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While working with quadratics that have real roots, I realized an interesting fact:

The slope of a quadratic at its roots is equal to $\pm \sqrt{D}$ where $D=b^2-4ac$

Proof:

$$f(x) = ax^2 + bx +c$$

$$f'(x) = 2ax+b$$

Roots:

$$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

So, if we try to find the slope at any root ($r$):

$$f’(r) = \pm \sqrt D$$

where the sign ($\pm$) can be determined by whether the root is on the right of the vertex or the left.

If the quadratic has only $1$ root (or $2$ roots that are the same) then it means the quadratic is at a stationary point so the slope must be $0$. This is backed by the fact that quadratics have only 1 distinct root when $b^2 - 4ac = 0$.

What geometric/intuitive approach can be applied to explain this interesting phenomenon?

KM101
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Vikhyat Agarwal
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1 Answers1

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I don’t see there is any special geometric related interpretation.

We will just discuss the case that $D \gt 0$ such that $\alpha$ and $\beta$ are the two real roots with $\alpha \lt \beta$.

For simplicity, we also drop the $\pm$ sign.

$\sqrt D = \dfrac aa \sqrt D$

$ = a\sqrt {(\dfrac {-b}{a})^2 – \dfrac {4c}{a}}$

$= a \sqrt {(\beta + \alpha)^2 – 4\alpha\beta}$

$= a \sqrt {(\beta - \alpha)^2}$

$= a \times (\beta - \alpha)$

At this point, we can just say that $f’(r)$ is just a times the length of the difference of the two x-intercepts cut-off by the quadratic function.

Mick
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