# How do you solve?

## $\sqrt[4]{194 - x} + \sqrt[4]{x} = 4$ ?

Apr 6, 2018

color(red)(x=97+-56sqrt3
color(red)(a^4+b^4=(a^2+b^2)^2-2a^2b^2

#### Explanation:

Here,

$\sqrt[4]{194 - x} + \sqrt[4]{x} = 4$

Let, $\textcolor{red}{a} = \sqrt[4]{194 - x}$ andcolor(red) b=root(4)x=>color(red)(a^4)=194-x andcolor(red)( b^4)=x

:.color(blue)(a+b=4...to(I) and a^4+b^4=194...to(II)

Squaring $\left(I\right) \implies {\left(a + b\right)}^{2} = {4}^{2} \implies {a}^{2} + 2 a b + {b}^{2} = 16$

i.e. color(blue)(a^2+b^2=16-2ab...to(III)

From $\left(I I\right) \to {a}^{4} + {b}^{4} = 194$,

=>color(brown)((a^2+b^2)^2-2a^2b^2=194

$\implies {\left(16 - 2 a b\right)}^{2} - 2 {a}^{2} {b}^{2} = 194. . . \to$using $\left(I I I\right)$

$\implies 256 - 64 a b + 4 {a}^{2} {b}^{2} - 2 {a}^{2} {b}^{2} = 194$

$\implies 2 {a}^{2} {b}^{2} - 64 a b + 256 = 194$

$\implies {a}^{2} {b}^{2} - 32 a b + 128 = 97$

$\implies {a}^{2} {b}^{2} - 32 a b + 256 = 97 + 128 = 225$

$\implies {\left(a b - 16\right)}^{2} = {15}^{2}$

$\implies a b - 16 = \pm 15$

=>ab=16+-15=>color(blue)(ab=1 or ab=31

If $a b = 1 , t h e n , a = \frac{1}{b} \to$ $\text{color(red)"Please see the comment below}$

From $\left(I\right) \to \frac{1}{b} + b = 4 \implies 1 + {b}^{2} = 4 b$

$\implies {b}^{2} - 4 b + 1 = 0 \implies {b}^{2} - 4 b + 4 = 3 \implies {\left(b - 2\right)}^{2} = {\left(\sqrt{3}\right)}^{2}$

$\implies b - 2 = \pm \sqrt{3} \implies b = 2 \pm \sqrt{3}$

${b}^{2} = {\left(2 \pm \sqrt{3}\right)}^{2} = 4 \pm 4 \sqrt{3} + 3 = 7 \pm 4 \sqrt{3}$

${b}^{4} = {\left(7 \pm 4 \sqrt{3}\right)}^{2} = 49 \pm 56 \sqrt{3} + 48$

color(red)(x=97+-56sqrt3

Note:If $a b = 31 \implies a = \frac{31}{b}$

From $\left(I\right) \to \frac{31}{b} + b = 4 \implies {b}^{2} - 4 b + 31 = 0$

triangle=16-124<0=>color(blue)(ab!=31