# The roots of the equation, x 6 − x 5 − 6 x 4 − x 2 + x + 10 = 0 .

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4.7, Problem 17E
To determine

## To calculate: The roots of the equation,  x6−x5−6x4−x2+x+10=0 .

Expert Solution

The roots are

x2=1.93822715

x3=1.93822883 .

### Explanation of Solution

Given information:

The equation is given as:

x6x56x4x2+x+10=0 .

Formula used:

Newton’s Method:

We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .

For x=xn , compute the next approximation xn+1 by

xn+1=xnf(xn)f'(xn) and so on.

Calculation:

Consider the equation,

x6x56x4x2+x+10=0

Now, the equation can be written as

x6x56x4=x2x10

The initial approximation x1 can be guessed by sketching both the graphs

y=x6x56x4 and y=x2x10 .

Both the graph intersect at three points whose x-coordinate can be seen in graph.

Let x1=1.938 where

f(x)=x6x56x4x2+x+10f'(x)=6x55x424x32x+1

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1x16x156x14x12+x1+106x155x1424x132x1+1

x2=(1.938)((1.938)6(1.938)56(1.938)4(1.938)2+(1.938)+106(1.938)55(1.938)424(1.938)32(1.938)+1)x2=(1.938)(52.98130316+27.3382337384.638184913.7558441.938+10164.029402470.53182076+174.6918161+3.876+1)x2=(1.938)(0.0124920254.99340706)

x2=1.938(0.0002271548658)x2=1.9380.0002271548658x2=1.938227155

The second approximation is x2=1.938227155 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2x26x256x24x22+x2+106x255x2424x232x2+1

x3=(1.938227155)(53.01857405+27.3541591484.677874013.7567245041.938227155+10164.124954870.56489501+174.7532508+3.87645431+1)x3=(1.938227155)(0.00009247955.0601447)

x3=1.938227155(0.000001679599654)x3=1.9382271550.000001679599654x3=1.938228835

The third approximation with six decimal places is x3=1.938228835 .

Therefore, the roots with eight decimal places are :-

x2=1.93822715

x3=1.93822883

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