Question

Let $\alpha ,\beta$ be real roots of the quadratic equation $x^2+kx+\left(k^2+2k-4\right)=0$, then the maximum value of $\left({\alpha }^2+{\beta }^2\right)$ is equal to

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

Answer: (D)

$\text{ For real roots }D\ge 0$
$k^2-4\left(k^2+2k-4\right)\ge 0\Rightarrow -3k^2-8k+16\ge 0$
$\Rightarrow 3k^2+8k-16\le 0\Rightarrow 3k^2+12k-4k-16\le 0$
$\Rightarrow 3k(k+4)-4(k+4)\le 0\Rightarrow k\in \left[-4,\frac{4}{3}\right]$
Also, ${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =k^2-2\left(k^2+2k-4\right)=-k^2-4k+8=12-(k+2{)}^2$
Maximum value is 12 when $k=-2$.