Question

If the equation (k-2)x²-(k-4)x-2=0 has difference of roots as 3 then the value of k is

Answer 1

If the equation $(k-2)x^{2}-(k-4)x-2=0$ has difference of roots as $3$, then the value of $k$ is $\dfrac{3}{2}$ or, $3$.

Concept before solution :

If $ax^{2}+bx+c=0\:(a\neq 0)$ be a quadratic equation with roots $\alpha$ and $\beta$, then

• $\alpha+\beta=-\dfrac{b}{a}$

• $\alpha\beta=\dfrac{c}{a}$

Step-by-step explanation :

The given equation is

$\quad (k-2)x^{2}-(k-4)x-2=0$

Let, its roots be $r_{1}$ and $r_{2}$

Then using the relation between roots and coefficients, we write:

• $r_{1}+r_{2}=-\dfrac{-(k-4)}{k-2}=\dfrac{k-4}{k-2}$

• $r_{1}r_{2}=\dfrac{-2}{k-2}$

Also given, the difference of the roots is $3$. Then

• $r_{1}-r_{2}=3$

We know, $(r_{1}-r_{2})^{2}=(r_{1}+r_{2})^{2}-4r_{1}r_{2}$

$\Rightarrow (3)^{2}=(\dfrac{k-4}{k-2})^{2}-4(\dfrac{-2}{k-2})$

$\Rightarrow 9=(\dfrac{k-4}{k-2})^{2}+\dfrac{8}{k-2}$

$\Rightarrow 9=\dfrac{(k-4)^{2}+8(k-2)}{(k-2)^{2}}$

$\Rightarrow 9(k-2)^{2}=k^{2}-8k+16+8k-16$

$\Rightarrow 9(k^{2}-4k+4)=k^{2}$

$\Rightarrow 8k^{2}-36k+36=0$

$\Rightarrow 2k^{2}-9k+9=0$

$\Rightarrow 2k^{2}-3k-6k+9=0$

$\Rightarrow k(2k-3)-3(2k-3)=0$

$\Rightarrow (2k-3)(k-3)=0$

Therefore, either $2k-3=0$ or, $k-3=0$

This gives $k=\dfrac{3}{2}$ or, $k=3$

$\Rightarrow \boxed{k=\dfrac{3}{2},3}$