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Find a root of x4x3+10x+7=0 correct up to three decimal places between 2 and 1 by Newten Raphson method.
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Solution:

f(x)=x4x3+10x+7=0

The root lies between 2 and 1,

Let x0=2

f(x)=4x33x2+10

By the Newton - Raphson method,

xn+1=xnf(xn)f(xn)

f(x0)=f(2)=11

f(x0)=f(2)=34

x1=x0f(x0)f(x0)=211(34)

x1=1.6765

f(x1)=f(1.6765)=2.8468

f(x1)=f(1.(765)=17.2802.

x2=x1f(x1)f(x1)

=1.67652.8468(17.2802)

x2=1.5118

f(x2)=f(1.5118)=0.561f(x2)=f(1.5118)=10.6777x3=x2f(x2)f(x2)=1.51180.561(10.6777)x3=1.4593f(x3)=f(1.4593)=0.0497f(x3)=f(1.4593)=8.8193 Now, x4=x3f(x3)f(x3)=1.45930.0497(8.8193)x4=1.4537f(x4)=t(1.4537)=0.0008f(x4)=f(1.4537)=8.6278x5=x4f(x4)f(x4)=1.45370.0008(8.6278)x5=1.4536

Since , X4 and X5 are same up to three decimal places, the root is -1.453.

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