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\begin{array}{l}f(x)=x^{3}+4 x^{2}-10 \\x \in[1,2]\end{array}1st Ioration\text { Here } f(1)=-5<0 \text { and } f(2)=14>0\therefore Now root lies behween land ?\begin{array}{l}x_{0}=\frac{1+2}{5}=1.5 \\f\left(x_{0}\right)=f(1.5)=1.5^{3}+4 \times 1.5^{2}-10=2.375>0\end{array}Ind GlerationHere. f(1)=-5<0 and f(1.5)=2.375>0\therefore now root lies between 1 and 1.5\begin{aligned}x_{1} & =\frac{1+1.5}{2}=1.25 \\f\left(x_{1}\right) & =f(1.25)=-1.79688<0\end{aligned}3rd GlerationHere f(1.25)=-1.79688<0 and f(1.5)=2.375>0\therefore now rout lies between 1.25 and 1.5\begin{aligned}x_{2} & =\frac{1.25+1.5}{2}=1.375 \\f\left(x_{2}\right) & =f(1.375)=0.16211>0\end{aligned}4rth IterationHere f(1.25)=-1.79688<0 and f(1.375)=0.16211>0\therefore now root lies between 1.25 and 1.275\begin{array}{l}x_{3}=\frac{1.25+1.375}{2}=1.3125 \\f\left(x_{s}\right)=f(1.3125)=-0.84839<0\end{array}Sth IterationHere f(1.3125)=-0.84839<0 and f(1.375)=0.16211>0\therefore row root lies between 1.3125 and 1.375\begin{aligned}x_{u} & =\frac{1.3125+1.375}{2}=1.34375 \\f\left(x_{4}\right) & =f(1.34375)=-0.35098<0\end{aligned}\text { Error (till } 5 \text { Steration) }=0.35098And for every gteration error =\left|f\left(c_{n}\right)\right|,\left(\begin{array}{c}n=\text { Iteration } \\ \text { number }\end{array}\right)\Rightarrow If we do gleration then It will guarantee to an accuracy of 10^{-5}. ...