Q.1 (f) Using Newton-Raphson Method, find the real root of x³-2x-5=0 correct to three decimal places.

f(x)=x³-2x-5,\ f'(x)=3x²-2 . f(2)=-1 0 : root lies in (2, 3), nearer 2. Take x₀=2 . x_(n+1)=x_n-x_n³-2x_n-5/3x_n²-2=2x_n³+5/3x_n²-2 x₁=2(8)+5/3(4)-2=21/10=2.1 x₂=2(9.261)+5/3(4.41)-2=23.522/11.23=2.09457 x₃=2(9.18932)+5/3(4.38722)-2=2.09455 x₂ and x₃ agree to three decimals. Root ≈2.095 (more precisely 2.0946).

Using Newton-Raphson Method, find the real root of \(x^3-2x-5=0\) correct to three decimal places.

Q.1 (f) Using Newton-Raphson Method, find the real root of \(x^3-2x-5=0\) correct to three decimal places.

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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\(f(x)=x^3-2x-5,\ f'(x)=3x^2-2\). \(f(2)=-1<0,\ f(3)=16>0\): root lies in (2, 3), nearer 2. Take \(x_0=2\).

\(x_{n+1}=x_n-\frac{x_n^3-2x_n-5}{3x_n^2-2}=\frac{2x_n^3+5}{3x_n^2-2}\)

\(x_1=\frac{2(8)+5}{3(4)-2}=\frac{21}{10}=2.1\)
\(x_2=\frac{2(9.261)+5}{3(4.41)-2}=\frac{23.522}{11.23}=2.09457\)
\(x_3=\frac{2(9.18932)+5}{3(4.38722)-2}=2.09455\)

\(x_2\) and \(x_3\) agree to three decimals. Root \(\approx2.095\) (more precisely 2.0946).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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Using Newton-Raphson Method, find the real root of \(x^3-2x-5=0\) correct to three decimal places.

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