Question

x^3-2x-5=0up to 3 decimal places by secant method​

Answer 1

Step-by-step explanation:

Here the equation is  x³ - 2·x - 5 = 0

Let f(x) = x³ - 2·x - 5

Now,

x       0              1          2          3

f(x)     -5           -6          -1         16

1st iteration :

Here f(2) = -1 < 0 and f(3) = 16 > 0

∴ Root lies between 2 and 3

\begin{gathered} x_0=\frac{2+3}{2}=2.5\\\\f(x_0)=f(2.5)=(2.5)\cdot 3-2\cdot (2.5)-5=5.625 > 0\end{gathered}x0=22+3=2.5f(x0)=f(2.5)=(2.5)⋅3−2⋅(2.5)−5=5.625>0

2nd iteration :

Here f(2) = -1 < 0 and f(2.5) = 5.625 > 0

∴ Now, Root lies between 2 and 2.5

\begin{gathered} x_1=\frac{2+2.5}{2}=2.25\\\\f(x_1)=f(2.25)=(2.25)\cdot 3-2\cdot (2.25)-5=1.89062 > 0\end{gathered}x1=22+2.5=2.25f(x1)=f(2.25)=(2.25)⋅3−2⋅(2.25)−5=1.89062>0

3rd iteration :

Here f(2) = -1 < 0 and f(2.25) = 1.89062 > 0

∴ Now, Root lies between 2 and 2.25

\begin{gathered} x_2=\frac{2+2.25}{2}=2.125\\\\f(x_2)=f(2.125)=(2.125)\cdot 3-2\cdot (2.125)-5=0.3457 > 0\end{gathered}x2=22+2.25=2.125

4th iteration :

Here f(2) = -1 < 0 and f(2.125) = 0.3457 > 0

∴ Now, Root lies between 2 and 2.125

\begin{gathered} x_3=\frac{2+2.125}{2}=2.0625\\\\f(x_3)=f(2.0625)=(2.0625)\cdot 3-2\cdot (2.0625)-5=-0.35132 > 0\end{gathered}x3=22+2.125=2.0625f(x3)=f(2.0625)=(2.0625)⋅3−2⋅(2.0625)−5=−0.35132>0

So, the root of the given equation upto 4 places of decimal is 2.0625

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