Question

find the real root of the equation x^3-2x-5=0 using bisection method ​

Answer 1

Answer:

Find the real root of the equation 0 2 5 3 x  x  

using bisection method.

Answer 2

Concept introduction:

Usually an algebraic formula or a number that represents the solution to an equation.

Given:

The equation is given by $x^3$$-2x-5=0$ .

To Find :

We have to find the real root of the equation $x^3-2x-5=0$ using bisection method. ​

Solution:

$x = 1 = > 1 - 2 - 5 = - 6$

$x = 2 = > 8 - 4 - 5 = - 1$

$x = 3 = > 27 - 6 - 5 = 16$

$0 lies between -1 & 16$

$Hence x lies between 2 & 3$

$(2 + 3)/2 = 2.5$

$x = 2.5 = >$   $x^{2} - 2x - 5$  =$5.625$

$0 lies between -1 & 5.625$

$= > x lies between 2 & 2.5$

$(2 + 2.5)/2 = 2.25$

$0 lies between -1 & 1.89$

=>  $x lies between 2 & 2.25$

=> $x = 2.125 = > x^{2} - 2x - 5 = 0.34$

$0 lies between -1 & 0.34$

=> $x lies between 2 * 2.125$

=> $x = 2.0625 = > x³ - 2x - 5 = -0.35$$0$

$lies between -0.35 & 0.34$

=> $x lies between 2.0625 & 2.125$

$x = 2.09375 = > x^{2} - 2x - 5 = -0.0089 ≈ 0$

$x = 2.094$    up to three decimal places

Final Answer:

$x = 2.094$ is the real root of the given equation.

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