[ Quadratic Equations. Vieta's Theorem]

The equation $x^{2} + p x + q = 0$ has roots $x_{1}$ and $x_{2}$. Write the equation whose roots will be the numbers $y_{1}, y_{2}$ equal to:
a) $x_{1}^{3}, x_{2}^{3}$;
b) $\frac{1}{x_{1}^{2}}, \frac{1}{x_{2}^{2}}$
c) $x_{1} + \frac{1}{x_{2}}, x_{2} + \frac{1}{x_{1}}$
d) $\frac{x_{2}}{x_{1}}, \frac{x_{1}}{x_{2}}$
All calculations are based on formulas from problem 60924.

a) $x_{1}^{3}+x_{2}^{3}=3 p q-p^{3}, x_{1}^{3} x_{2}^{3}=q^{3}$. According to the converse of Vieta's theorem, the required equation is: $x^{2}+\left(p^{3}-3 p q\right) x$ $+q^{3}=0$.
b) $\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{p^{2}-2 q}{q^{2}}, \frac{1}{x_{1}^{2}} \cdot \frac{1}{x_{2}^{2}}=\frac{1}{q^{2}}$. Therefore, the required equation is: $q^{2} x^{2}+\left(2 q-p^{2}\right) x+1=0$.

c) $\left(x_{1}+\frac{1}{x_{2}}\right)+\left(x_{2}+\frac{1}{x_{1}}\right)=\left(x_{1}+x_{2}\right)+\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)=-p-\frac{p}{q}$,

$\left(x_{1}+\frac{1}{x_{2}}\right)\left(x_{2}+\frac{1}{x_{1}}\right)=q+\frac{1}{q}+2=\frac{(q+1)^{2}}{q}$. Therefore, the required equation is: $q x^{2}+p(q+1) x+(q+1)^{2}=$ 0.

d) $\frac{x_{2}}{x_{1}}+\frac{x_{1}}{x_{2}}=\frac{x_{1}^{2}+x_{2}^{2}}{x_{1} x_{2}}=\frac{p^{2}-2 q}{q}, \frac{x_{2}}{x_{1}} \cdot \frac{x_{1}}{x_{2}}=1$. Therefore, the required equation is: $q x^{2}+\left(2 q-p^{2}\right) x+q=0$.

## Answer

a) $x^{2}+\left(p^{3}-3 p q\right) x+q^{3}=0$
b) $q^{2} x^{2}+\left(2 q-p^{2}\right) x+1=0$
c) $q x^{2}+p(q+1) x+(q+1)^{2}=0$;
d) $q x^{2}+\left(2 q-p^{2}\right) x+q=$ 0.

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