Find for which values of the real parameter $k$ one of the roots is with 3 greater than the other one I have to find for which values of the real parameter $k$ one of the roots of:
$$x^2-15x+k^2-10=0$$
is with $3$ greater than the other one.
So, demand $D=b^2-4ac=(-15)^2-4(k^2-10)=15^2-4k^2+40=225-4k^2+40=265-4k^2 > 0$ (I am not sure if it can be $0$, because if it is, the roots are equal?). Here, we get $k^2<\frac{265}{4}$. I don't know how to simplify this. Then I tried to use Vieta's formulas, but it seems useless at the end. $x_1=x_2+3$ and $x_1+x_2=\frac{-b}{a}=x_2+3+x_2=2x_2+3$ 
$x_1x_2=\frac{c}{a}=x_2(x_2+3)=x_2^2+3x^2$. 
 A: If the two roots of the polynomial are $\alpha$ and $\beta$ then the polynomial factors as
$$x^2-15x+k^2-10=(x-\alpha)(x-\beta).$$
Expanding the right hand side immediately shows that 
$$\alpha+\beta=15\qquad\text{ and }\qquad \alpha\beta=k^2-10.$$
Now you want one of the roots to be $3$ more than the other, i.e. without loss of generality you want
$$\alpha=\beta+3.$$
Then it follows that
$$15=\alpha+\beta=2\beta+3,$$
which shows that $\beta=6$ and $\alpha=9$. Then
$$k^2-10=\alpha\beta=9\times6=54,$$
and so $k^2=64$ which means that $k=\pm8$.
A: Using the quadratic formula:
$x = \dfrac{15\pm\sqrt{265-4k^2}}{2}$
We want one root to be $3$ away from the other one, so we let
$\dfrac{15+\sqrt{265-4k^2}}{2} = a$ 
$\dfrac{15-\sqrt{265-4k^2}}{2} = a+3$ 
Rearranging, we get
$2a-15=\sqrt{265-4k^2}$
and
$15-2(a+3) = \sqrt{265-4k^2}$
Equating these, we find $a$, and we can put this back into either equation to find our value(s) for $k$. Also note that we do not know which root is bigger than which, so we could also have tried $a-3$ for the RHS of the second equation. However, if you try both, can you see why this doesn't matter?
