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

2. 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 = −25
=−5
Change of sign implies a solution lies between 𝑥=2 and 𝑥=3
f 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

3. 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 = −2
=−2
Change of sign implies a solution lies between 𝑥=0 and 𝑥=1
f 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

4. 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
0.0024
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.

5. 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.