# $\sqrt{\Delta} =$ the distance between the midpoint and the two roots [closed]

I understand how the midpoint of a parabola is found when one of the roots is $$0$$ ( it is just $$-b/2a$$ )

but I can't understand why $$\sqrt{\Delta}$$ in the quadratic formula $$=$$ the distance between any of our real roots and our midpoint.

I want to know the proof behind this

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Commented Jul 28, 2023 at 12:06

Assuming what you are interested in is the derivation of $$\Delta$$ and a justification for its role as the distance between the solutions I think this might be helpful.

Consider the quadratic equation $$ax^2 + bx + c = 0$$ where $$a,b,c \in \mathbb{R}$$ and $$a \neq 0$$. Then, one can factor the $$a$$ out to obtain $$a(x^2 + \frac{b}{a}x) + c = 0$$. Since $$a \neq 0$$ one can also divide both sides of the equation by $$a$$ to obtain $$(x^2 + \frac{b}{a}x) + \frac{c}{a} = 0$$. The first non trivial manipulation is the recognition that $$(x^2 + \frac{b}{a}x)$$ is fairly close to $$(x + \frac{b}{2a})^2$$. Indeed, there is only a term of $$\frac{b^2}{4a^2}$$ of difference. Therefore, one can write

\begin{align} (x^2+\frac{b}{a}x) + \frac{c}{a} &= (x + \frac{b}{2a})^2 -\frac{b^2}{4a^2} + \frac{c}{a}. \end{align} Rearranging, one obtains $$$$(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}.$$$$ The numerator of the term on the right can be recognised as $$\Delta$$. Thus, $$$$(x + \frac{b}{2a})^2 = \frac{\Delta}{4a^2}.$$$$ The last step is to take square roots of both sides to find $$x$$ alone. However, we know that there are two real numbers such that their square equals $$\Delta$$ and that both these are solutoins. Thus, we find the usual quadratic formula $$$$x = \frac{-b \pm \sqrt{\Delta}}{2a}.$$$$ or more suggestively $$$$x = \frac{-b}{2a} \pm \frac{\sqrt{\Delta}}{2a}.$$$$ From this expression it is clear that the roots of the equations are going to be on either side of $$\frac{-b}{2a}$$ one being $$\frac{\sqrt{\Delta}}{2a}$$ towards one side and the other being $$\frac{\sqrt{\Delta}}{2a}$$ towards the other side of the vertex.

If the roots are $$r_1,r_2$$, then the sum of the roots $$r_1+r_2=\frac{-b}{a}$$ and the product of the roots is $$r_1r_2=\frac{c}{a}$$.

The distance between the two (real) roots is given by $$|r_1-r_2|=\sqrt{(r_1+r_2)^2-4r_1r_2}.$$

The distance between the mid-point and either of the roots is $$\frac{|r_1-r_2|}{2}$$. Now you can fill in the details.

• Sorry but isn't the sum of the roots -b/a? Commented Jul 28, 2023 at 12:38
• @Shadowsparkle you are right. I will fix the typo. Commented Jul 28, 2023 at 12:39

We know that a parabola is of the form $$f(x) =ax^2+bx+c=0$$. We can find the roots of the equation by using the quadratic formula: $$r_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a}$$ Suppose that a parabola $$\mathscr{P}$$ has two roots $$r_1, r_2$$ then the midpoint of the parabola is $$x_1 = -\frac{b}{2a}$$, which is also the axis of symmetry of the parabola. To see this, we rewrite the parabolaformula as: $$f(x) = a(x-e)^2 +k$$ where $$e = -\frac{b}{2c}, \:\:k= \frac{4ac-b^2}{4a}$$ (you can check for yourself that these are equivalent). Now it is clear, from the square, that $$f$$ is symmetric around the point $$e = -\frac{b}{2a}$$.

Because of the symmetry, the distance from the roots to $$x_{\text{1}}$$ must be constant.

We are considering a parabola with two roots, thus $$\Delta>0$$. The roots are then: $$r_1 = \frac{-b + \sqrt{\Delta}}{2a}, \:\: r_2 = \frac{-b - \sqrt{\Delta}}{2a}$$ so the distance from the midpoint $$x_1$$ to the roots is: \begin{align*} d_1 &= \left|x_1 - r_1 \right| = \left| - \frac{b}{2a} - \frac{-b + \sqrt{\Delta}}{2a}\right| = \frac{\sqrt{\Delta}}{2|a|} \\ d_2 &= \left| x_1 - r_2 \right| = \left| - \frac{b}{2a} - \frac{-b - \sqrt{\Delta}}{2a} \right| = \frac{\sqrt{\Delta}}{2|a|} \end{align*} so the distance between the roots and the midpoint is, in both cases, equal to $$\frac{\sqrt{\Delta}}{2a}$$

• Thank you so much Commented Jul 28, 2023 at 13:22