Question

The value of 'a' for which the sum of the squares of the roots of the equation $x^2-(a-2)x-a-1=0$ assume the least value is -

• A. $1$
• B. $0$
• C. $3$
• D. $2$

Answer 1

Answer: (A)

$1$

$x^2-(a-2)x-a-1=0$

formula for roots is

$\alpha ,\beta ==\frac{-b\sqrt{b^2-4ac}}{2a}$

$a=1,b=-(a-2),c=-a-1$

$\alpha ,\beta =\frac{-(a-2)\pm \sqrt{{-(a-2)}^2-4\left(1\right)(-a-1)}}{2\times 1}=\frac{(a-2)\pm \sqrt{{(a-2)}^2+4(a+1)}}{2}$

$=\frac{(a-2)\pm \sqrt{a^2+4-4a+4a+4}}{2}=\frac{a-2}{2}\pm \frac{\sqrt{a^2+8}}{2}$

$\alpha =\frac{a-2}{2}+\frac{\sqrt{a^2+8}}{2},\beta =\frac{a-2}{2}-\frac{\sqrt{a^2-8}}{2}$

${\alpha }^2+{\beta }^2={[\frac{a-2}{2}+\frac{\sqrt{a^2+8}}{2}]}^2+{[\frac{a-2}{2}-\frac{\sqrt{a^2-8}}{2}]}^2={\left(\frac{a-2}{2}\right)}^2+\frac{{(a^2+8)}^2}{{\left(2\right)}^2}+\frac{2\sqrt{a^2+8}}{2}\times \frac{a-2}{2}$

$=\frac{{(a-2)}^2}{2}+\frac{a^2+8}{2}=\frac{a^2-4a+4}{2}+\frac{a^2+8}{2}=\frac{a^2-4a+4+a^2+8}{2}=\frac{2a^2-4a+12}{2}=a^2-2a+6$

${\alpha }^2+{\beta }^2=a^2-2a+6$

This equation represents a parabola

Differentiate above equation and equate to zero we get minimum value

$\frac{d}{da}(a^2-2a+6)=0,2a-2=0,a=1$