4. Let $k$ be one of the quotients of the roots of the quadratic equation $\left.p x^{2}-q x\right\lrcorner q=0$, where $p, q>0$. Express the roots of the equation $\sqrt{p} x^{2}-\sqrt{q} x+\sqrt{p}=0$ in terms of $k$ (not in terms of $p$ and $q$).
Solution. If $a$ and $b$ are roots of the equation $p x^{2}-q x+q=0$, then from the condition of the problem we have $\frac{a}{b}=k$, and according to Vieta's formulas we have $a+b=\frac{q}{p}$, $a b=\frac{q}{p}$. Since $p$ and $q$ are positive real numbers, from the second equation we get that $a$ and $b$ have the same sign. From the first equation, we get that $a$ and $b$ are positive. According to this, if we take the square root of the second equation, we get $\sqrt{a b}=\sqrt{\frac{q}{p}}$. If we divide the equations $a+b=\frac{q}{p}$ by $\sqrt{a b}=\sqrt{\frac{q}{p}}$ (left by left, right by right side), we get $\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}=\sqrt{\frac{q}{p}}$. If we use the condition $\frac{a}{b}=k$, we get $\sqrt{k}+\sqrt{\frac{1}{k}}=\sqrt{\frac{q}{p}}$, i.e., $\sqrt{k}+\frac{1}{\sqrt{k}}=\frac{\sqrt{q}}{\sqrt{p}}$.

If we multiply the last obtained equation by $\sqrt{k} \sqrt{p}$, we get

$$
\sqrt{p}(\sqrt{k})^{2}-\sqrt{q} \sqrt{k}+\sqrt{p}=0
$$

Thus, $\sqrt{k}$ is a root of the equation

$$
\sqrt{p} x^{2}-\sqrt{q} x+\sqrt{p}=0
$$

From the fact that $\sqrt{k}$ is a root of equation (1) and the equality $\sqrt{k}+\frac{1}{\sqrt{k}}=\frac{\sqrt{q}}{\sqrt{p}}$, we get that $\frac{1}{\sqrt{k}}$ is the second root of equation (1).

If $a_{1}, b_{1}$ are roots of equation (1), then $a_{1}=\sqrt{k}$ and $b_{1}=\frac{1}{\sqrt{k}}$.

## III year