x-y-1-8-x-1-4-y-1-4-1-solve-system-

Question Number 14742 by Mr Chheang Chantria last updated on 04/Jun/17

\boldsymbol x + \boldsymbol y = \frac{1}{8}

\sqrt[4]{\boldsymbol x} + \sqrt[4]{\boldsymbol y} = 1

\boldsymbol s o l v e \boldsymbol s y s t e m

Commented by Tinkutara last updated on 04/Jun/17

It is easy to verify that \frac{1}{16} + \frac{1}{16} = \frac{1}{8}

and \sqrt[4]{\frac{1}{16}} + \sqrt[4]{\frac{1}{16}} = 1

Hence x = y = \frac{1}{16}

Commented by RasheedSoomro last updated on 04/Jun/17

Let \sqrt[4]{\boldsymbol x} = p and \sqrt[4]{y} = q

\Rightarrow {\begin{cases} p^{4} + q^{4} = 1 / 8 \\ p + q = 1 \end{cases}

\Rightarrow \frac{p^{4} + q^{4}}{p + q} = \frac{1}{8} \land (p^{4} + q^{4}) (p + q) = \frac{1}{8}

\Rightarrow \frac{p^{4} + q^{4}}{p + q} - (p^{4} + q^{4}) (p + q) = 0

\Rightarrow (p^{4} + q^{4}) {\frac{1}{p + q} - (p + q)} = 0

p^{4} + q^{4} = 0 \lor p + q - \frac{1}{p + q} = 0

p^{4} = - q^{4} \lor p + q = \frac{1}{p + q}

x = - y \lor {(p + q)}^{2} = 1

p + q = \pm 1

p + q = 1 is what we have been given! hahahaa . .

I reached at the starting point again!!!

p + q = - 1 is extraneous (?)

Oh no I^{'} ve gained something x = - y

But this is again in contradiction to the given

(x + y = 1 / 8)

!!!

Continue

Commented by prakash jain last updated on 04/Jun/17

\sqrt[4]{x} = u

\sqrt[4]{y} = v

u^{4} + v^{4} = \frac{1}{8}

u + v = 1

u^{2} + v^{2} + 2 u v = 1

u^{2} + v^{2} = 1 - 2 u v

u^{4} + v^{4} + 2 u^{2} v^{2} = {(1 - 2 u v)}^{2}

\frac{1}{8} + 2 u^{2} v^{2} = 1 - 4 u v + 4 u^{2} v^{2}

1 + 16 v^{2} u^{2} = 8 - 32 u v + 32 u^{2} v^{2}

16 u^{2} v^{2} - 32 u v + 7 = 0

(4 u v - 7) (4 u v - 1) = 0

u v = \frac{1}{4} o r u v = \frac{7}{4}

u v = \frac{1}{4}

u + v = 1

{(u - v)}^{2} = {(u + v)}^{2} - 4 u v = 1 - 1 = 0

u = v = \frac{1}{2} \Rightarrow x = \frac{1}{16}, y = \frac{1}{16}

u v = \frac{7}{4}

{(u - v)}^{2} = 1 - 7 = - 6

u - v = \pm i \sqrt{6}

u = \frac{1 \pm i \sqrt{6}}{2}, v = \frac{1 \mp i \sqrt{6}}{2}

x = \frac{1 \pm 20 i \sqrt{6}}{16}, y = \frac{1 \mp 20 i \sqrt{6}}{16}