- ahmad alhayek

 

Consider the following Polynomial f (x)=x* – 3x® – 10x² +x+2 Compute the root x=1 of the above function with actual error les

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    Find a root of an equation f(x) = x+ - 3r - 10x2 + x + 2 using Bisection method Solution: Here x4 - 3x3 - 10x2 + x + 2 = 0 Le

    2nd iteration : Here fo) - 2 > 0 and f(0.5) = -0.3125 <0 :: Now, Root lies between 0 and 0.5 x; -0705 , 0 +0.5 = 0.25 2 (21)

    4th iteration : Here f(0.375) = 0.8303 > 0 and 1(0.5) = -0.3125 < 0 : Now, Root lies between 0.375 and 0.5 X3 0.375 +0.5 2 0.

    6th iteration : Here f(0.4688) - 0.0108 > 0 and f(0.5) = -0.3125 <0 : Now, Root lies between 0.4688 and 0.5 X5 0.4688 +0.5 2

    8th iteration : Here f(0.4688) = 0.0108 > 0 and f(0.4766) = -0.0677 <0 : Now, Root lies between 0.4688 and 0.4766 X 0.4688 +0

    10th iteration : Here f(0.4688) - 0.0108 > 0 and f(0.4707) = -0.0087 <0 : Now, Root lies between 0.4688 and 0.4707 X9 0.4688

    12th iteration : Here f(0.4697) = 0.0011 > 0 and f(0.4702) = -0.0038 < 0 :: Now, Root lies between 0.4697 and 0.4702 0.4697 +

    Approximate root of the equation x+ - 3x3 - 10x2 + x + 2 = 0 using Bisection mehtod is 0.4698 (After 16 iterations) a fa) b f

    Find a root of an equation f(x) = x+ - 3x3 - 10x2 + x + 2 using Secant method Solution: Here x4 - 3x3 - 10x2 + x + 2 = 0 Let

    (x2) = R0.1818) = 0.18184 - 3.0.18183 - 10 - 0.18182 +0.1818 + 2 = 1.8343 2nd iteration : Xi = 1 and x2 = 0.1818 4(x1) = 11)

    3rd iteration : X2 = 0.1818 and x3 = 0.3203 1(x2) = R0.1818) = 1.8343 and f(x3) = R0.3203) = 1.2061 X3 - X2 :: X4 = x2-1 <2-4

    likewise, we get

    Approximate root of the equation x4 - 3x3 - 10x2 + x + 2 = 0 using Secant mehtod is 0.4698 (After 7 iterations) xo f(xo) 21 f

    Find a root of an equation f(x) = x* - 3x - 10x2 + x + 2 using False Position method (regula falsi method) Solution: Here x4

    F(x2) = R0.1818) = 0.18184-3.0.18183 - 10.0.18182 + 0.1818 +2 = 1.8343 > 0 2nd iteration : Here f(0.1818) = 1.8343 > 0 and f(

    *1-* X4 = Xo -f(x0) f(x1)--(20) X4 = 0.3203 - 1.2061 1 -0.3203 -9 - 1.2061 X4 = 0.4007 (**) = R0.4007) = 0.40074 - 3. 0.40073

    Axs) = R0.4398) = 0.43984- 3. 0.43983 - 10.0.43982 +0.4398 +2 = 0.2881 > 0 5th iteration : Here f(0.4398) = 0.2881 > 0 and f(

    likewise, we get iteration

    Approximate root of the equation x+ - 3x3 - 10x2 + x + 2 = 0 using False Position mehtod is 0.4698 (After 12 iterations) 71 x

    Find a root of an equation f(x) = x+ - 3x3 - 10x² + x + 2 using Newton Raphson method Solution: Here x4 - 3x3 - 10x2 + x + 2

    1iteration : Axo) = 100.5) = 0.54-3.0.5-10.0.52 +0.5 +2 = - 0.3125 f (x) = f (0.5) - 4.0.53-9.0.52 - 20-0.5 +1 = - 10.75 * =

    ܕܬ݁ = 0.4709 - -0.011 -9.9968 04698 - ܕ 3rd iteration : A(22) = R0.4698) = 0.46984- 3 - 0.46983 - 10. 0.46982 + 0.4698 +2 = 0

    likewise, we get

    Approximate root of the equation x4 - 3x - 10.x2 + x + 2 = 0 using Newton Raphson mehtod is 0.4698 (After 3 iterations) xo f(

    if you get like this

    thx

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