# The roots of the equation, x 2 ( 4 − x 2 ) = 4 x 2 + 1 .

### 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 18E
To determine

## To calculate: The roots of the equation,  x2(4−x2)=4x2+1 .

Expert Solution

The roots are

x2=0.84310821

x3=0.84310820 .

### Explanation of Solution

Given information:

The equation is given as:

x2(4x2)=4x2+1 .

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,

x2(4x2)=4x2+1

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

y=x2(4x2) and y=4x2+1 .

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

Let x1=0.843 where

f(x)=x6+3x4+4x24f'(x)=6x5+12x3+8x

Now, at n=1

x2=x1f(x1)f'(x1)

x2=x1{x16+3x14+4x1246x15+12x13+8x1}

x2=(0.843)((0.843)6+3(0.843)4+4(0.843)246(0.843)5+12(0.843)3+8(0.843))x2=(0.843)(0.35889338+1.515066004+2.84259642.554401282+7.188925284+6.744)x2=(0.843)(0.00123137611.378524)

x2=0.843(0.0001082193086)x2=0.843+0.0001082193086x2=0.843108219

The second approximation is x2=0.843108219 .

Let n=2 ,

x3=x2f(x2)f'(x2)

x3=x2{x26+3x24+4x2246x25+12x23+8x2}

x3=(0.843108219)((0.843108219)6+3(0.843108219)4+4(0.843108219)246(0.843108219)5+12(0.843108219)3+8(0.843108219))x3=(0.843108219)(0.359169903+1.515844132+2.84332587642.556041292+7.191694246+6.744865752)x3=(0.843108219)(0.00000010511.38051871)

x3=0.843108219(0.00000000922629299)x3=0.8431082190.00000000922629299x3=0.843108209

The third approximation with six decimal places is x3=0.843108209 .

Therefore, the roots with eight decimal places are :-

x2=0.84310821

x3=0.84310820

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