Solve 3.68x^3 + 92.3x^2 - 500 = 0. Find all three real roots to four decimal places.

Step-by-step solution

1. 1

   Note the missing linear term and the lack of rational roots. The polynomial $f(x)=3.68x^{3}+92.3x^{2}-500$ has no $x$ term, and the decimal coefficients rule out a tidy rational root. So the roots must be located numerically — but how many there are can be settled first.

2. 2

   Sweep for sign changes across a wide interval. Evaluating $f$ from $-40$ to $20$ in small steps reveals three sign changes, near $x=-24.86$, $x=-2.45$ and $x=2.23$. A cubic has at most three real roots, so all three are real and there are no complex ones. Sampling only the interval $[0,3]$ — where $f(2)=-101.36$ and $f(2.5)=+134.375$ — would find just one of them.

3. 3

   Bisect each bracket. Halving each interval until $|f|<10^{-9}$:

   $x_{1}\approx-24.861705,\qquad x_{2}\approx-2.450227,\qquad x_{3}\approx 2.230411.$

   The large negative root exists because the $92.3x^{2}$ term dominates until $x\approx-25$, where the cubic term finally overtakes it — a root easy to miss without a wide sweep.

4. 4

   Check against Vieta's formulas. For $ax^{3}+bx^{2}+cx+d$ the roots satisfy $\sum x_{i}=-\tfrac{b}{a}$ and $\prod x_{i}=-\tfrac{d}{a}$:

   $-24.861705-2.450227+2.230411=-25.081522=-\frac{92.3}{3.68} ✓,$

   $(-24.861705)(-2.450227)(2.230411)=135.8696=\frac{500}{3.68} ✓.$

   This is the decisive test: any proposed triple of roots whose sum is not $-25.0815$ is wrong, whatever the individual values look like.

5. 5

   Verify by substitution. $f(-24.861705)=0.0$, $f(-2.450227)=2\times10^{-13}$ and $f(2.230411)=6\times10^{-14}$ ✓. Rounded to four decimals the roots are $-24.8617$, $-2.4502$ and $2.2304$; note the positive root is $2.2304$, not $2.2307$, so quoting extra digits from a coarse bisection is risky.