JEE Advanced 2009 Paper 2Question 1
The smallest value of $k$, for which both the roots of the equation $$x^2-8kx+16(k^2-k+1)=0$$ are real, distinct and have values at least 4, is
Quadratic Equation · Location of Roots
Answer$2$
Solution
Concept. For an upward parabola, both roots are $\ge4$ when the roots are real and distinct, $f(4)\ge0$, and the axis is to the right of $4$. Formulas used. (1) Both roots $\ge k$ $\iff$ $D>0$, $f(k)\ge0$, $-\frac{b}{2a}>k$Step 1 - Distinct real roots. $\frac D4=16k^2-16(k^2-k+1)=16(k-1)>0$, so $k>1$. Step 2 - $f(4)\ge0$. $$f(4)=16-32k+16(k^2-k+1)=16(k-1)(k-2)\ge0,$$ so $k\le1$ or $k\ge2$. Step 3 - Axis. $4k>4$, so $k>1$. Step 4 - Smallest $k$. Together $k\ge2$; the smallest is $k=2$ (roots $4$ and $12$). Answer: 2