Showing that the roots of the quadratic are real

Question

If $x^2+bx+c=0$ has real roots, show that the roots of the equation $x^2+bx+c(x+a)(2x+b)=0$ are real for all real values of $a$.

I could do it by standard way by proving determinant is postive. But, I was looking for some better way because I think the two equations are related somehow. Can someone suggest me another smart way?

Answer 1

If $f(x) = x^2 + bx + c$ has two real roots, we know from it's derivative that it's minimum is obtained at $x = -\frac{b}{2}$, and as it has two real roots we have that $f(-\frac{b}{2}) \lt 0$

Observe that if we plug $x = -\frac{b}{2}$ into $g(x) = x^2 + bx + c(x+a)(2x+b)$ we get $g(-\frac{b}{2}) = -\frac{b^2}{4}$, and as $-\frac{b^2}{4} \leq 0 \forall b \in \mathbb{R}$

To conclude as we have a point which we can guarantee is always $\leq 0$ we know that $g(x)$ must always have at least one real root, which may be repeated.

Answer 2

Look at the value of the quadratic at $x=-b/2$. It is negative in the first case, because it is the vertex for that one, so it is negative in the second case.
EDIT: Sorry, that only works if you have a plus sign between $c$ and $(x+a)(2x+b)$.
If $a=2,b=4,c=-1$, the original quadratic is $x^2+4x-1$, the second one is $$x^2+4x-2(x+2)^2=-4-(x+2)^2$$