[JEE Advanced 2009] the smallest value of k for which both roots of x 2 8kx 16
The smallest value of $k$ for which both roots of $x^2-8kx+16(k^2-k+1)=0$ are real, distinct and at least $4$ is ____
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 2d ago
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Answer: $k=2$.
Distinct $\Rightarrow k>1$; $f(4)\ge0\Rightarrow(k-1)(k-2)\ge0$; vertex $4k\ge4\Rightarrow k\ge1$. Together $k\ge2$, smallest $k=2$.
JEE Advanced 2009 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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