# How to graphically depict the possible solutions of a quadratic equation

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?

• There is a formula for the two (positive) solutions $m_{1,2}$. – Wuestenfux Nov 28 '18 at 9:26
• @Wuestenfux I guess here, both the solutions might not be positive. Most likely one is positive and other one is negative. – newstudent Nov 28 '18 at 9:41
• Add the condition that the discriminant is positive. This should give you a condition on the coefficients. – Wuestenfux Nov 28 '18 at 9:42
• What do you mean by "depict possible solutions"? – Todor Markov Nov 28 '18 at 9:43
• @TodorMarkov I would like to plot the possible values of m as a function of $c_1$ and $c_2$ or $b$. – newstudent Nov 28 '18 at 9:47

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.