# No, You Don’t Need Quadratic Formula

On the MANSW Facebook page, a fellow teacher from another school threw this question to the brain’s trust. All the answers provided to the problem by others used the Quadratic Formula, but the Algebra from those solutions was not exactly clean. Personally, I don’t think getting the answer in Mathematics is always good enough but rather the elegance and ingenuity of the solution is something to be admired. I’ve posted a solution to this problem that avoids the quadratic formula and it ends up being cleaner.

The whole point of a question that uses the sum and product of roots to solve it is to avoid finding out exactly what those roots are.

# Question

Let $$f(x) = x^3+3bx^2 +3cx +d$$.

(1) Show that $$y=f(x)$$ has 2 distinct turning points if and only if $$b^2>c$$.

(2) If $$b^2 > c$$ show that the vertical distance between the turning points is $4(b^2-c)^\frac{3}{2}$

# Solution

Part 1 Solution

Differentiating $$f(x)$$:

$f'(x) = 3x^2 + 6bx+3c$

There are two distinct turning points if and only if $$f'(x)=0$$ has two distinct solutions, i.e. the discriminant of the quadratic is positive.

\begin{align*}(6b)^2-4(3)(3c) &> 0\\36b^2-36c&>0\\b^2&>c\end{align*}

[collapse]
Part 2 Solution

Since the leading term of $$f(x)$$ is $$x^3$$, then we know that the graph takes a cubic shape where if $$\alpha$$ and $$\beta$$ are the $$x$$ coordinates of the turning points and $$\alpha < \beta$$, then $$f(\alpha)>f(\beta)$$.

Hence, we need to calculate the value of $$f(\alpha)-f(\beta)$$ to find the vertical distance between the turning points.

Firstly, using Vieta’s formulas (otherwise known as sum and product of roots):

$\alpha + \beta = \frac{-6b}{3}=-2b$

$\alpha\beta = \frac{3c}{3}=c$

As a corollary, we can also derive the following result:

\begin{align*}(\alpha-\beta)^2&=\alpha^2-2\alpha\beta +\beta^2\\alpha-\beta)^2&=(\alpha + \beta)^2 -4\alpha\beta\\(\alpha-\beta)^2&=(-2b)^2-4(c)\\\alpha -\beta&=-\sqrt{4b^2-4c}\\&=-2\sqrt{b^2-c}\end{align*} Now the value of \(\alpha – \beta must be negative as we have already stated that $$\alpha <\beta$$.

Now the vertical distance of the turning points:

\begin{align*}f(\alpha)-f(\beta)&=\alpha^3-\beta^3+3b\alpha^2-3b\beta^2+3c\alpha-3c\beta\\&=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2)+3b(\alpha-\beta)(\alpha+\beta)+3c(\alpha-\beta)\\&=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2+3b(\alpha+\beta)+3c)\\&=(\alpha-\beta)\left((\alpha+\beta)^2-\alpha\beta+3b(\alpha+\beta)+3c\right)\\&=-2\sqrt{b^2-c}\left((-2b)^2-c+3b(-2b)+3c\right)\\&=-2\sqrt{b^2-c}(4b^2-c-6b^2+3c)\\&=-2\sqrt{b^2-c}(-2b^2+2c)\\&=-2\sqrt{b^2-c}(-2)(b^2-c)\\&=4(b^2-c)^\frac{3}{2}\end{align*}

[collapse]
One Comment