
4. Determine all pairs of positive real numbers $(a, b)$ with $a>b$ that satisfy the following equations:

$$
a \sqrt{a}+b \sqrt{b}=134 \quad \text { and } \quad a \sqrt{b}+b \sqrt{a}=126 .
$$


4. For convenience, write $y=\sqrt{a}$ and $z=\sqrt{b}$. The equations transform to

$$
y^{3}+z^{3}=134 \quad \text { and } \quad y^{2} z+y z^{2}=126
$$

Combining these two equations in a handy way, we find

$$
(y+z)^{3}=\left(y^{3}+z^{3}\right)+3\left(y^{2} z+y z^{2}\right)=134+3 \cdot 126=512=8^{3} .
$$

This immediately implies that $y+z=8$.

Rewrite the first equation as follows: $(y+z) y z=y^{2} z+y z^{2}=126$. Since $y+z=8$, we see that $y z=\frac{126}{8}=\frac{63}{4}$.

From $y+z=8$ and $y z=\frac{63}{4}$, we can determine $y$ and $z$ by solving a quadratic equation: $y$ and $z$ are precisely the roots of the equation $x^{2}-8 x+\frac{63}{4}=0$. The two solutions are $\frac{8 \pm \sqrt{64-4 \cdot \frac{63}{4}}}{2}$, that is $\frac{9}{2}$ and $\frac{7}{2}$.

Since $a>b$, also $y>z$ holds. Hence $y=\frac{9}{2}$ and $z=\frac{7}{2}$. We therefore find that $(a, b)=\left(\frac{81}{4}, \frac{49}{4}\right)$.

Because $(a, b)=\left(\frac{81}{4}, \frac{49}{4}\right)$ satisfies the given equations, as required, we conclude that this is the (only) solution.
