Question

Find the value of

$\sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } }$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } }$

Let assume that

$\rm \: x = \sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } }$

$\rm \:As \: \sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } } > 0$

$\rm\implies \:x > 0$

Now,

$\rm \: x = \sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } }$

So, on squaring both sides, we get

$\rm \: {x}^{2} = 1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + \sqrt{1 + - - - \infty } } } }$

$\rm \: {x}^{2} - 1 =\sqrt{7 + \sqrt{1 + \sqrt{7 + \sqrt{1 + - - - \infty } } } }$

On squaring both sides, we get

$\rm \: {( {x}^{2} - 1)}^{2} = 7 + \sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } }$

$\rm \: {x}^{4} + 1 - {2x}^{2} = 7 + x$

$\rm \: {x}^{4} - {2x}^{2} - x - 6 = 0$

Now, we have to use hit and trial method to factorize this biquadratic equation.

Let assume that x = 2, we get

$\rm \: {2}^{4} - {2(2)}^{2} - 2 - 6 = 0$

$\rm \: 16 - 8 - 2 - 6 = 0$

$\rm\implies \:x - 2 \: is \: a \: factor \: of \: {x}^{4} - {2x}^{2} - x - 6 = 0$

Now, using Synthetic Division, we have

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{}}}&{{\sf{\:\: \:\:}}}\\ {\underline{\sf{2}}}& {\sf{1 \: \: \: 0 \: \: \: - 2 \: \: \: - 1 \: \: \: - 6}} \\{\sf{}}& \underline{\sf{ \: \: \: \: \: 2 \: \: \: \: \: \: \: \: 4 \: \: \: \: \: \: \: \: 4\: \: \: \: \: \: \: 6}} \\ {{\sf{}}}& {\sf{1 \: \: \: \: 2 \: \: \: \: \: \: \: \: 2 \: \: \: \: \: \: \: \:3 \: \: \: \: \: \: \: [0}} \\ {\sf{\:}}  \end{array}\end{gathered}\end{gathered}\end{gathered} \\ \end{gathered}$

So, the biquadratic equation in factorize form, can be rewritten as

$\rm\implies \:(x - 2)( {x}^{3} + {2x}^{2} + 2x + 3) = 0$

$\rm\implies \:x = 2$

Or

$\rm \: {x}^{3} + {2x}^{2} + 2x + 3 = 0$

By Descartes Rule of Sign, as this cubic equation has all sign positive. It implies, equation has no positive real root.

As x > 0, no need to find the roots of this cubic equation.

$\rm\implies \: \sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } } = 2$

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Basic Concept Used :-

Descartes Rule of sign is used to find the possible number of positive real roots or negative real roots of the equation of more than one degree.

The number of positive real roots of the equation f(x) is atmost equals to number of sign changes.

The number of negative real roots of the equation f(x) is atmost equals to number of sign changes in f(- x).

Answer 2

$\tt\sqrt{1 + \sqrt{7 + \sqrt{1 + \sqrt{7 + - - - \infty } } } } \: \: = \: 2$