UNIT 1 • FINDING ROOTS OF EQUATIONS

The Bisection Method
Step-by-Step Friendly Guide

The Bisection Method is a straightforward and reliable numerical technique used to find roots (solutions) of continuous functions. It works by repeatedly dividing an interval in half and selecting the subinterval where the function changes sign, thereby narrowing down the location of the root.

HOW THE BISECTION METHOD WORKSMIDPOINT CALCULATION

C = \frac{a + b}{2}

1. Select Interval [a, b]Choose two points

a

and

b

such that

f(a)

and

f(b)

have opposite signs.

2. Calculate Midpoint (C)Compute

C = \frac{a+b}{2}

to split interval in half.

3. Evaluate f(C)Check sign and value of

f(C)

4. Update IntervalIf

f(C) \approx 0

C

is the root! Otherwise, set

a=C

b=C

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the root of $x^2 = 4$ using the Bisection Method with tolerance error margin $0.01$ .

STEP 01 • REARRANGING THE EQUATION

First, set the equation to zero to get our standard function $f(x) = 0$ :

f(x) = x^2 - 4 = 0

STEP 02 • CHOOSE INITIAL POINTS

Choose two initial points $a$ and $b$ such that $f(a)$ and $f(b)$ have opposite signs. Let's choose $a = 0$ and $b = 3$ :

f(0) = 0^2 - 4 = -4 \quad (\text{Negative})

f(3) = 3^2 - 4 = 5 \quad (\text{Positive})

Since $f(0) < 0$ and $f(3) > 0$ , a root must lie in $[0, 3]$ !

STEP 03 • CALCULATE FIRST MIDPOINT

Calculate the midpoint $C$ of the interval $[a, b] = [0, 3]$ :

C = \frac{a + b}{2} = \frac{0 + 3}{2} = 1.5

f(1.5) = (1.5)^2 - 4 = 2.25 - 4 = -1.75 \quad (\text{Negative})

STEP 04 • UPDATE THE INTERVAL

Since $f(0) < 0$ and $f(1.5) < 0$ , both left points are negative. The sign change happens between $C = 1.5$ and $b = 3$ .

Set new left bound:

a = C = 1.5

. New interval is

[1.5, 3.0]

STEP 05 • STEP-BY-STEP ITERATIONS WALKTHROUGH

Iteration 2:

Midpoint:

C_2 = \frac{1.5 + 3.0}{2} = 2.25

Function Value:

f(2.25) = (2.25)^2 - 4 = 5.0625 - 4 = +1.0625

(Positive)

Update: Since

f(1.5) < 0

and

f(2.25) > 0

, set right bound

b = 2.25

New Interval:

[1.5, 2.25]

Iteration 3:

Midpoint:

C_3 = \frac{1.5 + 2.25}{2} = 1.875

Function Value:

f(1.875) = (1.875)^2 - 4 = 3.515625 - 4 = -0.484375

(Negative)

Update: Since

f(1.875) < 0

and

f(2.25) > 0

, set left bound

a = 1.875

New Interval:

[1.875, 2.25]

Iteration 4:

Midpoint:

C_4 = \frac{1.875 + 2.25}{2} = 2.0625

Function Value:

f(2.0625) = (2.0625)^2 - 4 = +0.2539

(Positive)

Update: Set

b = 2.0625

. New Interval:

[1.875, 2.0625]

STEP 06 • ITERATION SUMMARY TABLE

Step (k)	Left Bound (a)	Right Bound (b)	Midpoint (C)	Value f(C)	Sign of f(C)	Next Interval
1	0.0000	3.0000	1.5000	-1.7500	f(1.5) < 0	[1.5000, 3.0000]
2	1.5000	3.0000	2.2500	+1.0625	f(2.25) > 0	[1.5000, 2.2500]
3	1.5000	2.2500	1.8750	-0.4844	f(1.875) < 0	[1.8750, 2.2500]
4	1.8750	2.2500	2.0625	+0.2539	f(2.0625) > 0	[1.8750, 2.0625]
5	1.8750	2.0625	1.9688	-0.1240	f(1.9688) < 0	[1.9688, 2.0625]
6	1.9688	2.0625	2.0156	+0.0627	f(2.0156) > 0	[1.9688, 2.0156]
7	1.9688	2.0156	1.9922	-0.0311	f(1.9922) < 0	[1.9922, 2.0156]
8	1.9922	2.0156	2.0039	+0.0156	f(2.0039) > 0	[1.9922, 2.0039]

FINAL CONCLUSION

Exact Root Converged: x = 2

\sqrt{4} = 2, \quad f(2) = 2^2 - 4 = 0

The Bisection Method guarantees convergence as long as you start with points that bracket the root. Notice how the interval width $|b - a|$ cut in half at every single step ( $3 \to 1.5 \to 0.75 \to 0.375 \to 0.1875 \dots$ ) until it dropped below the target error margin $0.01$ , yielding our root estimate of $x = 2.00$ !

LABORATORY

Interactive Live Calculator & Visualizer

ROOT FINDER LABORATORY

Bisection Method Interactive Engine

Unit 1•1 of 28

NextFalse Position Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 1 • FINDING ROOTS OF EQUATIONS

The Bisection Method
Step-by-Step Friendly Guide

HOW THE BISECTION METHOD WORKSMIDPOINT CALCULATION

C = \frac{a + b}{2}

1. Select Interval [a, b]Choose two points

a

and

b

such that

f(a)

and

f(b)

have opposite signs.

2. Calculate Midpoint (C)Compute

C = \frac{a+b}{2}

to split interval in half.

3. Evaluate f(C)Check sign and value of

f(C)

4. Update IntervalIf

f(C) \approx 0

C

is the root! Otherwise, set

a=C

b=C

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the root of $x^2 = 4$ using the Bisection Method with tolerance error margin $0.01$ .

STEP 01 • REARRANGING THE EQUATION

First, set the equation to zero to get our standard function $f(x) = 0$ :

f(x) = x^2 - 4 = 0

STEP 02 • CHOOSE INITIAL POINTS

Choose two initial points $a$ and $b$ such that $f(a)$ and $f(b)$ have opposite signs. Let's choose $a = 0$ and $b = 3$ :

f(0) = 0^2 - 4 = -4 \quad (\text{Negative})

f(3) = 3^2 - 4 = 5 \quad (\text{Positive})

Since $f(0) < 0$ and $f(3) > 0$ , a root must lie in $[0, 3]$ !

STEP 03 • CALCULATE FIRST MIDPOINT

Calculate the midpoint $C$ of the interval $[a, b] = [0, 3]$ :

C = \frac{a + b}{2} = \frac{0 + 3}{2} = 1.5

f(1.5) = (1.5)^2 - 4 = 2.25 - 4 = -1.75 \quad (\text{Negative})

STEP 04 • UPDATE THE INTERVAL

Since $f(0) < 0$ and $f(1.5) < 0$ , both left points are negative. The sign change happens between $C = 1.5$ and $b = 3$ .

Set new left bound:

a = C = 1.5

. New interval is

[1.5, 3.0]

STEP 05 • STEP-BY-STEP ITERATIONS WALKTHROUGH

Iteration 2:

Midpoint:

C_2 = \frac{1.5 + 3.0}{2} = 2.25

Function Value:

f(2.25) = (2.25)^2 - 4 = 5.0625 - 4 = +1.0625

(Positive)

Update: Since

f(1.5) < 0

and

f(2.25) > 0

, set right bound

b = 2.25

New Interval:

[1.5, 2.25]

Iteration 3:

Midpoint:

C_3 = \frac{1.5 + 2.25}{2} = 1.875

Function Value:

f(1.875) = (1.875)^2 - 4 = 3.515625 - 4 = -0.484375

(Negative)

Update: Since

f(1.875) < 0

and

f(2.25) > 0

, set left bound

a = 1.875

New Interval:

[1.875, 2.25]

Iteration 4:

Midpoint:

C_4 = \frac{1.875 + 2.25}{2} = 2.0625

Function Value:

f(2.0625) = (2.0625)^2 - 4 = +0.2539

(Positive)

Update: Set

b = 2.0625

. New Interval:

[1.875, 2.0625]

STEP 06 • ITERATION SUMMARY TABLE

Step (k)	Left Bound (a)	Right Bound (b)	Midpoint (C)	Value f(C)	Sign of f(C)	Next Interval
1	0.0000	3.0000	1.5000	-1.7500	f(1.5) < 0	[1.5000, 3.0000]
2	1.5000	3.0000	2.2500	+1.0625	f(2.25) > 0	[1.5000, 2.2500]
3	1.5000	2.2500	1.8750	-0.4844	f(1.875) < 0	[1.8750, 2.2500]
4	1.8750	2.2500	2.0625	+0.2539	f(2.0625) > 0	[1.8750, 2.0625]
5	1.8750	2.0625	1.9688	-0.1240	f(1.9688) < 0	[1.9688, 2.0625]
6	1.9688	2.0625	2.0156	+0.0627	f(2.0156) > 0	[1.9688, 2.0156]
7	1.9688	2.0156	1.9922	-0.0311	f(1.9922) < 0	[1.9922, 2.0156]
8	1.9922	2.0156	2.0039	+0.0156	f(2.0039) > 0	[1.9922, 2.0039]

FINAL CONCLUSION

Exact Root Converged: x = 2

\sqrt{4} = 2, \quad f(2) = 2^2 - 4 = 0

LABORATORY

Interactive Live Calculator & Visualizer

ROOT FINDER LABORATORY

Bisection Method Interactive Engine

Unit 1•1 of 28

NextFalse Position Method

The Bisection Method Step-by-Step Friendly Guide

Step-by-Step Worked Example

Exact Root Converged: x = 2

Interactive Live Calculator & Visualizer

Bisection Method Interactive Engine

NETZ Engine

The Bisection Method Step-by-Step Friendly Guide

Step-by-Step Worked Example

Exact Root Converged: x = 2

Interactive Live Calculator & Visualizer

Bisection Method Interactive Engine

The Bisection Method
Step-by-Step Friendly Guide

The Bisection Method
Step-by-Step Friendly Guide