Q.1 Given that one root of a nonlinear equation x3 − 9 x+ 1 = 0 lies between 2 and 4. Find the root correct to four significant digits by bisection method. Ans. 2.9428

Iteration x0 x1 x2 y0 = f(x0) f(x1) f(x2) Replacement0 2.000000 3.000000 2.500000 -9.0000 1.0000 -5.8750 x0 by x2 1 2.500000 3.000000 2.750000 -5.8750 1.0000 -2.9531 x0 by x2 2 2.750000 3.000000 2.875000 -2.9531 1.0000 -1.1113 x0 by x2 3 2.875000 3.000000 2.937500 -1.1113 1.0000 -0.0901 x0 by x2 4 2.937500 3.000000 2.968750 -0.0901 1.0000 0.4463 x1 by x2 5 2.937500 2.968750 2.953125 -0.0901 0.4463 0.1759 x1 by x2 6 2.937500 2.953125 2.945313 -0.0901 0.1759 0.0424 x1 by x2 7 2.937500 2.945313 2.941406 -0.0901 0.0424 -0.0240 x0 by x2 8 2.941406 2.945313 2.943359 -0.0240 0.0424 0.0092 x1 by x2 9 2.941406 2.943359 2.942383 -0.0240 0.0092 -0.0074 x0 by x2

10 2.942383 2.943359 2.942871 -0.0074 0.0092 0.0009 x1 by x2 11 2.942383 2.942871 2.942627 -0.0074 0.0009 -0.0033 x0 by x2 12 2.942627 2.942871 2.942749 -0.0033 0.0009 -0.0012 x0 by x2 13 2.942749 2.942871 2.942810 -0.0012 0.0009 -0.0002 x0 by x2 14 2.942810 2.942871 2.942841 -0.0002 0.0009 0.0003 x1 by x2

Q.2 Given that one root of a nonlinear equation x3 − x − 1 = 0 lies between 1 and 2. Find the root correct to four significant digits by bisection method. Ans. 1.3247 Iteration x0 x1 x2 y0 = f(x0) f(x1) f(x2) Replacement

0 1.000000 2.000000 1.500000 -1.00000000 5.00000000 0.87500000 x1 by x2 1 1.000000 1.500000 1.250000 -1.00000000 0.87500000 -0.29687500 x0 by x2 2 1.250000 1.500000 1.375000 -0.29687500 0.87500000 0.22460938 x1 by x2 3 1.250000 1.375000 1.312500 -0.29687500 0.22460938 -0.05151367 x0 by x2 4 1.312500 1.375000 1.343750 -0.05151367 0.22460938 0.08261108 x1 by x2 5 1.312500 1.343750 1.328125 -0.05151367 0.08261108 0.01457596 x1 by x2 6 1.312500 1.328125 1.320313 -0.05151367 0.01457596 -0.01871061 x0 by x2 7 1.320313 1.328125 1.324219 -0.01871061 0.01457596 -0.00212795 x0 by x2 8 1.324219 1.328125 1.326172 -0.00212795 0.01457596 0.00620883 x1 by x2 9 1.324219 1.326172 1.325195 -0.00212795 0.00620883 0.00203665 x1 by x2

10 1.324219 1.325195 1.324707 -0.00212795 0.00203665 -0.00004659 x0 by x2 11 1.324707 1.325195 1.324951 -0.00004659 0.00203665 0.00099479 x1 by x2 12 1.324707 1.324951 1.324829 -0.00004659 0.00099479 0.00047404 x1 by x2 13 1.324707 1.324829 1.324768 -0.00004659 0.00047404 0.00021371 x1 by x2 14 1.324707 1.324768 1.324738 -0.00004659 0.00021371 0.00008355 x1 by x2

Q.3 Given that one root of a nonlinear equation x3 − 2.5 x2 −2.46 x + 3.96 = 0 lies between 2 and 3. Find the root correct to four significant digits by bisection method. Ans. 2.876 Iteration x0 x1 x2 y0 = f(x0) f(x1) f(x2) Replacement

1 2.0000 3.0000 2.5000 -2.9600 1.0800 -2.1900 x0 by x2 2 2.5000 3.0000 2.7500 -2.1900 1.0800 -0.9144 x0 by x2 3 2.7500 3.0000 2.8750 -0.9144 1.0800 -0.0129 x0 by x2 4 2.8750 3.0000 2.9375 -0.0129 1.0800 0.5089 x1 by x2 5 2.8750 2.9375 2.9063 -0.0129 0.5089 0.2419 x1 by x2 6 2.8750 2.9063 2.8906 -0.0129 0.2419 0.1130 x1 by x2 7 2.8750 2.8906 2.8828 -0.0129 0.1130 0.0497 x1 by x2 8 2.8750 2.8828 2.8789 -0.0129 0.0497 0.0183 x1 by x2 9 2.8750 2.8789 2.8770 -0.0129 0.0183 0.0027 x1 by x2

10 2.8750 2.8770 2.8760 -0.0129 0.0027 -0.0051 x0 by x2 11 2.8760 2.8770 2.8765 -0.0051 0.0027 -0.0012 x0 by x2 12 2.8765 2.8770 2.8767 -0.0012 0.0027 0.0007 x1 by x2 13 2.8765 2.8767 2.8766 -0.0012 0.0007 -0.0002 x0 by x2 14 2.8766 2.8767 2.8766 -0.0002 0.0007 0.0002 x1 by x2