Which of these statements is true regarding the bisection method in solving $f(x) = x^2 + x - 5$ with $x_l= -3$ and $x_u = 3$?

Question

BlackTom AI · Accepted Answer

A sign change across the endpoints is required for the bisection method; here f(-3) and f(3) are both positive, so the method cannot be applied with these bounds.

1) Relevant knowledge points and concepts
- Bisection method requires f(x_l) and f(x_u) to have opposite signs (f(x_l)·f(x_u) < 0) so the Intermediate Value Theorem guarantees at least one root in the interval.
- A quadratic ax^2 + bx + c has real roots if the discriminant Δ = b^2 − 4ac > 0; the roots are (-b ± sqrt(Δ)) / (2a).

2) Step-by-step reasoning
- Evaluate the function at the given endpoints to test the bisection bracketing condition.
- If the product f(x_l)·f(x_u) > 0, no sign change occurs and the standard bisection cannot proceed on that interval.

3) Calculations (all intermediate steps)
- f(x) = x^2 + x − 5.
- f(−3) = (−3)^2 + (−3) − 5 = 9 − 3 − 5 = 1.
- f(3) = 3^2 + 3 − 5 = 9 + 3 − 5 = 7.
- Product f(−3)·f(3) = 1 × 7 = 7 > 0 → no sign change → bisection condition fails.
- For completeness, discriminant Δ = 1^2 − 4(1)(−5) = 1 + 20 = 21 > 0, so two real roots exist:
  roots = (−1 ± sqrt(21))/2 ≈ (−1 ± 4.5826)/2 → x1 ≈ −2.7913, x2 ≈ 1.7913 (both lie inside [−3, 3]).

4) Why the correct answer is correct
- Even though two real roots exist inside the interval, both endpoints evaluate to positive values, so there is no single guaranteed sign change across the whole [−3, 3]. The bisection method cannot be applied directly to this interval because it requires f(x_l) and f(x_u) to have opposite signs.

5) Why the other options are incorrect
- (b) “A single root will be returned”: incorrect because the bisection method cannot start without a sign change; it will not return a root from these bounds.
- (c) “Two roots will be returned”: incorrect because the standard bisection method finds one root per bracketing interval; it cannot return two roots from a single interval and, in any case, cannot start here.
- (d) “This equation has no real roots”: incorrect because Δ = 21 > 0, so two real roots exist (computed above).

6) Conclusion
- The correct statement is that the bisection method cannot be used with these bounds, which matches the provided answer.

Which of these statements is true regarding the bisection method in solving \(f(x) = x^2 + x - 5\) with \(x_l= -3\) and \(x_u = 3\)?单项选择题

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