Bracketing and bisecting a root
Begin with an interval whose endpoint values have opposite signs. Halve the interval repeatedly and keep the half that still contains a sign change
- Check endpoint signs For a continuous polynomial, opposite signs at the endpoints guarantee at least one root strictly between them \(\displaystyle P(1)=-2,\quad P(2)=1\quad\Rightarrow\quad P(1)P(2)<0\)
- Find each midpoint The next trial input is the average of the current endpoints \(\displaystyle M_1=\frac{1+2}{2}=\frac{3}{2}\)
- Keep the sign-changing half Retain the half whose endpoints still have opposite signs \(\displaystyle P\left(\frac{3}{2}\right)=-\frac{3}{4}<0\quad\Rightarrow\quad x\in\left[\frac{3}{2},2\right]\)
- Use horizontal width For the midpoint estimate, half the interval width bounds the input error. This alone does not certify rounding of the true root: the root must lie within one rounding cell \(\displaystyle |M-r|\le\frac{R-L}{2}\)
Worked example
Use bisection to approximate a root of the displayed polynomial on the displayed interval to one decimal place
- \(\displaystyle P(2)=-1,\quad P(3)=4\)The endpoint values have opposite signs, so the interval brackets at least one root
- \(\displaystyle M_1=\frac{5}{2},\quad P(M_1)=\frac{5}{4}>0\quad\Rightarrow\quad x\in\left[2,\frac{5}{2}\right]\)The midpoint is positive, so pair it with the negative endpoint at x=2.
- \(\displaystyle M_2=\frac{9}{4},\quad P(M_2)=\frac{1}{16}>0\quad\Rightarrow\quad x\in\left[2,\frac{9}{4}\right]\)The new midpoint is positive, so the sign-changing half remains the left half
- \(\displaystyle M_3=\frac{17}{8},\quad P(M_3)=-\frac{31}{64}<0\quad\Rightarrow\quad r\in\left(\frac{17}{8},\frac94\right)\)The next midpoint estimate has an error of at most one sixteenth, but this bracket crosses 2.15. It does not yet certify the true root’s one-decimal rounding
- \(\displaystyle M_4=\frac{35}{16},\quad P(M_4)=-\frac{55}{256}<0<P\left(\frac94\right)=\frac1{16}\)Retain the sign-changing half shown by the midpoint and endpoint values above. The root is strictly inside because neither endpoint has polynomial value zero
- \(\displaystyle 2.15=\frac{43}{20}<\frac{35}{16}<r<\frac94=2.25\quad\Rightarrow\quad r\approx2.2\text{ (1 d.p.)}\)Every possible root value is strictly within the rounding cell for 2.2. This certifies the true-root rounding, unlike the earlier half-width claim. The square function is strictly increasing on the displayed interval, so the bracket contains only one root
Common mistakes
Difficulty levels
Questions about this topic
What does a sign-changing interval prove?+
For a continuous polynomial, opposite endpoint signs prove that at least one real root lies between the endpoints. They do not prove that there is only one root
Why is the midpoint used?+
The midpoint divides the current bracket into two equal horizontal intervals, so each step halves the possible root interval
What determines when bisection stops?+
Match the question: use the specified step count for a fixed-step estimate, half-width for a midpoint error bound, and a rounding-cell proof for a correctly rounded true root. A small residual alone is not the bisection error guarantee
Must both endpoints round to the same decimal?+
That is a convenient sufficient check, but a strict bracket can also certify rounding when an endpoint is exactly a halfway boundary. First prove that boundary cannot be the root. If it can be the root, apply the stated tie convention; do not silently exclude it
Can I use a calculator?+
No. This topic is practised without a calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, Mathematical Methods, chapter 5: Higher-degree polynomials