UNIT 1 • FINDING ROOTS OF EQUATIONS

False Position Method
Regula Falsi Strategy

The False Position Method (Regula Falsi) improves upon the Bisection method by connecting boundary points $(a, f(a))$ and $(b, f(b))$ with a straight secant line. Instead of cutting blindly in the middle, it places the next guess right at the x-intercept of this secant line!

HOW REGULA FALSI WORKSSECANT INTERCEPT

C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}

1. Pick Starting Bracket [a, b]Choose points where

f(a)

and

f(b)

have opposite signs.

2. Draw Secant Line (C)Find x-intercept

C

where secant line crosses

y=0

3. Update Bound & RepeatReplace

a

b

with

C

keeping root bracketed.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the real root of $f(x) = x^3 - 4x - 9 = 0$ in the interval $[a, b] = [2, 3]$ , stopping when two successive intercepts differ by less than $0.0001$ .

STEP 01 • INITIAL BRACKET EVALUATION

Evaluate $f(x)$ at initial points $a = 2$ and $b = 3$ :

f(2) = (2)^3 - 4(2) - 9 = 8 - 8 - 9 = -9 \quad (\text{Negative})

f(3) = (3)^3 - 4(3) - 9 = 27 - 12 - 9 = +6 \quad (\text{Positive})

Since $f(2) < 0$ and $f(3) > 0$ , a root exists in $[2, 3]$ .

STEP 02 • COMPUTE FIRST INTERCEPT (C₁)

Substitute values into the Regula Falsi secant formula:

C_1 = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)} = \frac{2(6) - 3(-9)}{6 - (-9)} = \frac{12 + 27}{15} = \frac{39}{15} = 2.6

f(2.6) = (2.6)^3 - 4(2.6) - 9 = 17.576 - 10.4 - 9 = -1.824 \quad (\text{Negative})

STEP 03 • UPDATE INTERVAL BOUNDS

Since $f(2.6) < 0$ and $f(3) > 0$ , replace the old left bound $a = 2$ with $a = 2.6$ .

New Search Bracket:

[a, b] = [2.6, 3]

STEP 04 • SUBSEQUENT ITERATIONS WALKTHROUGH

Iteration 2:

Secant Formula:

C_2 = \frac{2.6(6) - 3(-1.824)}{6 - (-1.824)} = \frac{21.072}{7.824} \approx 2.693252

Function Evaluation:

f(2.693252) = -0.237227

(Negative)

Update: Set

a = 2.693252

. Change from last intercept:

|C_2 - C_1| = 0.093252

Iteration 3:

Secant Formula:

C_3 \approx 2.704918

Function Evaluation:

f(2.704918) = -0.028912

(Negative)

Update: Set

a = 2.704918

|C_3 - C_2| = 0.011667

Iteration 4:

Secant Formula:

C_4 \approx 2.706333

Function Evaluation:

f(2.706333) = -0.003495

(Negative)

Update: Set

a = 2.706333

|C_4 - C_3| = 0.001415

Iteration 5:

Secant Formula:

C_5 \approx 2.706504

Function Evaluation:

f(2.706504) = -0.000422

(Negative)

Update: Set

a = 2.706504

|C_5 - C_4| = 0.000171

Iteration 6:

Secant Formula:

C_6 \approx 2.706525

Function Evaluation:

f(2.706525) = -0.000051

(Negative)

Stop:

|C_6 - C_5| = 0.000021 < 0.0001

STEP 05 • ITERATION SUMMARY TABLE

Step (k)	Left Bound (a)	Right Bound (b)	Intercept (C)	Value f(C)	Sign of f(C)	Next Bracket
1	2.0000	3.0000	2.6000	-1.8240	f(c) < 0	[2.6000, 3.0000]
2	2.6000	3.0000	2.6933	-0.2372	f(c) < 0	[2.6933, 3.0000]
3	2.6933	3.0000	2.7049	-0.0289	f(c) < 0	[2.7049, 3.0000]
4	2.7049	3.0000	2.7063	-0.0035	f(c) < 0	[2.7063, 3.0000]
5	2.7063	3.0000	2.7065	-0.0004	f(c) < 0	[2.7065, 3.0000]
6	2.7065	3.0000	2.706525	-0.00005	\|C₆ − C₅\| = 0.00002 < 0.0001	Stop

FINAL CONCLUSION

Approximated Root: x ≈ 2.7065

\text{Exact Root } \alpha \approx 2.706528, \quad f(2.706525) = -0.00005 \approx 0

Six chord steps pin the root to 4 decimal places. Bisection needs 14 halvings for the same accuracy on the same interval.

LABORATORY

Interactive Live Calculator & Visualizer

REGULA-FALSI LABORATORY

False Position Interactive Engine

PreviousIteration Method

Unit 1•4 of 63

NextNewton-Raphson Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 1 • FINDING ROOTS OF EQUATIONS

False Position Method
Regula Falsi Strategy

HOW REGULA FALSI WORKSSECANT INTERCEPT

C = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)}

1. Pick Starting Bracket [a, b]Choose points where

f(a)

and

f(b)

have opposite signs.

2. Draw Secant Line (C)Find x-intercept

C

where secant line crosses

y=0

3. Update Bound & RepeatReplace

a

b

with

C

keeping root bracketed.

Step-by-Step Worked Example

HAND-HOLDING TUTORIAL

OUR PROBLEM TO SOLVE

Find the real root of $f(x) = x^3 - 4x - 9 = 0$ in the interval $[a, b] = [2, 3]$ , stopping when two successive intercepts differ by less than $0.0001$ .

STEP 01 • INITIAL BRACKET EVALUATION

Evaluate $f(x)$ at initial points $a = 2$ and $b = 3$ :

f(2) = (2)^3 - 4(2) - 9 = 8 - 8 - 9 = -9 \quad (\text{Negative})

f(3) = (3)^3 - 4(3) - 9 = 27 - 12 - 9 = +6 \quad (\text{Positive})

Since $f(2) < 0$ and $f(3) > 0$ , a root exists in $[2, 3]$ .

STEP 02 • COMPUTE FIRST INTERCEPT (C₁)

Substitute values into the Regula Falsi secant formula:

C_1 = \frac{a \cdot f(b) - b \cdot f(a)}{f(b) - f(a)} = \frac{2(6) - 3(-9)}{6 - (-9)} = \frac{12 + 27}{15} = \frac{39}{15} = 2.6

f(2.6) = (2.6)^3 - 4(2.6) - 9 = 17.576 - 10.4 - 9 = -1.824 \quad (\text{Negative})

STEP 03 • UPDATE INTERVAL BOUNDS

Since $f(2.6) < 0$ and $f(3) > 0$ , replace the old left bound $a = 2$ with $a = 2.6$ .

New Search Bracket:

[a, b] = [2.6, 3]

STEP 04 • SUBSEQUENT ITERATIONS WALKTHROUGH

Iteration 2:

Secant Formula:

C_2 = \frac{2.6(6) - 3(-1.824)}{6 - (-1.824)} = \frac{21.072}{7.824} \approx 2.693252

Function Evaluation:

f(2.693252) = -0.237227

(Negative)

Update: Set

a = 2.693252

. Change from last intercept:

|C_2 - C_1| = 0.093252

Iteration 3:

Secant Formula:

C_3 \approx 2.704918

Function Evaluation:

f(2.704918) = -0.028912

(Negative)

Update: Set

a = 2.704918

|C_3 - C_2| = 0.011667

Iteration 4:

Secant Formula:

C_4 \approx 2.706333

Function Evaluation:

f(2.706333) = -0.003495

(Negative)

Update: Set

a = 2.706333

|C_4 - C_3| = 0.001415

Iteration 5:

Secant Formula:

C_5 \approx 2.706504

Function Evaluation:

f(2.706504) = -0.000422

(Negative)

Update: Set

a = 2.706504

|C_5 - C_4| = 0.000171

Iteration 6:

Secant Formula:

C_6 \approx 2.706525

Function Evaluation:

f(2.706525) = -0.000051

(Negative)

Stop:

|C_6 - C_5| = 0.000021 < 0.0001

STEP 05 • ITERATION SUMMARY TABLE

Step (k)	Left Bound (a)	Right Bound (b)	Intercept (C)	Value f(C)	Sign of f(C)	Next Bracket
1	2.0000	3.0000	2.6000	-1.8240	f(c) < 0	[2.6000, 3.0000]
2	2.6000	3.0000	2.6933	-0.2372	f(c) < 0	[2.6933, 3.0000]
3	2.6933	3.0000	2.7049	-0.0289	f(c) < 0	[2.7049, 3.0000]
4	2.7049	3.0000	2.7063	-0.0035	f(c) < 0	[2.7063, 3.0000]
5	2.7063	3.0000	2.7065	-0.0004	f(c) < 0	[2.7065, 3.0000]
6	2.7065	3.0000	2.706525	-0.00005	\|C₆ − C₅\| = 0.00002 < 0.0001	Stop

FINAL CONCLUSION

Approximated Root: x ≈ 2.7065

\text{Exact Root } \alpha \approx 2.706528, \quad f(2.706525) = -0.00005 \approx 0

Six chord steps pin the root to 4 decimal places. Bisection needs 14 halvings for the same accuracy on the same interval.

LABORATORY

Interactive Live Calculator & Visualizer

REGULA-FALSI LABORATORY

False Position Interactive Engine

PreviousIteration Method

Unit 1•4 of 63

NextNewton-Raphson Method

False Position Method Regula Falsi Strategy

Step-by-Step Worked Example

Approximated Root: x ≈ 2.7065

Interactive Live Calculator & Visualizer

False Position Interactive Engine

NETZ Engine

False Position Method Regula Falsi Strategy

Step-by-Step Worked Example

Approximated Root: x ≈ 2.7065

Interactive Live Calculator & Visualizer

False Position Interactive Engine

False Position Method
Regula Falsi Strategy

False Position Method
Regula Falsi Strategy