# How do you solve 4= sqrt x - 3root4x?

Mar 14, 2018

$x = {4}^{4} = 256$

#### Explanation:

$4 = \sqrt{x} - 3 \sqrt[4]{x}$

Set $y = \sqrt[4]{x}$, $\sqrt{x}$ became ${y}^{2}$. Hence,

${y}^{2} - 3 y = 4$

${y}^{2} - 3 y - 4 = 0$

$\left(y - 4\right) \cdot \left(y + 1\right) = 0$

So ${y}_{1} = - 1$ and ${y}_{2} = 4$

But $x = {\left(- 1\right)}^{4} = 1$ doesn't provide equation. Thus,

$x = {4}^{4} = 256$ is solution of it.

Mar 14, 2018

$\text{ The Soln. Set "sub RR" is } \left\{256\right\}$.

#### Explanation:

We will solve the eqn. in $\mathbb{R}$.

Letting, $\sqrt[4]{x} = t , x = {t}^{4} \therefore \sqrt{x} = {t}^{2}$.

Substituting in the given eqn., we get,

$4 = {t}^{2} - 3 t , \mathmr{and} , {t}^{2} - 3 t - 4 = 0$.

$\therefore \underline{{t}^{2} - 4 t} + \underline{t - 4} = 0$.

$\therefore t \left(t - 4\right) + 1 \left(t - 4\right) = 0$.

$\therefore \left(t - 4\right) \left(t + 1\right) = 0$.

$\therefore t = 4 , \mathmr{and} , t = - 1$.

$\therefore x = {t}^{4} = {4}^{4} = 256 , \mathmr{and} , x = {\left(- 1\right)}^{4} = 1$.

But, we observe that, $x = 1$ does not satisfy the original eqn.

$\therefore \text{ The Soln. Set "sub RR" is } \left\{256\right\}$.