Determine the value of ${k}$ for quadratic roots

Solve for ${k}$

${kx^2-2kx+5=0}$; one root exceeds the other by 3.

I tried using the ${x_1 \times x_2 = c/a}$ and the ${x_1+x_2 = -b/a}$, I don't know if it has something to do with the discriminant somehow but I doubt it. Any help towards my understanding is greatly appreciated.

Solution 1:

Since one root exceeds the other by $3$, we have $$|x_1-x_2|=3\implies9=|x_1-x_2|^2=(x_1+x_2)^2-4x_1x_2$$ I think you can take it from here.

Also, indeed, the difference between the roots of a quadratic equation (say $ax^2+bx+c=0$, with roots $x_1$ and $x_2$) is related to the discriminant, and the following holds: $$|a|\cdot|x_1-x_2|=\sqrt{b^2-4ac}$$

Solution 2:

Let the two roots be $r$ and $r+3$. Then the sum of roots is $2r+3=-\frac{-2k}{k}=2$. Hence $r=-\frac12$ and the product of roots is $(-\frac12)(-\frac12+3)=\frac5k$. Therefore $k=-4$.

Solution 3:

The difference between the roots of a quadratic equation is

$$\pm\frac{\sqrt{b^2-4ac}}a.$$

Hence

$$\frac{\sqrt{4k^2-20k}}k=\pm3$$

or

$$-20k-5k^2=0.$$

The only solution is $k=-4$.