JEE Main 2018 — Quadratic Equation Question with Solution

$x^2+\left(2-\lambda \right)x+\left(10-\lambda \right)=0$

$\Rightarrow \alpha +\beta =\lambda -2\& \alpha \beta =10-\lambda ..........\left(i\right)$

Let roots are $\alpha \mathrm{a}\mathrm{n}\mathrm{d} \beta$.

$\Rightarrow {\alpha }^3+{\beta }^3={\left(\alpha +\beta \right)}^3-3\alpha \beta (\alpha +\beta )$

$={\left(\lambda -2\right)}^3-3\left(10-\lambda \right)(\lambda -2)$

$= {\lambda }^3-{6\lambda }^2+12\lambda -8-3(10\lambda -{\lambda }^2-20+2\lambda )$

$= {\lambda }^3-{3\lambda }^2-24\lambda +52$

$\frac{dz}{d\lambda }=3{\lambda }^2-6\lambda -24=3\left({\lambda }^2-2\lambda -8\right)$ (where, $z={\alpha }^3+{\beta }^3$)

For maximum and critical points, derivative must be zero.

$\Rightarrow {\lambda }^2-2\lambda -8=0$

$\Rightarrow \left(\lambda -4\right)\left(\lambda +2\right)=0$

$\Rightarrow \lambda =-\mathrm{2,} 4$

Now, $\frac{d^{2}z}{d{\lambda }^2}=6\lambda -6$

For $\left(\lambda =-2\right), \frac{d^{2}z}{d{\lambda }^2}<0\Rightarrow {\alpha }^3+{\beta }^3$ is maximum and

For $(\lambda =4), \frac{d^{2}z}{d{\lambda }^2}>0\Rightarrow {\alpha }^3+{\beta }^3$ is minimum.

$\Rightarrow$Equation will be $x^2-2x+6=0$.

Using quadratic formula, we get

$x=\frac{2\pm \sqrt{{\left(-2\right)}^2-4\times 1\times 6}}{2\times 1}=\frac{2\pm \sqrt{-20}}{2}=\frac{2\pm 2\sqrt{5}i}{2}=1\pm \sqrt{5}i$

Thus, difference of roots is $\left|\alpha -\beta \right|=2\sqrt{5}$