What are the real pairs $(a,\,b)$ that satisfy this system of equations?

[anemone]

Determine all real paris $(a,\,b)$ that satisfy the system of equations below:

$4(b^2-a^2)=3\sqrt[3]{a^2b^5}$

$b^2+a^2=5\sqrt[3]{a^4b}$

 

[mente oscura]

Determine all real paris $(a,\,b)$ that satisfy the system of equations below:

$4(b^2-a^2)=3\sqrt[3]{a^2b^5}$

$b^2+a^2=5\sqrt[3]{a^4b}$

Hello.

Spoiler

$$4(b^2-a^2)(b^2+a^2)=3\sqrt[3]{a^2b^5} \ 5\sqrt[3]{a^4b}$$

$$4(b^4-a^4)=15a^2b^2$$

$$4a^4+15a^2b^2-4b^4=0$$

$$a^2=\dfrac{-15b^2 \pm \sqrt{289b^4}}{8}$$

Real solution:

$$a^2=\dfrac{b^2}{4} \rightarrow{}a=\pm \dfrac{b}{2}$$

Regards.

 

[anemone]

Hello.

Spoiler

$$4(b^2-a^2)(b^2+a^2)=3\sqrt[3]{a^2b^5} \ 5\sqrt[3]{a^4b}$$

$$4(b^4-a^4)=15a^2b^2$$

$$4a^4+15a^2b^2-4b^4=0$$

$$a^2=\dfrac{-15b^2 \pm \sqrt{289b^4}}{8}$$

Real solution:

$$a^2=\dfrac{b^2}{4} \rightarrow{}a=\pm \dfrac{b}{2}$$

Regards.

That is a good "partial" solution, mente oscura! :pHehehe...and thanks for participating!

 

[mente oscura]

That is a good "partial" solution, mente oscura! :pHehehe...and thanks for participating!

Hello.

Yes. It is that I am a bit lazy.

Spoiler

$$4(b^2-\dfrac{b^2}{4})=3 \sqrt[3]{\dfrac{b^7}{4}}$$

Solutions: (0,0), (2,4), (-2,4)

Regards.

 

[CellCoree]

To solve this system of equations, we can start by isolating one variable in each equation. From the first equation, we can rearrange it to get $b^2-a^2=\frac{3\sqrt[3]{a^2b^5}}{4}$. From the second equation, we can rearrange it to get $b^2=5\sqrt[3]{a^4b}-a^2$.

Next, we can substitute the second equation into the first equation to get:

$5\sqrt[3]{a^4b}-a^2-a^2=\frac{3\sqrt[3]{a^2b^5}}{4}$

Simplifying this, we get:

$5\sqrt[3]{a^4b}-2a^2=\frac{3\sqrt[3]{a^2b^5}}{4}$

We can then cube both sides to get rid of the cube root:

$125a^4b-100a^6+20a^4=\frac{27a^2b^5}{64}$

Simplifying this, we get:

$2000a^6-128a^4b+27a^2b^5=0$

This can be factored into:

$(20a^2-3b)(100a^4-9b^4)=0$

From this, we can see that the solutions will be either when $20a^2=3b$ or when $100a^4=9b^4$.

If we let $x=a^2$ and $y=b^2$, we can rewrite these equations as $20x=3y$ and $100x^2=9y^2$.

Solving these equations simultaneously, we get two solutions: $(a,b)=(\frac{\sqrt{15}}{5},\frac{\sqrt{20}}{5})$ and $(a,b)=(-\frac{\sqrt{15}}{5},-\frac{\sqrt{20}}{5})$.

Therefore, the real pairs that satisfy this system of equations are $(\frac{\sqrt{15}}{5},\frac{\sqrt{20}}{5})$ and $(-\frac{\sqrt{15}}{5},-\frac{\sqrt{20}}{5})$.