Question

Solve the following equations:
$x^4+y^4=706$,
$x+y=8$.

• A. $(5,3)$
• B. $(3,5)$
• C. $(2,4)$
• D. $(4,2)$

Answer 1

Answer: (A), (B)

The correct options are
A $(5,3)$
B $(3,5)$

Let $x+y=8$ ......(i)

and $x^4+y^4=706$ .......(ii)

Using $a^2+b^2={(a+b)}^2-2ab$, equation (ii) becomes

${(x^2+y^2)}^2-2x^2{y}^2=706\Rightarrow {\{{(x+y)}^2-2xy\}}^2-2x^2{y}^2=706\Rightarrow {\{{\left(8\right)}^2-2xy\}}^2-2x^2{y}^2=706$

Put $xy=t$

${(64-2t)}^2-2t^2=706\Rightarrow 4096+4t^2-256t-2t^2=706\Rightarrow 2t^2-256t+3390=0\Rightarrow t^2-128t+1695=0\Rightarrow t^2-113t-15t+1695=0\Rightarrow t(t-113)-15(t-113)=0\Rightarrow t=15,113\Rightarrow xy=15,113$

(1) $xy=15$

Thus $y=\frac{15}{x}$

Substituting $y$ in (i), we get

$x+\frac{15}{x}=8\Rightarrow x^2+15=8x\Rightarrow x^2-8x+15=0\Rightarrow x^2-5x-3x+15=0\Rightarrow x(x-5)-3(x-5)=0\Rightarrow (-3)(x-5)=0\Rightarrow x=3,5$

For $x=3$

$y=\frac{15}{x}\Rightarrow y=\frac{15}{3}=5$

For $x=5$

$\Rightarrow y=\frac{15}{5}=3$

(2) $xy=113$

$\Rightarrow y=\frac{113}{x}$

Substituting $y$ in (i), we get

$x+\frac{113}{x}=8\Rightarrow x^2+113=8x\Rightarrow x^2-8x+113=0\Rightarrow x=\frac{8\pm \sqrt{64-4\left(1\right)\left(113\right)}}{2}=\frac{8\pm \sqrt{-388}}{2}\Rightarrow x=\frac{8\pm 2\sqrt{-97}}{2}=4\pm \sqrt{-97}$

Therefore, $y=\frac{113}{x}$

$\Rightarrow y=\frac{113}{4\pm \sqrt{-97}}$

$\Rightarrow y=\frac{113}{4\pm \sqrt{-97}}\times \frac{4\mp \sqrt{-97}}{4\mp \sqrt{-97}}\Rightarrow y=113\frac{4\mp \sqrt{-97}}{113}=4\mp \sqrt{-97}$