I have the following quadratic equation :
$$am^2 + bm + (c_1^2 +c_2^2) =0,$$ where the solution is given by
$$m = \frac{-b\pm\sqrt{b^2-4a(c_1^2+c_2^2)}}{2a}.$$ Here, $\Delta>0$.
Thus I have two real roots What I would like to do is to graphically depict possible solutions based on the values of $c_1$ and $c_2$. Here, $c_1$ and $c_2$ are two components of a vector and are real numbers.
Can anyone help me out with this? And is it possible to do that?
Let $f(x) = ax^2 + bx + (c_1^2 +c_2^2)$.
If $\Delta \gt 0$, then
(1) $f(x) = 0$ has two real roots ($\alpha$ and $\beta$, say) and this implies the graph of $y = f(x)$ will cut the x-axis at two distinct points [$(\alpha, 0)$ and $(\beta, 0)$].
(2) $b^2 \gt 4a(c_1^2 + c_2^2)$. From $\dfrac {b^2 }{4a }\gt c_1^2 + c_2^2$, we can say $a \gt 0$. This further means the graph of $y = f(x)$ concaves upward. Since $\alpha \times \beta = \dfrac {c_1^2 + c_2^2}{a}$, which is positive, the two points will lie on the same side of the y-axis (either left or right but not both).
Since $f(0) = (c_1^2 +c_2^2) \gt 0$, $y = f(x)$ passes through $(0, k)$ where $k = c_1^2 +c_2^2$, which is positive.
Because we know nothing about the sign of b, there are two possible solutions for the plot as shown.
Even from the fact that the minimum occurs at $(\dfrac {-b}{2a}, f[\dfrac {-b}{2a}])$ with $f[\dfrac {-b}{2a}] < 0$, we can at the most deduce $\dfrac {(c_1^2 + c_2^2)}{a} < (\dfrac {-b}{2a})^2$. But we still cannot make any further deduction.