Set down:
p1.5 + p0.5/p-0.5 = 4.
p1.5 + p(0.5 − -0.5) = 4.
p1.5 + p1 = 4 or p1.5 + p − 4 = 0.
Methods Of Differential Calculus give the First Derivative of p1.5 + p1 − 4
as 1.5p(1.5 − 1) + 1p(1 − 1) − 0 or 1.5p0.5 + 1p0 or 1.5p0.5 + 1.
Placing an estimate of p equal to 1.5 into p1.5 + p gives a result of 3.337117307. This is reasonably close to 4, so adjust p = 1.5 to p = 1.7 and gain 1.71.5 + 1.7 or 3.916528818 which lands much closer to 4.
Now place "Function Over Derivative" according to Newton's Method Of Root Approximation and
build the expression p − [p1.5 + p − 4]/[1.5p0.5 + 1]. The table below shows the evaluation of this expression for p = 1.7. Each evaluation is fed back into the expression as "the new p".
p----------------------------------------------p − [p1.5 + p − 4]/[1.5p0.5 + 1]
1.7------------------------------------------------------1.728240169
1.728240169-----------------------------------------1.728163202
1.728163202-----------------------------------------1.728163201
1.728163201-----------------------------------------1.728163201
The table finally shows an evaluation of the expression equal to the input p and Newton's Method
can go no further.
Test p = 1.728163201 in p1.5 + p to obtain 3.999999999, which verifies an extremely accurate approximation of p for the equation given.