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.
 A: 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:
$$|a|\cdot|x_1-x_2|=\sqrt{b^2-4ac}$$
A: 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 $k=-4$.
A: 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 $k=-4$.
A: This does not appear to need the quadratic equation if we are solving for $k$.
$$kx^2-2kx+5=0\quad\implies k(x^2-2x)=-5\quad\implies k=\frac{5}{x(2-x)}\quad\implies x = 1 \land k = 5$$
These are the only integer solutions.
A: Working from the "vertex form" of the equation of a parabola is also helpful. For the vertex at $ \ (h \ , \ p) \ \ , $ we have
$$ k·(x \ -  \ h)^2 \ + \ p \ \ = \ \ kx^2 \ - \ 2khx \ + \ (kh^2 \ + \ p) \ \ \ = \ \ kx^2 \ - \ 2kx \ + \ 5 \ \ . $$
This tells us at once that $ \ h \ = \ 1 \ \ , $ and thus $ \ k + p \ = \ 5 \ \  . $
The zeroes of the quadratic polynomial are separated by a distance of $ \ 3 \ \ , $ so they are $ \ \frac32 \ $ units to either side of the axis of symmetry of the parabola, or at  $ \ r \ = \ 1 \ \pm \ \frac32 \ = \ -\frac12 \ , \ \frac52 \ \ . $
One may proceed in various ways from here.  A fairly direct one would be to write $ \ r - h \ = \ \pm \frac32 \ \Rightarrow \ (r - h)^2 \ = \ \frac94 \ \ $ and $ \ p \ = \ 5 - k \ \ , $ which leads to
$$ k·\frac94 \ + \ (5-k) \ \ = \ \ 0 \ \ \Rightarrow \ \ \frac54·k \ = \ -5 \ \ \Rightarrow \ \ k \ = \ -4 \ \ . $$
ADDENDUM --
A reasonable question to ask would be:  if $ \ p \ = \ 5 - k \ $ and the parabola will have two $ \ x-$intercepts when $ \ p < 0 \   $ for positive $ \ k \ \ , $ why isn't there a solution with $ \ k > 5 \ $ ?  The distance of the $ x-$intercepts of the parabola $ \ y \ = \ ax^2 + bx + c \ $ (zeroes of the polynomial) from the axis of symmetry is given by $$ \ |h - r| \ \ = \ \ \frac{\sqrt{\Delta}}{2a} \ \Rightarrow \ \ (h - r)^2 \ = \ \frac{\Delta}{4a^2} \ \  $$
(as noted by user65203).  With $ \ a \ = \ k \ $  and $ \ \Delta \ = \ (-2k)^2 \ - \ 4·k·5 \ = \ 4k^2 \ - \ 20k \ \ , $ we find
$$ (1 - r)^2 \ \ = \ \ \frac{4k^2 \ - \ 20k}{4k^2} \ \ \Rightarrow \ \ |1 - r| \ \ = \ \ \sqrt{1 \ - \ \frac{5}{k}} \ \ . $$
So if we take any increasingly large values with $ \ k > 5 \ \ , $ the zeroes cannot be farther than $ \ 1 \ $ from $ \ x \ = \ 1 \ $ (or the separation is never larger than $ \ 2 \   ) \ . $
