Vertical length of two roots

Question

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

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

(b) If $b^2 > c$, show that the vertical distance between the turning points is $4(b^2 − c)^{\frac{3}{2}}$ . [Hint: Use the sum and product of the roots of the derived function.]

I'm really confused on part B of the question! I worked out part A using the discriminant. Using products of sum and roots: $x^2+2bx+c$

$α+β=-2b$

$αβ=c$

$(α^2-β^2)^{\frac{1}{2}}$ would equal to the $x $length right? I tried other 'dodgy' methods hoping to find the value of $y$ but I don't understand the question. Any advice or help will be greatly appreciated!

Answer 1

Answering the question (a) we found: $$ y'=0 \iff x^2+2bx+c=0 $$ that has distinct real solutions iff $\Delta '=b^2-c >0$.Let $\alpha, \beta$ the solutions of this equation, so that $ \alpha +\beta= -2b$ and $\alpha \beta=c$ and $\alpha-\beta=2\sqrt{b^2-c}$.

For the question (b) we have to find $|f(\alpha)-f(\beta)|$ and we find: $$ |f(\alpha)-f(\beta)|=\left|(\alpha^3-\beta^3)+3b(\alpha^2-\beta^2)+3c(\alpha-\beta) \right|= \left| (\alpha-\beta)[ \alpha^2+\alpha \beta +\beta^2+3b(\alpha+\beta)+3c]\right|= (\alpha-\beta)[(\alpha+\beta)^2-2\alpha\beta+\alpha\beta+3b(\alpha+\beta)+3c]=\left|(\alpha-\beta)(2c-b^2)\right|=4\sqrt{b^2-c}\left|(c-b^2) \right|=4\sqrt{(b^2-c)^3} $$

Answer 2

Let $\alpha,\beta\ (\alpha\lt \beta)$ be the roots of $x^2+2bx+c$.

You want to show that $$f(\alpha)-f(\beta)=(\alpha^3-\beta^3)+3b(\alpha^2-\beta^2)+3c(\alpha-\beta)=4(b^2-c)^{\frac 32}.$$

Here use the followings : $$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta\Rightarrow \alpha-\beta=-\sqrt{(\alpha+\beta)^2-4\alpha\beta}$$ $$\alpha^3-\beta^3=(\alpha-\beta)(\alpha^2+\alpha\beta+\beta^2)=(\alpha-\beta)((\alpha+\beta)^2-\alpha\beta)$$ $$\alpha^2-\beta^2=(\alpha-\beta)(\alpha+\beta)$$