Question

Find the roots of the equation `x^4+x^3 -7x^2-x+5 = 0` which lies between 2 and 2.1 correct to 3 places of decimals using Regula Falsi method.

Answer 1

Given that a=2 and b=2.1.
`f(2) = (2)^4+(2)^3-7(2)^2-2+5 = -1.`

`f(2.1)=(2.1)^4+(2.1)^3-7(2.1)^2-(2.1)+5` = 0.739100.

`x_1=(af(b)-bf(a))/(f(b)-f(a))` = (2×0.73910—1))×(2.1)0.739100—1 = 2.05750. ……………(1)

`f(x_1) = (2.05750)^4 + (2.05750)^3 - 7(2.05750)^2 - 2.05750+5`

= - 0.05973.

`x_2=(af(x_1)-x_1f(a))/(f(x_1)-f(a))`= (2×(-0.05973)—1))×(2.05750)0.05973—1 = 2.061152 ……………(2)

`f(x_2) = (2.061152)^4 + (2.061152)^3 -7(2.061152)^2-2.061152 + 5`

= 0.005326.

`x_3=(af(x_2)-x_2f(a))/(f(x_2)-f(a))` = (2×(0.005326)—1))×(2.061152)0.005326—1 = 2.06082

`f(x_3) =(2.06082)^4 + (2.06082)^3 - (2.06082)^2-2.06082+5`

= - 0.000582.

`x_4=(af(x_3)-x_3f(a))/(f(x_3)-f(a))` = (2×(-0.000582)—1))×(2.06082)0.000582—1 = 2.0608. ………………(4)

Hence from (4) and (3) iteration we get that value of x is coinciding.

Therefore the final value of x is 2.0608.