Let f(x) = ax²+bx+c with b²-4ac>0. Show that between

Let $f(x) = ax^{2}+bx+c$ with $b^{2}-4ac>0$. Show that between the two roots of $f$ there lies a root of $f'(x) = 0$.

Step 1: The two roots

The condition $b^{2}-4ac>0$ guarantees two distinct real roots

\[ x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}, \]

which we label so that $x_{1}\lt x_{2}$. By definition $f(x_{1}) = f(x_{2}) = 0$.

Step 2: Check Rolle's hypotheses on $\bigl[x_{1},x_{2}\bigr]$

$f$ is a polynomial, hence continuous on $\bigl[x_{1},x_{2}\bigr]$ and differentiable on $\bigl(x_{1},x_{2}\bigr)$; and the endpoint values agree, both being $0$. All three conditions hold.

Step 3: Apply the theorem

There exists $\xi\in\bigl(x_{1},x_{2}\bigr)$ with $f'(\xi) = 0$ — a root of $f'$ strictly between the roots of $f$, which is what was to be shown.

Step 4: Identifying $\xi$ explicitly

Here the abstract statement can be made concrete. Since $f'(x) = 2ax+b$, $f'(\xi) = 0 \;\Longrightarrow\; \xi = -\frac{b}{2a}$.

the axis of symmetry of the parabola — the arithmetic mean $\tfrac12\bigl(x_{1}+x_{2}\bigr)$ of the two roots. It lies strictly between them because

\[ \frac{-b-\sqrt{b^{2}-4ac}}{2a} \ \ \text{and} \ \ \frac{-b+\sqrt{b^{2}-4ac}}{2a} \]

straddle $-\dfrac{b}{2a}$ symmetrically, the offset $\dfrac{\sqrt{b^{2}-4ac}}{2a}$ being non-zero.

Step 5: Verification

\[f(x) = x^{2}-5x+6 = \bigl(x-2\bigr)\bigl(x-3\bigr):\ x_{1} = 2,\ x_{2} = 3\]

\[f'(x) = 2x-5 = 0 \;\Longrightarrow\; \xi = 2.5 \in \bigl(2,3\bigr) \;\;\checkmark\]

\[\text{And } \xi = \frac{x_{1}+x_{2}}{2} = \frac{2+3}{2} = 2.5 \;\;\checkmark\]