Hello,
Surely one of
f <- function(a) a^2 - a - 8.313
curve(f, from=0, to=1)
# zeros of f
root <- polyroot(c(-8.313, -1, 1))
ifelse(Im(root) == 0, Re(root), root)
# minimum of f
optimize(f, interval=c(0, 1))
Hope this helps,
Rui Barradas
Em 26-07-2012 22:13, Nordlund, Dan (DSHS/RDA) escreveu:
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Diviya Smith
Sent: Thursday, July 26, 2012 1:50 PM
To: r-help@r-project.org
Subject: Re: [R] Solving quadratic equation in R
Sorry it is important for me to constrain the value of 'a' between
c(0,1)
On Thu, Jul 26, 2012 at 4:48 PM, Diviya Smith
<diviya.sm...@gmail.com>wrote:
Hi there,
I would like to solve a simple equation in R
a^2 - a = 8.313
There is no real solution to this problem but I would like to get an
approximate numerical solution. Can someone suggest how I can set
this up?
Thanks in advance,
Diviya
What is your definition of approximate? For all values of a in the interval
[0,1], a^2-a is less than or equal to 0. Or am I missing something?
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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