Question

Solve the following equations:
$x^2+xy+y^2=84$,
$x-\sqrt{xy}+y=6$.

• A. $x=8,y=2$
• B. $x=3,y=4$
• C. $x=2,y=8$
• D. $x=3,y=-4$

Answer 1

Answer: (A), (C)

$x=8,y=2$
$x=2,y=8$

$x-\sqrt{xy}+y=6x+y=6+\sqrt{xy}$ ........(i)

$x^2+y^2+xy=84$ ......(ii)

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

${(x+y)}^2-xy=84$ ........(iii)

Substituting (i) in (iii), we get

${(6+\sqrt{xy})}^2-xy=8436+xy+12\sqrt{xy}-xy=8412\sqrt{xy}=48\sqrt{xy}=4\Rightarrow xy=16y=\frac{16}{x}$

Substituting $y$ in (ii), we get

$x^2+{\left(\frac{16}{x}\right)}^2+x.\frac{16}{x}=84$

$\Rightarrow x^2+\frac{256}{x^2}=68$

$\Rightarrow x^4+256=68x^2$

$\Rightarrow x^4-68x^2+256=0$

$\Rightarrow x^4-4x^2-64x^2+256=0$

$\Rightarrow x^2(x^2-4)-64(x^2-4)=0$

$\Rightarrow (x^2-4)(x^2-64)=0$

$\Rightarrow x^2=4,64$

$\Rightarrow x=\sqrt{4},\sqrt{64}$

$\Rightarrow x=\pm 2,\pm 8$

Negative values do not satisfy equation (i), so they are not included.

So, $x=2,8$

Now $y=\frac{16}{x}$

$\Rightarrow x=2$

$\Rightarrow y=\frac{16}{2}=8$

$\Rightarrow x=8$

$\Rightarrow y=\frac{16}{8}=2$

So, the values of $x$ are $2,8$ and corresponding values of $y$ are $8,2$.