Question

how to simplify underroot solution and how can we get the roundoff answer of any rootSquare.

Answer 1

hmmm, maybe assume\[\sqrt{p}=\sqrt{n^2+\frac 1x}\]
\[p=n^2+\frac 1x \]and solve for x

Answer 2

i think i remembered that a little off

\[\sqrt{p}=n+\frac{1}{x}\]
\[p=n^2+\frac{2n}{x}+\frac{1}{x^2}\]
ignore the 1/x^2 and solve for x

Answer 3

spose we want say:\[\sqrt{123}\]that is close to 10^2 right?
\[\sqrt{123}=10+\frac{1}{x}\]
\[123=100+\frac{20}{x}\]
\[\frac{20}{23}=x\]therefore\[\sqrt{123}=abt~10+\frac{23}{20}\]

Answer 4

if i take 27

Answer 5

lets try to define the x in general by this method:
\[x=\frac{2n}{p-n^2}\]\[\sqrt{27}=5+\frac1x\]\[x=\frac{10}{27-25}=5\]
\[\sqrt{27}~=abt~5.2\]
we can keep refining i\[\sqrt{27}=5.2+\frac{1}{x}\]
\[x=\frac{2(5.2)}{27-(5.2)^2}\]

Answer 6

from where 5 comes

Answer 7

5 is a guess. since 5^2 = 25 and 6^2 = 36, we know that sqrt(27) is somewhere between 5 and 6, so 5+1/x is a way to refine it

Answer 8

this point i din't understand

Answer 9

we could have well of started with sqrt(27) = 1 + 1/x and did some iterations

x = 2/26; 1 + 26/2 = 14

let the new guess be 14

x = 28/(27-14^2) = -28/169

14 - 169/28 =abt 8

refine again

Answer 10

if we can get a good guess to begin with, the refinements converge quicker

Answer 11

another method is to approximate the sqrt(x) function with a linear function by determing the slope at say x=25

y = sqrt(x)
y' = 1/2sqrt(x) ; at x=25, y=1/10

using the point (25,5), and the slope of 1/10. we can determine an approximation for x=27 ... which is just 2 away from 25

y-5=1/10 (x-25)

y-5=1/10 (27-25)

y-5=1/10 (2)

y = 5.2