If the sum of the squares of the reciprocals of the roots and of the equation 3x^2+ x-1=0 …

Question

If the sum of the squares of the reciprocals of the roots $\alpha$ and $\beta$ of the equation $3x^2+\lambda x-1=0$ is 15, then $6(\alpha^3+\beta^3)^2$ is equal to:

Accepted Answer

$$\alpha+\beta=-\frac\lambda3, \qquad \alpha\beta=-\frac13.$$

$$\frac1{\alpha^2}+\frac1{\beta^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}=15.$$

$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{\lambda^2}9+\frac23.$$

$$\frac{\frac{\lambda^2}9+\frac23}{1/9}=15 \implies \lambda^2+6=15 \implies \lambda^2=9 \implies \lambda=\pm3.$$

Taking $\lambda=3$: $\alpha+\beta=-1$.
$$\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=-1-3\left(-\frac13\right)(-1)=-1-1=-2.$$

$$6(\alpha^3+\beta^3)^2=6(4)=24.$$

(The other value $\lambda=-3$ gives $\alpha^3+\beta^3=2$, same squared value.)

Solutions · 1