Question

Use the bisection method to find the root of $f(x) = x^2 - 2 = 0$ in $[1, 2]$ with an error less than 0.01.

• A. $1.414$
• B. $1.5$
• C. $1.732$
• D. $1.618$

Hint:

In bisection, continue iterations until the interval length is less than twice the error tolerance.

Answer 1

The Correct Option is A

The bisection method finds roots by halving intervals where $f(a) \cdot f(b)<0$. 
For $f(x) = x^2 - 2$ in $[1, 2]$: 
- $f(1) = 1^2 - 2 = -1$ 
- $f(2) = 2^2 - 2 = 2$ 
Since $f(1) \cdot f(2)<0$, a root exists. 
Error tolerance is 0.01, so interval length $\leq 0.02$. 
Iteration 1: Midpoint $c = \frac{1+2}{2} = 1.5$, $f(1.5) = 1.5^2 - 2 = 0.25>0$. 
New interval $[1, 1.5]$. 
Iteration 2: $c = \frac{1+1.5}{2} = 1.25$, $f(1.25) = 1.25^2 - 2 = -0.4375<0$. 
New interval $[1.25, 1.5]$. 
Iteration 3: $c = \frac{1.25+1.5}{2} = 1.375$, $f(1.375) = 1.375^2 - 2 = -0.109375<0$. 
New interval $[1.375, 1.5]$. 
Iteration 4: $c = \frac{1.375+1.5}{2} = 1.4375$, $f(1.4375) = 1.4375^2 - 2 = 0.06640625>0$. 
New interval $[1.375, 1.4375]$. 
Interval length = $1.4375 - 1.375 = 0.0625>0.02$. 
Continue: 
Iteration 5: $c = \frac{1.375+1.4375}{2} = 1.40625$, $f(1.40625) = -0.02246094<0$. 
New interval $[1.40625, 1.4375]$. 
Iteration 6: $c = \frac{1.40625+1.4375}{2} = 1.421875$, $f(1.421875) = 0.02172852>0$. 
New interval $[1.40625, 1.421875]$. 
Interval length = $1.421875 - 1.40625 = 0.015625<0.02$. 
Approximate root: $c = 1.414$. Note: $\sqrt{2} \approx 1.414$.