Question

Solve the following equations:
$x{y}^{\frac{1}{2}}+y{x}^{\frac{1}{2}}=20$,
$x^{\frac{3}{2}}+y^{\frac{3}{2}}=65$.

• A. $(1,16)$
• B. $(2,10)$
• C. $(3,15)$
• D. $(16,1)$

Answer 1

Answer: (A), (D)

$(1,16)$
$(16,1)$

$x\sqrt{y}+y\sqrt{x}=20x^{\frac{3}{2}}+y^{\frac{3}{2}}=65$

Put $x=u^2,y=v^2$

$u^{2}v+v^{2}u=20\Rightarrow uv(u+v)=20\Rightarrow u+v=\frac{20}{uv}$ ........(i)

We have$u^3+v^3=65$

Using $a^3+b^3=(a+b)(a^2+b^2-ab)$

$\Rightarrow (u+v)(u^2+v^2-uv)=65$

Substituting value of equation (i) in above equation, we get

$\frac{20}{uv}(u^2+v^2-uv)=65\Rightarrow 20(u^2+v^2)=85uv\Rightarrow u^2+v^2=\frac{17}{4}uv$

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

$\Rightarrow u^2+v^2+2uv=\frac{17}{4}uv+2uv\Rightarrow {(u+v)}^2=\frac{25}{4}uv$

Substituting (i) in above equation, we get

${\left(\frac{20}{uv}\right)}^2=\frac{25}{4}uv\Rightarrow u^3{v}^3=\frac{400\times 4}{25}=64\Rightarrow uv=4\Rightarrow v=\frac{4}{u}$

Substituting $v$ in (i), we get

$4(u+\frac{4}{u})=20\Rightarrow \frac{u^2+4}{u}=5\Rightarrow u^2-5u+4=0\Rightarrow u^2-4u-u+4=0\Rightarrow u(u-4)-1(u-4)=0\Rightarrow (u-4)(u-1)=0\Rightarrow u=1,4$

Now, $v=\frac{4}{u}$

$u=1\Rightarrow v=\frac{4}{1}=4u=4\Rightarrow v=\frac{4}{4}=1\therefore v=4,1$

According to our assumption, we have

$x=u^2\Rightarrow x=1^2,4^2\Rightarrow x=1,16$

and $y=v^2$

$\Rightarrow y=4^2,1^2\Rightarrow y=16,1$