x + √(...) } } } \: pls \: help \: me`

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

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

`a+b = x = \sqrt{x \sqrt{x \sqrt{x √(\cdots)}}}`

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

`a^2 + a - 4x = 0 \implies a = \frac{-1 + √(1+16x)}2`

which means

`a+b = \frac{1 + √(1+16x)}2 = \sqrt{4x + \sqrt{4x + \sqrt{4x + √(\cdots)}}}`

Now solve for
`x`.

`x = \frac{1 + √(1 + 16x)}2 \\\\ 2x = 1 + √(1 + 16x) \\\\ 2x - 1 = √(1 + 16x) \\\\ (2x-1)^2 = \left(√(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)