Algebra
Quadratic Equations
MMTS_Full_Test_21
Grade 12
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 $6(\alpha^3+\beta^3)^2=$
Step-by-Step Solution
Key Concept: Use $\alpha+\beta=-\lambda/3$, $\alpha\beta=-1/3$; find $\lambda$ from condition
$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2}=\frac{\lambda^2/9+2/3}{1/9}=\lambda^2+6=15\Rightarrow\lambda=3$. $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=(-1)^3+3\cdot(-1/3)\cdot(-1)=0$... Actually $6(\alpha^3+\beta^3)^2=24$.
Correct Answer: 24