Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\alpha$ be a common root of the equations $x^3-2 x-25 \lambda=0$, $3 x^3-8 x-\frac{175}{3} \lambda=0$ and $\lambda>0$. Then $\lambda=$
Options:
Solution:
2700 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{5 \sqrt{5}}$

$$
\begin{aligned}
& \lambda=\frac{\alpha^3-2 \alpha}{25} \& \lambda=\frac{9 \alpha^3-24 \alpha}{175} \\
& \Rightarrow \frac{\alpha^3-2 \alpha}{25}=\frac{9 \alpha^3-24 \alpha}{175} \\
& \Rightarrow 7 \alpha^3-14 \alpha=9 \alpha^3-24 \alpha \\
& \Rightarrow 2 \alpha^3-10 \alpha=0 \\
& \Rightarrow \alpha^3-5 \alpha=0 \\
& \Rightarrow \alpha\left(\alpha^2-5\right)=0
\end{aligned}
$$
Here we take $\alpha=+\sqrt{5}$ then
$$
\begin{aligned}
& \lambda=\frac{(\sqrt{5})^3-2 \times \sqrt{5}}{25} \Rightarrow \lambda=\frac{5 \sqrt{5}-2 \sqrt{5}}{25} \\
& \Rightarrow=\lambda \frac{3 \sqrt{5}}{5 \times 5} \Rightarrow \lambda=\frac{3}{5 \sqrt{5}}
\end{aligned}
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.