Question

Find a root of the equation x³-5x-7 that lies between 2 and 3
using
Regula-Falsi Method correct up to four decimal places.​

Answer 1

Given:-

•  x³ -  2x  - 5  = 0  

To Find:-

• root of the equation  by Bisection method correct up to three decimal places

Solution:-

x³ -  2x  - 5  = 0

• x = 1  => 1  - 2 - 5  = - 6

• x = 2 => 8 - 4 -  5 = - 1

• x = 3 =>  27 - 6 - 5 = 16

0 lies between -1  & 16

Hence x lies between 2 & 3

• (2 + 3)/2 = 2.5

• x = 2.5  =>     x³ -  2x  - 5  = 5.625

0 lies between -1  & 5.625

• => x lies between 2  &  2.5

• (2 + 2.5)/2 = 2.25

• x = 2.25  =>  x³ -  2x  - 5  = 1.89

0 lies between -1  & 1.89

• =>  x lies between 2  &  2.25

• => x = 2.125  => x³ -  2x  - 5  = 0.34

0 lies between -1  & 0.34

• => x lies between 2  × 2.125

• => x = 2.0625   => x³ -  2x  - 5  = -0.35

0 lies between  -0.35 & 0.34

• => x lies between  2.0625   & 2.125

• x = 2.09375    => x³ -  2x  - 5  =    -0.0089 ≈  0

• x =  2.094    up to three decimal places

x =  2.094  

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Learn more:

→Show the function f of x is equal to x cube minus 3 x square + 6 x ...

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→a+b+c=3 a^2+b^2+c^2=6 1/a+1/b+1/c=1 find values of a , b and c

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Answer 2

Answer:

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