Numerical Methods for solutions of equations Decimal Search method Tuesday, 10 September 2019

Example Show that the equation 𝑥 2 +8𝑥−25=0 has a solution between 𝑥=2 and 𝑥=3. Use Decimal search to obtain values of this solution correct to 2 decimal places. Let f 𝑥 = 𝑥 2 +8𝑥−25 f 2 = 2 2 +8 2 −25 =−5 Change of sign implies a solution lies between 𝑥=2 and 𝑥=3 f 3 = 3 2 +8 3 −25 =8 𝑥 𝑓(𝑥) 2.1 2.2 2.3 2.4 2.5 Change of sign implies a solution lies between 𝑥=2.4 and 𝑥=2.5 -3.79 -2.56 -1.31 -0.04 1.25 𝑥 𝑓(𝑥) 2.40 2.41 Change of sign implies a solution lies between 𝑥=2.40 and 𝑥=2.41 -0.04 0.0881 Solution 𝑥=2.40 to 2 decimal places

Has a solution between 𝑥=0 and 𝑥=1 Example Show that the equation 𝑥 3 +4𝑥−2=0 Has a solution between 𝑥=0 and 𝑥=1 Hence, using the decimal search method find this solution correct to 3 decimal places. Let f 𝑥 = 𝑥 3 +4𝑥−2 f 0 = 0 3 +4 0 −2 =−2 Change of sign implies a solution lies between 𝑥=0 and 𝑥=1 f 1 = 1 3 +4 1 −2 =3 𝑥 𝑓(𝑥) 0.1 0.2 0.3 0.4 0.5 -1.599 -1.192 -0.773 -0.336 0.125 𝑥 𝑓(𝑥) 0.40 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48 -0.336 -0.291 -0.245 -0.200 -0.154 -0.108 -0.062 -0.016 0.030

Solution 𝑥=0.474 to 3 decimal places 𝑓(𝑥) 0.470 0.471 0.472 0.473 0.474 -0.016 -0.011 -0.006 -0.0021 0.0024 Solution 𝑥=0.474 to 3 decimal places Question1 Show that the equation 4𝑥 3 −2𝑥−5=0 Has a solution between 𝑥=1 and 𝑥=2 Hence, using the decimal search method find this solution correct to 3 decimal places.

Question2 Show that the equation 𝑥 3 +𝑥−4=0 Has a solution between 𝑥=1 and 𝑥=2 Hence, using the decimal search method find this solution correct to 2 decimal places.