Answer:
The two numbers are a = 13. 5 and b = 6.51
Step-by-step explanation:
Data:
Let the two numbers be a and b respectively.
sum of two numbers is 20, so a + b = 20 ...1
sum plus square roots multiplied gives 155.55 so (a+√a) (b+√b) = 155.55 ...2
let f(y) = [20-b+√(20-b)]×[(b+√b)]-155.55 = 0
Multiplying out the terms gives;
f(y) = 20y-b²+20√b - b√b+b√(20-b)+√b·√(20-b) - 155.55 = 0
differentiating:
f'(y) = [tex]20-2y+\frac{10}{\sqrt{b} } -(\frac{b}{\sqrt{b} } +\sqrt{b)} + \frac{b}{2\sqrt{20-b} } +\sqrt{20-b} +\frac{\sqrt{20-b} }{2\sqrt{b} } +\frac{\sqrt{b} }{2\sqrt{20-b} }[/tex]
applying the Newton-Ralphson iteration method with the initial estimate of real zero roots gives P₀ = 10
That is [tex]P_{n+1} = P_{n} - \frac{f(P_{n} }{f'(P_{n}) }[/tex]
letting P₀ = 10 and iterating, the convergence occurs for b to 4 decimal places at P₇ = 6.51 = b
we also obtain from equation (1)
a + b = 20
a = 20 - b
a = 20 - 6.51
a = 13.5
Two numbers therefore are a= 13.5 and b = 6.51
