Question

$\texttt{\textsf{\Large{\underline{Maths Question:}}}}$

Solve for x. Given that —
$\tt \implies \sqrt{x + 2 \sqrt{x + 2 \sqrt{x + 2 \sqrt{3x} } } } = x$
​

Answer 1

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

Given equation is

$\rm :\longmapsto\:\sqrt{x + 2 \sqrt{x + 2 \sqrt{x + 2 \sqrt{3x} } } } = x$

Let assume that,

$\red{\rm :\longmapsto\:x + 2 = y}$

So, given equation can be rewritten as

$\rm :\longmapsto\:\sqrt{y\sqrt{y \sqrt{y\sqrt{3(y - 2)} } } } = y - 2$

Now, Squaring both sides, we get

$\rm :\longmapsto\:\ \: y\sqrt{y \sqrt{y\sqrt{3(y - 2)} } } = (y - 2)^{2}$

$\rm :\longmapsto\:\ \: \sqrt{y \sqrt{y\sqrt{3(y - 2)} } } = \dfrac{ {(y - 2)}^{2} }{y}$

On squaring both sides, we get

$\rm :\longmapsto\:\ \:y \sqrt{y\sqrt{3(y - 2)} } = \dfrac{ {(y - 2)}^{4} }{ {y}^{2} }$

$\rm :\longmapsto\:\ \:\sqrt{y\sqrt{3(y - 2)} } = \dfrac{ {(y - 2)}^{4} }{ {y}^{3} }$

Again, On squaring both sides, we get

$\rm :\longmapsto\:\ \:y\sqrt{3(y - 2)} = \dfrac{ {(y - 2)}^{8} }{ {y}^{6} }$

$\rm :\longmapsto\:\ \:\sqrt{3(y - 2)} = \dfrac{ {(y - 2)}^{8} }{ {y}^{7} }$

Again, on squaring both sides,

$\rm :\longmapsto\:\ \:3(y - 2) = \dfrac{ {(y - 2)}^{16} }{ {y}^{14} }$

$\rm :\implies\:3(y - 2) {y}^{14} = {(y - 2)}^{16}$

$\rm :\implies\:3(y - 2) {y}^{14} - {(y - 2)}^{16} = 0$

$\rm :\longmapsto\:(y - 2)\bigg(3 {y}^{14} - {(y - 2)}^{15} \bigg) = 0$

$\rm :\longmapsto\:y - 2 = 0 \: \: or \: \: \bigg(3 {y}^{14} - {(y - 2)}^{15} \bigg) = 0$

$\rm :\implies\:y = 2$

$\rm :\implies\:x + 2= 2$

$\bf\implies \:x = 0$

Now, Consider

$\rm :\longmapsto\: 3 {y}^{14} - {(y - 2)}^{15}= 0$

Now, by hit and trial method,

If y = 5, then

$\rm :\longmapsto\: 3 \times {5}^{14} - {(5 - 2)}^{15}= 0$

$\rm :\longmapsto\:18310546875 - 18310546875 = 0$

$\rm :\implies\:0 = 0$

$\rm :\implies\:y = 5$

$\rm :\implies\:x + 2 = 5$

$\rm :\implies\:x = 5 - 2$

$\bf :\implies\:x = 3$

So,

Solution of

$\rm :\longmapsto\:\sqrt{x + 2 \sqrt{x + 2 \sqrt{x + 2 \sqrt{3x} } } } = x$

is

$\boxed{ \bf{ \: \: \: x = 0 \: \: \: or \: \: \: x = 3 \: \: \: \: }}$

Verification :-

When x = 3

$\rm :\longmapsto\:\sqrt{3 + 2 \sqrt{3 + 2 \sqrt{3 + 2 \sqrt{9} } } } = 3$

$\rm :\longmapsto\:\sqrt{3 + 2 \sqrt{3 + 2 \sqrt{3 + 6 } } } = 3$

$\rm :\longmapsto\:\sqrt{3 + 2 \sqrt{3 + 2 \sqrt{9 } } } = 3$

$\rm :\longmapsto\:\sqrt{3 + 2 \sqrt{3 + 6 } } = 3$

$\rm :\longmapsto\:\sqrt{3 + 2 \sqrt{9 } } = 3$

$\rm :\longmapsto\:\sqrt{3 + 6} = 3$

$\rm :\longmapsto\:\sqrt{9} = 3$

$\rm :\longmapsto\:3 = 3$

Verified

Answer 2

Answer:

If x=0 which seems very trivial, let us put in the left side to find if we arrive at the right side answer working from the inner most square root.

3*0=0 and its square root is 0. 0*2=0 and square root again 0. 2*0=0 and so on.

LHS becomes 0 which is =0 on the RHS.

If x=3, 3*3=9, its square root is 3, considering only the positive root throughout.

3+(2*3) =9 whose square root is 3.

3+(2*3) once again =9 square root is 3

once again, 3+(2*3) =9 square root is 3 which is the right hand side.

Next one which might suffice is x=12,

36 square root is 6. 6*2=12+12+24 and its square root is 2Sqaure root of 6. Though innermost square root gets simplified outer one lands up in a square root where as the RHS must be devoid of square root. If we proceed further too, we may not succeed. Likewise, one can keep working with other numbers to find other possible solutions.

Thus, x must be 0&3.

THANKS!