Let and be two real roots of the equation (k+1)^2x-√2 x=(1-k), where k(-1) and are real nu…

Question

Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1)	an^2x-\sqrt2\lambda	an x=(1-k)$, where $k(
e-1)$ and $\lambda$ are real numbers. If $	an^2(\alpha+\beta)=50$, then a value of $\lambda$ is:

Accepted Answer

$\tan\alpha+\tan\beta=\dfrac{\sqrt2\lambda}{k+1}$, $\tan\alpha\tan\beta=\dfrac{k-1}{k+1}$.

$$\tan(\alpha+\beta)=\frac{\frac{\sqrt2\lambda}{k+1}}{1-\frac{k-1}{k+1}}=\frac{\sqrt2\lambda}2=\frac\lambda{\sqrt2}.$$

$$\tan^2(\alpha+\beta)=\frac{\lambda^2}2=50 \implies \lambda=10.$$

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