Question

Solve the following equations:
$y^2(4x^2-108)=x(x^3-9y^3),2x^2+9xy+y^2=108$.

• A. $(\pm 2\sqrt{3},\mp 2\sqrt{3})$
• B. $(\pm \sqrt{3},\mp \sqrt{3})$
• C. $(\pm 2\sqrt{5},2\mp \sqrt{5})$
• D. $(2,5)$

Answer 1

The correct option is A $(\pm 2\sqrt{3},\mp 2\sqrt{3})$

$2x^2+9xy+y^2=\mathrm{108......}\left(i\right)y^2(4x^2-108)=x(x^3-9y^3)$

substituting $\left(i\right)$ in place of $108$

$y^2(4x^2-2x^2-9xy-y^2)=x^4-9x{y}^{3}2x^2{y}^2-9x{y}^3-y^4=x^4-9x{y}^3{x}^4+y^4-2x^2{y}^2=0{(x^2-y^2)}^2=0x^2=y^2\Rightarrow y=\pm x$

substituting $y$ in $\left(i\right)$

(1) $y=x$

$2x^2+9x\left(x\right)+x^2=10812x^2=108x^2=9x=\pm 3\Rightarrow y=\pm 3$

(2) $y=-x$

$2x^2+9x(-x)+x^2=108-6x^2=108x^2=-18x=\pm \sqrt{-18}=\pm 2\sqrt{3}y=-x\Rightarrow y=\mp 2\sqrt{3}$

So the values of $x$ are $\pm 3,\pm 2\sqrt{3}$ and corresponding values of $y$ are $\pm 3,\mp 2\sqrt{3}$