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Find a root of x4−x3+10x+7=0 correct up to three decimal places between −2 and −1 by Newten Raphson method.
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written 2.9 years ago by | • modified 2.9 years ago |
Solution:
⇒f(x)=x4−x3+10x+7=0
The root lies between −2 and −1,
Let x0=−2
f′(x)=4x3−3x2+10
By the Newton - Raphson method,
xn+1=xn−f(xn)f′(xn)
f(x0)=f(−2)=11
f′(x0)=f′(−2)=−34
∴x1=x0−f(x0)f(x0)=−2−11(−34)
⇒x1=−1.6765
f(x1)=f(−1.6765)=2.8468
f′(x1)=f′(−1.(765)=−17.2802.
x2=x1−f(x1)f′(x1)
=−1.6765−2.8468(−17.2802)
⇒x2=−1.5118
f(x2)=f(−1.5118)=0.561f′(x2)=f′(−1.5118)=−10.6777∴x3=x2−f(x2)f(x2)=−1.5118−0.561(−10.6777)⇒x3=−1.4593f(x3)=f(−1.4593)=0.0497f′(x3)=f′(−1.4593)=−8.8193 Now, x4=x3−f(x3)f′(x3)=−1.4593−0.0497(−8.8193)⇒x4=−1.4537f(x4)=t(−1.4537)=0.0008f′(x4)=f′(−1.4537)=−8.6278∴x5=x4−f(x4)f′(x4)=−1.4537−0.0008(−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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