Question

The value of ${}^{\prime }{a}^{\prime }$ for which the sum of the square of the roots of $2x^2-2(a-2)x-a-1=0$ is least is

• A. $\frac{3}{2}$
• B. $-\frac{3}{2}$
• C. $2$
• D. $-2$

Answer 1

The correct option is A $\frac{3}{2}$

${2x}^2-2(a-2)x-(a+1)=0letrootsbe\alpha \&\beta \alpha +\beta =a-2\alpha \beta =-\left(\frac{a+1}{2}\right)sumofsquares={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta ={(a-2)}^2+2\frac{(a+1)}{2}=a^2-4a+4+a+1=a^2-3a+5a^2-3a+5isleastatthevertexwhichisequalto\frac{3}{2}$