EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 5, Problem 1P

Determine the real roots of f ( x ) = 0.5 x 2 + 2.5 x + 4.5 :

(a) Graphically.

(b) Using the quadratic formula.

(c) Using three iterations of the bisection method to determine the highest root. Employ initial guesses of x l = 5  and x u = 10 .

Compute the estimated error ε a and the true error ε t after each iteration.

(a)

Expert Solution
Check Mark
To determine

The real roots of the equation f(x)=0.5x2+2.5x+4.5 using the graphical method.

Answer to Problem 1P

Solution:

The real roots of the equation are 1.4 and 6.4.

Explanation of Solution

Given Information:

The equation f(x)=0.5x2+2.5x+4.5.

Calculation:

The graph of the function can be plotted using MATLAB.

Code:

clc;

x=linspace(-2,8);

y1=-0.5*x.^2+2.5*x+4.5;

y2=0;

plot(x,y1);

hold on

line([-2,8],[y2,y2])

Output:

This gives the following plot:

EBK NUMERICAL METHODS FOR ENGINEERS, Chapter 5, Problem 1P

The roots of an equation can be represented graphically by the x-coordinate of the point where the graph cuts the x-axis. From the plot, the two zeros of the equation can be approximated as 1.4 and 6.4.

(b)

Expert Solution
Check Mark
To determine

To calculate: The real roots of the equation f(x)=0.5x2+2.5x+4.5 using the quadratic formula.

Answer to Problem 1P

Solution:

The roots of the equation are 1.40512484 and 6.40512484.

Explanation of Solution

Given Information:

The equation f(x)=0.5x2+2.5x+4.5.

Formula Used:

The roots of an equation ax2+bx+c can be obtained using the quadratic formula as:

x=b±b24ac2a

Calculation:

Consider the provided equation,

f(x)=0.5x2+2.5x+4.5

Now substitute 0.5 for a, 2.5 for b and 4.5 for c in the quadratic formula to get the roots as:

x=2.5±2.524(0.5)(4.5)2(0.5)=2.5±3.905124841=1.40512484,6.40512484

Thus, the roots of the equation are 1.40512484 and 6.40512484.

(c)

Expert Solution
Check Mark
To determine

To calculate: The highest root of the equation f(x)=0.5x2+2.5x+4.5 using three iterations of the bisection method by assuming xl and xu to be 5 and 10 respectively while computing the true and estimated errors after each iteration.

Answer to Problem 1P

Solution:

The highest root of the equation can be approximated as 6.875. The true and approximate errors are as follows:

nεaεt117.1%220%2.42%39.09%7.32%

Explanation of Solution

Given Information:

The equation f(x)=0.5x2+2.5x+4.5.

Formula Used:

A root of an equation can be obtained using the bisection method as follows:

1. Choose 2 values x, say a and b such that f(a)f(b)<0.

2. Now, estimate the root by x1=a+b2.

3. If, f(a)f(x1)<0, the root would lie between a and x1. Now assume the next root to be x2=a+x12. If, f(a)f(x1)>0, the root would lie between b and x1. Now assume the next root to be x2=b+x12 and if f(a)f(x1)=0, the root is x1.

Calculation:

For the provided function:

f(5)=0.5(5)2+2.5(5)+4.5=4.5f(10)=0.5(10)2+2.5(10)+4.5=20.5

Hence,

f(5)f(10)<0

Now take the first root to be,

x1=5+102=7.5

As, the true root computed from part (b) was 6.40512484. Now, the true relative percentage error would be:

εt=(|6.405124847.56.40512484|×100)%=17.1%

The true error is 17.1%. There would be no approximate error for the first iteration.

Now,

f(7.5)=0.5(7.5)2+2.5(7.5)+4.5=4.875

Thus, f(5)f(7.5)<0. This implies that the root would be between 5 and 7.5.

Now, the second root would be:

x2=5+7.52=6.25

As, the true root computed from part (b) was 6.40512484. Now, the true relative percentage error would be:

εt=(|6.405124846.256.40512484|×100)%=2.42%

The true error is 2.42%.

The approximate error can be computed as:

εa=(|6.257.56.25|×100)%=20%

The approximate error is 2%.

Now,

f(6.5)=0.5(6.5)2+2.5(6.5)+4.5=0.59375

Thus, f(5)f(6.25)>0. This implies that the root would be between 6.25 and 7.5.

Now, the third root would be:

x3=6.25+7.52=6.875

As, the true root computed from part (b) was 6.40512484. Now, the true relative percentage error would be:

εt=(|6.405124846.8756.40512484|×100)%=7.34%

The true error is 7.34%.

The approximate error can be computed as:

εa=(|6.8756.256.875|×100)%=9.09%

The approximate error is 9.09%.

Thus, the highest root can be approximated as 6.875.

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Chapter 5 Solutions

EBK NUMERICAL METHODS FOR ENGINEERS

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