Question

The set of values of $k$ for which the equation $(k+2)x^2-2kx-k=0$ has two roots on the number line symmetrically placed about $1$ is

• A. $(-\infty ,-2)\cup (1,\infty )$
• B. $(1,\infty )$
• C. $\{0,2\}$
• D. $\varphi$

Answer 1

The correct option is D $\varphi$


$x-$coordinate of line of symmetry
$=1$
So,
$-\frac{b}{2a}=1$
$\Rightarrow \frac{2k}{2(k+2)}=1$
$\Rightarrow k=k+2$
Therefore, no solution exists.