solve-for-x-R-3-3-3-3-3-x-x-

Question Number 177873 by mr W last updated on 10/Oct/22

s o l v e f o r x \in R

\sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + x}}}}} = x

Answered by Frix last updated on 10/Oct/22

for given a ⩾ 0 we need x so that

\sqrt{a + x} = x \Rightarrow x = \frac{1 + \sqrt{4 a + 1}}{2}

in this case a = 3 \Rightarrow x = \frac{1 + \sqrt{13}}{2}

Commented by mr W last updated on 10/Oct/22

t h a n k s f o r s o l v i n g!

Commented by Frix last updated on 10/Oct/22

you are welcome

at first I had been confused to get the same

solution for all these

\sqrt{3 + x} = x

\sqrt{3 + \sqrt{3 + x}} = x

\dots

\sqrt{3 + \sqrt{3 \underset{(n times)}{+ \dots +} \sqrt{3 + x}}} = x

but then I saw {it}^{'} s obvious

Answered by a.lgnaoui last updated on 11/Oct/22

3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + x}}}} = x^{2}

3 + \sqrt{3 + \sqrt{3 + \sqrt{3 + x}}} = {(x^{2} - 3)}^{2}

\sqrt{3 + \sqrt{3 + \sqrt{3 + x}}} = {(x^{2} - 3)}^{2} - 3

3 + \sqrt{3 + \sqrt{3 + x})} = {[{(x^{2} - 3)}^{2} - 3]}^{2}

\sqrt{3 + \sqrt{3 + x}} = {[{(x^{2} - 3)}^{2} - 3]}^{2} - 3

{3 + \sqrt{x + 3} = {[{(x^{2} - 3)}^{2} - 3)}^{2} - 3]}^{2}

{x {+ 3 = {[{(x^{2} - 3)}^{2} - 3)}^{2} - 3]}^{2} - 3]}^{2}

{x^{2} {{= {[{(x^{2} - 3)}^{2} - 3)}^{2} - 3)}^{2} - 3)}^{2} - 3]}^{2}

{x^{2} - {3 {= {[{(x^{2} - 3)}^{2} - 3)}^{2} - 3)}^{2} - 3)}^{2} - 3)]}^{2} - 3

p o s o n s t = {(x^{2} - 3)}^{2}

{{({t {- 3)}^{2} - 3)}^{2} - 3)}^{2} - 3)}^{2}

{{({t^{2} - 6 t + 6)}^{2} - 3)}^{2} - 3)}^{2} - 3

{t^{4} {- 24 t^{3} + 48 t^{2} - 72 t + 30)}^{2} - 3)}^{2} - 3

{[{(t - 0, 57541)}^{2} - 3)}^{2}] - \overset{}{3}

{x^{2} - 3 = \sqrt{t} = {(t - 0, 57541)}^{4} - 6 (t - 0, 57541)]}^{2} + 6

{t = {[t - 0, 57541)}^{4} - 6 {(t - 0, 57541)}^{2} + 6]}^{2}

\dots \dots .

t o c o m p l e t e \dots

Commented by a.lgnaoui last updated on 11/Oct/22

{[{(t - 0, 56541)}^{4} - 6 {(t - 0, 57541)}^{2} + 6]}^{2} - (t - 0, 57541) + 0, 57541 = 0

(Z^{4} - 6 Z^{2}) - Z + 0, 57541 = 0

{(x^{2} - 3)}^{2} = t x^{2} = \sqrt{t^{}} + 3