Question #180294

x2-2x-3=0 for the positive root by iteration method.


1
Expert's answer
2021-05-10T14:15:19-0400

Given equation is x2x3=0......(1)To initiate the iterative method ,we find the interval in which (1) has a root by using Intermetiate value theorem.Consider value of (1) at x=1 which is (1)213=14=3<0At x=2 ,2223=45=1<0At x=3, 3233=3>0By Intermediate value theorem there which states that, if f(x) iscontinuous in an interval[a,b]andf(a)f(b)<0,then  there exists c in (a,b) such that f(c)=0.Hence,equation(1) has a root in (2,3).Let us set up the iterative equation for xFrom (1), we have x2=x+3x=x+3 The iterative equation for the root is xn=xn1+3, n=1,2,3,....Let x0=Midvalue of (2,3)=2+32=1.5x0=1.5x1=1.5+3=4.5=2.1213203436x1=2.1213203436x2=x1+3=2.1213203436+3=5.1213203436=2.2630334385 x2=2.2630334385x3=x2+3=2.2630334385+3=5.2630334385=2.2941302139 x3=2.2941302139x4=x3+3=2.2941302139+3=5.2941302139=2.3008976974 x4=2.3008976974x5=x4+3=2.3008976974+3=5.3008976974=2.3023678458 x5=2.3023678458x6=x5+3=2.3023678458+3=5.3023678458=2.3026870925 x6=2.3026870925x7=x6+3=2.3026870925+3=5.3026870925=2.3027564119 x7=2.3027564119x8=x7+3=2.3027564119+3=5.3027564119=2.3027714632 x8=2.3027714632x9=x8+3=2.3027714632+3=5.3027714632=2.3027747313 x9=2.3027747313Since, x8,and  x9 are equal upto 5th decimalthe approximate root of (1) is x=2.30277(upto 5 decimals)Given\ equation \ is \ x^2-x-3=0......(1)\\ To \ initiate \ the \ iterative \ method \ , we \ find \ the \ interval \ \\ in \ which\ (1) \ has\ a\ root \ by \ using \ Intermetiate \ value \ theorem.\\ Consider \ value \ of \ (1) \ at \ x= 1\ which \ is \ (1)^2 -1-3=1-4=-3<0\\ At \ x= 2\ , 2^2 -2-3=4-5=-1<0\\ At \ x=3, \ 3^2 -3-3=3>0\\ By \ Intermediate \ value \ theorem \ there \ which\ states\ that, \ if \ f(x) \ is \\continuous \ in \ an \ interval [a,b] and f(a)f(b)<0, then \ \ there \ exists \ c \ in \ (a, b) \ such \ that \ f(c)=0.\\ Hence, equation ( 1) \ has \ a \ root \ in \ (2, 3). Let\ us \ set \ up \ the \ iterative \ equation \ for \ x\\ From \ (1),\ we \ have \ x^2 =x+3\Rightarrow x=\sqrt{x+3}\\ \Rightarrow\ The \ iterative \ equation \ for \ the \ root \ is \ x_n=\sqrt{x_{n-1}+3},\ n=1, 2,3 , ....\\ Let \ x_{0}=Midvalue \ of\ (2,3)=\frac{2+3}{2}=1.5\Rightarrow x_{0}=1.5\\ x_{1}=\sqrt{1.5+3}\\ =\sqrt{4.5}\\ =2.1213203436 \Rightarrow x_{1}=2.1213203436\\ x_2=\sqrt{x_{1}+3}\\ =\sqrt{2.1213203436+3}\\ =\sqrt{5.1213203436}\\ =2.2630334385 \Rightarrow \ x_{2}= 2.2630334385\\ x_3=\sqrt{x_{2}+3}\\ =\sqrt{2.2630334385+3}\\ =\sqrt{5.2630334385}\\ =2.2941302139 \Rightarrow \ x_{3}= 2.2941302139\\ x_4=\sqrt{x_{3}+3}\\ =\sqrt{2.2941302139+3}\\ =\sqrt{5.2941302139}\\ =2.3008976974 \Rightarrow \ x_{4}= 2.3008976974 \\ x_5=\sqrt{x_{4}+3}\\ =\sqrt{2.3008976974+3}\\ =\sqrt{5.3008976974}\\ =2.3023678458 \Rightarrow \ x_{5}= 2.3023678458 \\ x_6=\sqrt{x_{5}+3}\\ =\sqrt{2.3023678458+3}\\ =\sqrt{5.3023678458}\\ =2.3026870925 \Rightarrow \ x_{6}= 2.3026870925\\ x_7=\sqrt{x_{6}+3}\\ =\sqrt{ 2.3026870925+3}\\ =\sqrt{ 5.3026870925}\\ =2.3027564119 \Rightarrow \ x_{7}= 2.3027564119\\ x_8=\sqrt{x_{7}+3}\\ =\sqrt{2.3027564119+3}\\ =\sqrt{5.3027564119}\\ =2.3027714632 \Rightarrow \ x_{8}= 2.3027714632\\ x_9=\sqrt{x_{8}+3}\\ =\sqrt{2.3027714632+3}\\ =\sqrt{5.3027714632}\\ =2.3027747313\Rightarrow \ x_{9}=2.3027747313\\ Since, \ x_8, and \ \ x_9 \ are\ equal \ upto \ 5th \ decimal\\ the\ approximate \ root\ of \ (1) \ is \ x= 2.30277(upto\ 5 \ decimals)


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