Nested radical mania

As we can see that the RHS of the equation is of the power 1/4 and the LHS is of the power 1/2. So, we need to see that if we raise the equation to power 4, it would not do any harm as the LHS will get its too "under roots" removed.

$\sqrt { \sqrt { \sqrt { x } } } =\quad \sqrt [ 4 ]{ \sqrt [ 4 ]{ \sqrt [ 4 ]{ 3{ x }^{ 4 }+4 } } }$

$Raising\quad both\quad the\quad sides\quad to\quad power\quad '4'$

$\sqrt { x } \quad =\quad \sqrt [ 4 ]{ \sqrt [ 4 ]{ 3{ x }^{ 4 }+4 } }$

$Raising\quad both\quad the\quad sides\quad again\quad to\quad power\quad '4'$

${ x }^{ 2 }\quad =\quad \sqrt [ 4 ]{ 3{ x }^{ 4 }+4 }$

$Raising\quad both\quad the\quad sides\quad again\quad to\quad power\quad '4'$

${ x }^{ 8 }\quad =\quad 3{ x }^{ 4 }+4$

This equation has now become quadratic in ' ${ x }^{ 4 }$ '. Solving the roots of this equation, we get

${ x }^{ 4 }\quad =\quad \frac { 3\pm 5 }{ 2 }$

Or, ${ x }^{ 4 }\quad =\quad 4,-1$

But, since value of ${ x }^{ 4 }$ can't be negative, Therefore,

$x\quad =\quad 4$

Cheers!!:)

x^1/8=(3x^4+4)^1/64....raise both the sides to power 64 to get x^8-3x^4-4=0 ....(x^4+1)(x^4-4)=0... x^4=-1,4 negative no. is not possible hence 4

Similar to Sonali Srivastava's:

$\sqrt {\sqrt{\sqrt {x}}} = \sqrt [ 4 ] { \sqrt [ 4 ] { \sqrt [ 4 ] {3x^4 +4 }}} \quad \Rightarrow ((x^{\frac{1}{2}})^{\frac{1}{2}})^{\frac{1}{2}} = (( (3x^4+4)^{\frac{1}{4}})^{\frac{1}{4}})^{\frac{1}{4}}$

$x^{\frac{1}{8}} = (3x^4+4)^{\frac{1}{64}} \quad \Rightarrow x^{\frac{64}{8}} = (3x^4+4)^{\frac{64}{64}} \quad \Rightarrow x^8 = 3x^4+4$

$(x^4)^2 - 3x^4 - 4 = 0 \quad \Rightarrow x^4 = \frac {3\pm \sqrt{9-4(-4)} } {2} = \frac {3+5}{2} = \boxed {4}$ (since $x > 0$ )

By taking off all radicals by rising all to the power of 64, we get a quadratic in x^4 and 4 is one of answers