Question

If the difference of the roots of the equation $(k-2)x^2-(k-4)x-2=0,k\ne 2$ is $3$, then the sum of all the values of $k$ is

• A. $0$
• B. $-\frac{3}{2}$
• C. $\frac{7}{2}$
• D. $\frac{9}{2}$

Answer 1

The correct option is D $\frac{9}{2}$

Let $\alpha ,\beta$ be the roots of the equation $(k-2)x^2-(k-4)x-2=0$, then
$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta$
Now, we know that,
$\alpha +\beta =\frac{(k-4)}{(k-2)},\alpha \beta =\frac{-2}{k-2}$
So,
$(\alpha -\beta {)}^2={\left(\frac{(k-4)}{(k-2)}\right)}^2+\frac{8}{k-2}\Rightarrow 9={\left(\frac{(k-4)}{(k-2)}\right)}^2+\frac{8}{k-2}[\because |\alpha -\beta |=3]$
$\Rightarrow 9=\frac{(k-4{)}^2+8(k-2)}{(k-2{)}^2}\Rightarrow 9(k-2{)}^2=k^2-8k+16+8k-16\Rightarrow 8k^2-36k+36=0\Rightarrow 2k^2-9k+9=0\Rightarrow (k-3\left)\right(2k-3)=0$
$\therefore k=3\text{or}k=\frac{3}{2}$
Hence, the sum of values of $k$ is $\frac{9}{2}$