Numerical Methods: Bisection Method with 2 Solved Examples and Practice Problems

In this article, we discuss the Solution of Algebraic and Transcendental Equations. Then we provide the definition of algebraic and transcendental equations, discuss the Bisection Method, provide the solved examples, and list questions for practice.

We will talk about a couple of numerical methods for the solution of algebraic and transcendental equations. For that, we should recognise what algebraic and transcendental equations are.

We consider the equation \[ a_{0}\,x^{n} + a_{1}\,x^{\,n-1} + a_{2}\,x^{\,n-2} + \dots + a_{n}=0 \] or the compact form \(f(x)=0\), where \(n\neq0\).

For example, if \(n=1\) it becomes a linear equation; if \(n=3\) it is cubic. Hence the expression is called an algebraic equation. When it contains logarithmic, exponential, or trigonometric terms it is called a transcendental equation, e.g. \[ x\,e^{3x}+2 = 0,\quad x^{2}\tan x + 3 = 0,\quad \tan x + \log_{10}x = 2 . \]

To solve algebraic and transcendental equations we use direct and iterative methods, illustrated below.

To solve the algebraic and transcendental equations by directive and iterative methods. We can understand by flow chart and specify a few methods.

So, we will examine some numerical techniques for the solution of `f(x)=0`, where `f(x)` is an algebraic or transcendental equation.

BISECTION METHOD

For a continuous function \(f(x)\) with \(f(a)f(b)<0\), a real root of \(f(x)=0\) lies in \([a,b]\).

Formula used:

\[ x_{1}= \frac{a_{1}+a_{2}}{2}. \]

If \(f(x_{1})=0\) then \(x_{1}\) is the root; otherwise the root is in \([a_{1},x_{1}]\) or \([x_{1},a_{2}]\) according to the sign of \(f(x_{1})\). Iterate until the desired accuracy is reached.

Example – Find a root of the equation
\[x^{3}-4x-9=0\] Bisection up to four iterations

Take \(f(x)=x^{3}-4x-9\).

Step 1  (test integers until sign change): \[ f(0)=-9\;(−ive),\; f(1)=-12\;(−ive),\; f(2)=-9\;(−ive),\; f(3)=+6\;(＋ive) \] ⇒ Now we stop as the sign changes. Root in \([2,3]\).

1st iteration

\[ x_{1}= \frac{2+3}{2}=2.5,\qquad f(2.5)=-3.375\;(−) \] ⇒ new interval \([2.5,3]\).

2nd iteration

\[ x_{2}= \frac{2.5+3}{2}=2.75,\qquad f(2.75)=0.7969\;(＋) \] ⇒ new interval \([2.5,2.75]\).

3rd iteration

\[ x_{3}= \frac{2.5+2.75}{2}=2.625,\qquad f(2.625)=-1.4121\;(−) \] ⇒ new interval \([2.625,2.75]\).

4th iteration

\[ x_{4}= \frac{2.625+2.75}{2}=2.6875 \] Approximate root ≈ 2.6875.

Example 2: Find a root of the equation x log10x = 1.2, using the bisection method correct up to four decimal places.

Given: f(x) = x log10x - 1.2

Step 1: Try different values of x to locate sign change:

• f(1) = 1 × log₁₀(1) - 1.2 = -1.2 (negative)
• f(2) = 2 × log₁₀(2) - 1.2 ≈ -0.5979 (negative)
• f(3) = 3 × log₁₀(3) - 1.2 ≈ 0.2314 (positive)

Sign changes between 2 and 3 → root lies between 2 and 3.

Try narrower range:

• f(2.74) ≈ -0.000563 (negative)
• f(2.75) ≈ 0.008164 (positive)

Root lies between 2.74 and 2.75

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Step 2: First Iteration
x₁ = (2.74 + 2.75) / 2 = 2.745
f(2.745) = 0.003798 (positive)
→ Root lies between 2.74 and 2.745

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Second Iteration
x₂ = (2.74 + 2.745) / 2 = 2.7425
f(2.7425) = 0.001617 (positive)
→ Root lies between 2.74 and 2.7425

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Third Iteration
x₃ = (2.74 + 2.7425) / 2 = 2.74125
f(2.74125) = 0.0005267 (positive)
→ Root lies between 2.74 and 2.74125

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Fourth Iteration
x₄ = (2.74 + 2.74125) / 2 = 2.740625
f(2.740625) = -0.0000183 (negative)
→ Root lies between 2.740625 and 2.74125

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Fifth Iteration
x₅ = (2.740625 + 2.74125) / 2 = 2.7409375
f(2.7409375) = 0.000254 (positive)
→ Root lies between 2.740625 and 2.7409375

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Sixth Iteration
x₆ = (2.740625 + 2.7409375) / 2 = 2.74078125
f(2.74078125) = 0.0001178 (positive)
→ Root lies between 2.740625 and 2.74078125

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Seventh Iteration
x₇ = (2.740625 + 2.74078125) / 2 = 2.740703125
→ Root correct up to four decimal places is 2.7407.

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Conclusion: The approximate root of the equation x log10x = 1.2 is 2.7407.

For Practice, few numerical are given.

Example 1: Find the real root lying in the interval (1,2) up to four decimal places for the equation x⁶ - x⁴ - x³ - 1 = 0 by the bisection method.

Example 2: Find the real root of the equation cos(x) - x·eˣ = 0 correct to three decimal places by bisection method.

Example 3: Find the real root of the equation f(x) = 3x³ + 10x² + 10x + 7 = 0 correct to three decimal places by bisection method.

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