Question

Solve the following equations:
$x^4+y^4=272$,
$x-y=2$.

• A. $(4,2)$
• B. $(3,4)$
• C. $(-2,-4)$
• D. $(\frac{\sqrt{3}+2}{4},\frac{1}{2})$

Answer 1

Answer: (A), (C)

The correct options are
A $(4,2)$
C $(-2,-4)$

Let $x-y=2$ ......(i)

and $x^4+y^4=272$ .......(ii)

Using $a^2+b^2={(a+b)}^2-2ab$

${(x^2+y^2)}^2-2x^2{y}^2=272$

Using $a^2+b^2={(a-b)}^2+2ab$

${\{{(x-y)}^2+2xy\}}^2-2x^2{y}^2=272\Rightarrow {\{{\left(2\right)}^2+2xy\}}^2-2x^2{y}^2=272$

Put $xy=t$

${(4+2t)}^2-2t^2=272\Rightarrow 16+4t^2+8t-2t^2=272\Rightarrow {2t}^2+8t-256=0\Rightarrow t^2+4t-128=0\Rightarrow t^2-8t+16t-128=0\Rightarrow t(t-8)+16(t-8)=0\Rightarrow (t+16)(t-8)=0\Rightarrow t=-16,8$

Thus $xy=-16,8$

(a) $xy=8$

$\Rightarrow y=\frac{8}{x}$

On substituting $y$ in (i), we get
$x-\frac{8}{x}=2\Rightarrow x^2-8=2x\Rightarrow x^2-2x-8=0\Rightarrow x^2-4x+2x-8=0\Rightarrow x(x-4)+2(x-4)=0\Rightarrow x=-2,4$

We know $y=\frac{8}{x}$

Thus when $x=-2$,

$\Rightarrow y=\frac{8}{-2}=-4$

When $x=4$, then $y=\frac{8}{4}=2$

(b) $xy=-16$

$\Rightarrow y=\frac{-16}{x}$

On substituting $y$ in (i), we get

$x+\frac{16}{x}=2\Rightarrow x^2+16=2x\Rightarrow x^2-2x+16=0\Rightarrow x=\frac{2\pm \sqrt{4-4\left(1\right)\left(16\right)}}{2}=\frac{2\pm \sqrt{-60}}{2}\Rightarrow x=\frac{2\pm 2\sqrt{-15}}{2}=1\pm \sqrt{-15}$

Therefore, $y=\frac{-16}{1\pm \sqrt{-15}}$

$\Rightarrow y=\frac{-16}{1\pm \sqrt{-15}}\times \frac{1\mp \sqrt{-15}}{1\mp \sqrt{-15}}\Rightarrow y=-1\pm \sqrt{-15}$