Let 0 be in R.

Question

Let $\lambda
e0$ be in $\mathbb R$. If $\alpha,\beta$ are the roots of $x^2-x+2\lambda=0$ and $\alpha,\gamma$ are the roots of $3x^2-10x+27\lambda=0$, then $\dfrac{\beta\gamma}\lambda$ is equal to

Answer: 18

Accepted Answer

The common root $\alpha$ satisfies both equations; eliminating gives $-7\alpha+21\lambda=0\Rightarrow\alpha=3\lambda$. Substituting into $x^2-x+2\lambda=0$: $9\lambda^2-3\lambda+2\lambda=0\Rightarrow\lambda=\dfrac19$ (as $\lambda\ne0$), so $\alpha=\dfrac13$.

From $\alpha+\beta=1$: $\beta=\dfrac23$. From $\alpha+\gamma=\dfrac{10}3$: $\gamma=3$.

$$\frac{\beta\gamma}\lambda=\frac{\frac23\times3}{\frac19}=18.$$

Solutions · 1