Question

If $\alpha$ and $\beta$ are the real roots of the equation $x^2-(k-2)x+\left(k^2+3k+5\right)=0(k\in R)$ Find the maximum and minimum values of $\left({\left(\alpha \right)}^2+{\left(\beta \right)}^2\right)$

• A. $18,50/9$
• B. $18,25/9$
• C. $27,50/9$
• D. None of these

Answer 1

Answer: (A)

For real roots $D\ge 0(k-2{)}^2-4$
$k^2+3k+5\ge 0$
$k^2+4-4k-4k^2-12k-20\ge 0$
$-3k^2-16k-16\ge 0;3k^2+16k+16\le 0$
$k+\frac{4}{3}(k+4)\le 0$
Now $E={\alpha }^2+{\beta }^2$
$E=(\alpha +\beta {)}^2-2\alpha \beta$
$E=(k-2{)}^2-2k^2+3k+5=-k^2-10k$
$-6$
$E=-k^2+10k+6=-(k+5{)}^2-19$
$=19-(k+5{)}^2$
$\therefore E_{\min }\text{ Occurs when }k=-4/3$
$\therefore E_{\min }=19-\frac{121}{9}=\frac{171-121}{9}=\frac{50}{9}$
$E_{\max }\text{ Occurs when }k=-4$
$E_{\max }=19-1=18$