Square root formula in Calculated Characteristics
Have a great days.
There are two MIC in my inspection plan. first (0010) is for entering number and another (0020) is calculated characteristics. I want to calculate the square root of 0010 charateristics in 0020.
What should be teh formula to put in 0020 char.
Please guide. thanks in advance.
Regards,
Dipesh Bhavsar
Edited by: dipeshbhavsar1982 on Jul 31, 2011 9:41 AM
HI
0020 MIC should be created with control indicator for formula tab(calc.characteristic)
and in quality plan assign the first MIC as normal
and assign second MIC then system will ask a formula ...
there you can give the formula in terms of character
example (0010)2
Faisal
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I have a program that has three columns, in the first column it lists numbers 0-10, in the second column I need to get the square root of each number in the first column. How would I get those results. What is the formula for the square root and how would I code it in? Here is my code so far
public class ThreeColumn {
public static void main(String[] args) {
System.out.println("Number \t Square root \t Even/Odd \t \n");
int Ans1 = 0, Ans2 = 0, Ans3 = 0;
for (int i = 0; i <= 10; i++)
System.out.println(Ans1 + "\t" );
Ans1++;
}ThanksHere's something to play around with
import java.text.*;
class Testing
public Testing()
DecimalFormat df = new DecimalFormat("0.00");
String[] header = {"Number", "Sqr Root","Even/Odd"};
String pad = " ";
for(int x = 0; x < header.length; x++) System.out.print(header[x]+" ");
System.out.println("\n========================");
String sqrNum;
String[] evenOdd = {"Even","Odd"};
for(int x = 1; x <= 10; x++)
sqrNum = df.format(Math.sqrt(x));
System.out.println((pad+x).substring(pad.length()-header[0].length()+(""+x).length())+
(pad+sqrNum).substring(pad.length()-header[1].length()+sqrNum.length()-1)+
(pad+evenOdd[x%2]).substring(pad.length()-header[2].length()+evenOdd[x%2].length()-1));
System.exit(0);
public static void main(String[] args){new Testing();}
} -
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A Java Tutorial is here: http://java.sun.com/docs/books/tutorial/ -
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ForLoop Square Root Calculator Program
* Programmer: R McBride
* Date: November 29, 2006
* Filename: SquareRootForLoop.java
* Purpose: This program uses for()loops to
* find the square root of a number
* up to four decimal places long.
import APCS.Keyboard;
class SquareRootForLoopTest1
public static void main (String[]args)
//define variables
double SquareRoot = 0; //holds initial value
double num1 = 0;
boolean done = false;
double tempReg1 = 0; //holds initial value of i
double tempReg2 = 0; //holds
double tempReg3 = 0;
//while loop start
// while (!done)
//program input
System.out.println("\t\t\tSQUARE ROOT CALCULATOR PROGRAM");
System.out.println(""); //skip line
System.out.println("Enter the number you would like to be square rooted or [0] to end the program.");
System.out.print("");
SquareRoot = Keyboard.readInt();
System.out.println(""); //skip line
System.out.println("You entered "+SquareRoot+".");
System.out.println(""); //skip line
//for loop function
for(int i = 0; i < 999999; i++)
tempReg1 = i * i;
tempReg2 = i;
// while (!done)
if(tempReg1 == SquareRoot)
System.out.println("The square root is "+tempReg2);
System.out.println("");
else
System.out.print("");
//while loop check to finish
// if(SquareRoot == 0)
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}Here is my code for and of you who ever wants to use it. Our assignment was to make a square root calculator using ForLoop (a.k.a. no square root functions).
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* find the square root of a number
* up to four decimal places long.
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//define variables
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double num1 = 0;
boolean done = false;
double tempReg1 = 0; //holds initial value of i
double tempReg2 = 0; //holds
double tempReg3 = 0;
//while loop start
// while (!done)
//program input
System.out.println("\t\t\tSQUARE ROOT CALCULATOR PROGRAM");
System.out.println(""); //skip line
System.out.println("Enter the number you would like to be square rooted or [0] to end the program.");
System.out.print("");
SquareRoot = Keyboard.readInt();
System.out.println(""); //skip line
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System.out.println(""); //skip line
//for loop function
for(double i = 0; i < 999999; i+=.0001)
tempReg1 = i * i;
tempReg2 = i;
// while (!done)
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System.out.println("The square root is "+tempReg2);
System.out.println("");
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}Still doesn't work. =[ -
Calculating the square root...
Hi all,
let me shortly sketch the situation.
I have a number of fields that I select from the database (coming from different queries)
that I need to do calculations on...
so far, I was able to do most of those calculations, but now I'm stuck.
One of my calculations contains a square root.
So I got something like:
VARIABLE1 + VARIABLE2 + SQRT(3*VARIABLE1)
with VARIABLE1 being a multiplication of several other fields.
my question: is there an xsl or xdofx function to calculate the square root? Or any other solutions for my problem?
Any help would be greatly appreciated :)
Wkr,
FilipYou can use like this.
<?xdoxslt:pow(10,2)?>
<?xdoxslt:pow(10,3)?>
<?xdoxslt:pow(10,4)?> -
Square root calculation method
I'm trying to create java code that displays the square root of a number that the user enters, as I am not a great mathematician i cannot work out the logic for this. Any help would be great.
This is the part of my code that I need assistance with: n1 is the number that the user enters and n2 is where the square root is returned to. I'm guessing that i need to have a loop of some sort to divide n1, but am unsure of the conditions. The n1 is 0 so that after the total is calculated from n1, it is cleared for another number to be entered.
else if (source == btnSquareRoot)
n2 = ;
n1 = 0;
Please help!An adequate method for finding a square root of a number x is this:
Choose a number that is too low to be the actual square root. (How you do this, I don't know, but zero is pretty small and might be low enough)
Call that number lo:
Choose a number that is too high to be the actual square root. (Same comment. well almost. Don't use zero for this - it's too small)
Call it hi:
Note, it is easy to check and see that the numbers you have choosen have the required properties. because lo*lo < x and hi*hi > x
Now you have two candidates for the square root, one of them too low and one of them two high. The actual square root lies somewhere between those two bounds.
What you want to do now is squeeze those bounds tighter. You do that by choosing a number that is between lo and hi. The mid point would be a nice choice. You may need to figure out how to compute that.
call that number a:
well now, either a was itself too low, too high or just right. If it was too low, why replace lo with a and you have just improved the lower bound, if it was too high...
surely you get the idea.
Now the question is: How long do you keep this up? Do you ever get the actual square root this way? The answer is: Of course you don't. Most of the time the actual square root requires an infinite number of decimals to represent it. All you are looking for is something that is good enough.
What does good enough mean? Did you want the number correct to 2 decimal places, to 4 decimals. What do you want? How can you tell if lo and hi are practically the same number?
If this was a homework problem, that of course goes back to what the teacher wants. If the teacher did not clearly specify, you need to go back and ask what they actually wanted.
On the other hand if you want to earn a reputations as a smart ass, you figure out some way to detect that the number that you are working on does not have an exact square root and when it does not, you simply print out the message "Sorry - the square root of the number you are looking for cannot be represented in a finite number of digits without resorting to the notation of continued fractions but here it is to 3 decimals..."
This is the sort of stuff that wins big bonus points with professors.
On the other hand the way you lose big points with the professors if you say something like that and CAN NOT explain to him what a continued fraction is and explain what you meant by the qualification.
The other thing that loses big is to cut a chunk of text directly off of a web page and try to fob it off as your own program. Since most professors know how to google, they will stuff some unlikely looking phrase like "notation of continued fractions" and boom they are immediately at this page.
My point is that if you are going to be a smart ass, you MUST do it carefully or you lose all credibility.
Sorry for the digression there. Hope this algorithm outline is enough to get you thinking.
Wow! Killer. I previewed my question and it put asterisks in where I wrote the word, "ass". That is Fucking awesome! They've put in some kind of bad word filter so that when I get fucking abusive with my language and call you a no good shit for brains mother fucker. it cleans it all up for me. That is just too fucking sweet! It sure is a load off of my mind. I guess they just couldn't figure out how to put in !@%#^ which is the way that you actually are supposed to replace vulgarity in the printed word. Oh well, we always knew that this site was maintained by a bunch of **** ***** * **** ***** ** ***** * * *********! -
Hi, I'm suppose to write a programme that asks the user to enter a number. Find the square root of that number. Format theoutput to 4 decimal places. Use Math.sqrt(double a). i.e. to find the square root of 9.56, use: double sq = Math.sqrt(9.56);
This is what I have:
import java.io.*;
import java.math.*;
public class MethodEx2
static BufferedReader keyboard = new BufferedReader (new
InputStreamReader(System.in));
public static void main (String[] args) throws IOException
double num = 0;
System.out.print("Please enter a number");
System.out.flush();
num = Integer.parseInt(keyboard.readLine());
calsqrt(num);
public static double sqrt (double number);
double result = 0;
result = sqrt(number);
System.out.print(result);
}I keep getting an error saying that it cannot return a value from method whose result type is void at line 33, col 11.
How do I make the program to work?Oh okay, thanks. I also found another typo. I fixed it but it still doesn't work. Anyway, this is my new code:
import java.io.*;
import java.math.*;
public class MethodEx2
static BufferedReader keyboard = new BufferedReader (new
InputStreamReader(System.in));
public static void main (String[] args) throws IOException
double num = 0;
System.out.print("Please enter a number");
System.out.flush();
num = Integer.parseInt(keyboard.readLine());
sqrt(num);
public static double sqrt (double number)
double result = 0;
result = sqrt(number);
System.out.print(result);
}It says that MethodEx2.java is missing return statement at line 17, column 3. I'm not sure which variable I should return. I tried 'number' and 'result' but something like this happens:
"C:\Program Files\JBuilder9\jdk1.4\bin\javaw" -classpath "E:\Lili\School\GR12 Comp Sci\Period 1\Programs\MethodExs\Ex2\Ex2\classes;C:\Program Files\JBuilder9\jdk1.4\demo\jfc\Java2D\Java2Demo.jar;C:\Program Files\JBuilder9\jdk1.4\demo\plugin\jfc\Java2D\Java2Demo.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\charsets.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\dnsns.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\ldapsec.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\localedata.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\sunjce_provider.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\im\indicim.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jaws.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jce.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jsse.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\rt.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\sunrsasign.jar;C:\Program Files\JBuilder9\jdk1.4\lib\dt.jar;C:\Program Files\JBuilder9\jdk1.4\lib\htmlconverter.jar;C:\Program Files\JBuilder9\jdk1.4\lib\tools.jar" MethodEx2
Please enter a number9
java.lang.StackOverflowError
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
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at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19)
at MethodEx2.sqrt(MethodEx2.java:19) -
How to find square root, log recursively???
I need to find the square root of a number entered recursively and log as well. Your help would be greatly appreciated. Thanks in advance!
import java.io.*;
/**Class provides recursive versions
* of simple arithmetic operations.
public class Ops2
private static BufferedReader in = null;
/**successor, return n + 1*/
public static int suc(int n)
return n + 1;
/**predecessor, return n - 1*/
public static int pre(int n)
if (n == 0)
return 0;
else
return n - 1;
/**add two numbers entered*/
public static int add(int n, int m)
if (m == 0)
return n;
else
return suc(add(n, pre(m)));
/**subtract two numbers entered*/
public static int sub(int n, int m)
if (n < m)
return 0;
else if (m == 0)
return n;
else
return pre(sub(n, pre(m)));
/**multiply two numbers entered*/
public static int mult(int n, int m)
if (m == 0)
return 0;
else
return add(mult(n, pre(m)), n);
/**divide two numbers entered*/
public static int div(int n, int m)
if (n < m)
return 0;
else
return suc(div(sub(n, m), m));
/**raise first number to second number*/
public static int exp(int n, int m)
if (m == 0)
return 1;
else
return mult(exp(n, pre(m)), n);
/**log of number entered*/
public static int log(int n)
if (n < 2)
return 0;
else
return suc(log(div(n, 2)));
/**square root of number entered*/
public static int sqrt(int n)
if (n == 0)
return 0;
else
return sqrt(div(n, ));
/**remainder of first number entered divided by second number*/
public static int mod(int n, int m)
if (n < m)
return 0;
else
return mod(div(n, pre(m)), m);
public static void prt(String s)
System.out.print(s);
public static void prtln(String s)
System.out.println(s);
public static void main(String [ ] args)
prtln("Welcome to the amazing calculator");
prtln("It can add, multiply and do powers for");
prtln("naturals (including 0). Note that all the");
prtln("HARDWARE does is add 1 or substract 1 to any number!!");
in = new BufferedReader(new InputStreamReader ( System.in ) );
int It;
while ( (It = getOp()) >= 0)
prt("" + It + "\n");
private static int getOp( )
int first, second;
String op;
try
System.out.println( "Enter operation:" );
do
op = in.readLine( );
} while( op.length( ) == 0 );
System.out.println( "Enter first number: " );
first = Integer.parseInt( in.readLine( ) );
System.out.println( "Enter second number: " );
second = Integer.parseInt( in.readLine( ) );
prtln("");
prt(first + " " + op + " " + second + " = ");
switch( op.charAt( 0 ) )
case '+':
return add(first, second);
case '-':
return sub(first, second);
case '*':
return mult(first, second);
case '/':
return div(first, second);
case '^':
return exp(first, second);
case 'v':
return log(first);
case 'q':
return sqrt(first);
case '%':
return mod(first, second);
case 's':
return suc(first);
case 'p':
return pre(first);
default:
System.err.println( "Need +, *, or ^" );
return -1;
catch( IOException e )
System.err.println( e );
return 0;
}Hi,
Is there any one to make a program for me in Turbo
C++ for Dos, which can calculate the square root of
any number without using the sqrt( ) or any ready
made functions.
The program should calculate the s.root of the number
by a formula or procedure defined by the user
(programmer).
Thanks.This is a Java forum!
If you want Java help:
1. Start your own thread.
2. Use code tags (above posting box) if you post code.
3. No one will write the program for you. We will help by answering your questions and giving advice on how to fix problems in code you wrote.
4. The formula you need to implement is given above by dizzy. -
Sampling procedure-Square root n+1
Hi All,
I need to calculate the sample size based on the Containers & Square root n+1.
How to configure the system & how to maintain the master data to get the sample size.
User also insisted only the sample size has to be calculated, but he will enter only one composite result entry
Regards
SubbuHi Subbu,
you can use Physical Sample to control the sample Size.
You must to define a Sample Drawing Procedure (QPV2) as relevant for Container Number, then assign the formula 'Square Root n1´ ( TRUNC(SQRT(P2)1) ) in the partial sample. In this scenario, I suggest you to configure the "Container number' as a mandatory field during the goods receipt once the system will need this information to calculate the sample. For a single result recording, you can define a pooled sample.
I hope it can help you.
Best regards,
Robson -
I expect no replies to this thread - because there are no
answers, but I want to raise awareness of a faculty of other
languages that is missing in Flash that would really help 3D and
games to be built in Flash.
Below is an optimisation of the Quake 3 inverse square root
hack. What does it do? Well in games and 3D we use a lot of vector
math and that involves calculating normals. To calculate a normal
you divide a vector's parameters by it's length, the length you
obtain by pythagoras theorem. But of course division is slow - if
only there was a way we could get 1.0/Math.sqrt so we could just
multiply the vector and speed it up.
Which is what the code below does in Java / Processing. It
runs at the same speed as Math.sqrt, but for not having to divide,
that's still a massive speed increase.
But we can't do this in Flash because there isn't a way to
convert a Number/float into its integer-bits representation. Please
could everyone whinge at Adobe about this and give us access to a
very powerful tool. Even the guys working on Papervision are having
trouble with this issue.that's just an implementation of newton's method for finding
the zeros of a differentiable function. for a given x whose inverse
sq rt you want to find, the function is:
f(y) = 1/(y*y) - x;
1. you can find the positive zero of f using newton's method.
2. you only need to consider values of x between 1 and 10
because you can rewrite x = 10^^E * m, where 1<=m<10.
3. the inverseRt(x) = 10^^(-E/2) * inverseRt(m)
4. you don't have to divide E by 2. you can use bitwise shift
to the right by 1.
5. you don't have to multiply 10^^(-E/2) by inverseRt(m): you
can use a decimal shift of inverseRt(m);
6. your left to find the positive zero of f(y) = 1/(y*y) - m,
1<=m<10.
and at this point i realized what, i believe, is a much
faster way to find inverse roots: use a look-up table.
you only need a table of inverse roots for numbers m,
1<m<=10.
for a given x = 10^^E*m = 10^^(e/2) *10^^(E-e/2)*m, where e
is the largest even integer less than or equal to E (if E is
positive, e is the greatest even integer less than or equal to E,
if E is negative), you need to look-up, at most, two inverse roots,
perform one multiplication and one decimal shift:
inverseRt(x) = 10^^(-e) * inverseRt(10) *inverseRt(m), if
E-e/2 = 1 and
inverseRt(x) = 10^^(-e) * inverseRt(m), if E-e/2 = 0. -
Does anyone have an clever ways to yield a square root result. I can use the ^ feature a business rule or calc script and it validates fine, but then I don't get a result
ex: I can say 100^.5 and should get a value of 10, but no go.....anyone having a similar type issue? Thanks for any assistance.Hyperacle,
You do not need a positive integer for the power argument in the @POWER(expession, power) function -- it can be any numeric expression. It sounds like you might have a block creation issue if you are seeing #missing. You can enter a value at the particular intersection to create the block. Your calc or business rule should calcluate correctly at this point. I just created a test business rule using the example @POWER(100, .5) and the formula returned 10 in the desired intersection.
Regards,
csdjr -
ok, I used the search function and had a hard time understanding how to do things and I don't think any of the posts answered my question.
Right now, I am working on an assignment that makes me figure out the quadratic formula. I am having trouble with the square root part of the program.
For instance that part is: sqrt( b*b - 4*a*c)
I want my program to be able to tell if the square root is a perfect square then go ahead an calculate it, but if it is not, then we just want to reduce the square root function, or even if it is a perfect square, I need to make the program reduce fractions when possible.
For instance, here is an example that is a perfect square: sqrt(25) = 5 , the program would just spit out five automatically.
BUT
If it were: sqrt(24), then I would want it to reduce it to 2 sqrt(6) and that is what the program would pop out.
Also, I am having trouble on how to tell the program how to "reduce fractions" , because if I had the square root in the last example, I might be able to reduce fractions for the whole quadratic function.
example. ( 2 - 2 sqrt(6) ) / 2 would reduce to 1 - sqrt(6) and that is what the program would say.
basically it seems like we are avoiding actually dealing with decimals.
If you could just point me in the right direction or have a good link, it would be appreciated!!!It seems to me your real problem is taking a number and finding all prime factors of that number. Once you have a list of those, you know you can pull duplicates out of the square root. If all of them were duplicates, you had a perfect square.
Here's a simple program I scrawled out to factor and reassemble into the appropriate form:
import java.util.*;
class FactorTest {
public static void main(String[] args) {
int value = 96;
LinkedList<Integer> factors = new LinkedList<Integer>();
LinkedList<Integer> primeNumbers = new LinkedList<Integer>();
populatePrimeList(primeNumbers);
while (value > 1) {
boolean factorFound = false;
for (int prime : primeNumbers) {
if (value % prime == 0) {
value = value / prime;
factors.add(prime);
factorFound = true;
break;
if (!factorFound) {
expandPrimeList(primeNumbers);
int left = 1;
int right = 1;
int lastValue = 1;
for (int factor : factors) {
if (lastValue == factor) {
left *= factor;
lastValue = 1;
else {
right *= lastValue;
lastValue = factor;
right *= lastValue;
System.out.println(left + " sqrt " + right);
public static void populatePrimeList(LinkedList<Integer> primes) {
primes.add(2);
primes.add(3);
primes.add(5);
primes.add(7);
primes.add(11);
primes.add(13);
public static void expandPrimeList(LinkedList<Integer> primes) {
int testPrime = primes.get(primes.size() - 1);
boolean foundFactors;
do {
foundFactors = false;
testPrime += 2;
for (int prime : primes) {
if (testPrime % prime == 0) {
foundFactors = true;
break;
} while (foundFactors);
primes.add(testPrime);
}Edited by: jboeing on Jan 15, 2008 11:18 AM -
I am trying to use Diadem and still learning to do basic calculations.
Given a channel containing several values. How do I calculate that will give me another channel giving corresponding square root of values in original channel..?
Thanks. Himanshu.
Solved!
Go to Solution.In the Analysis panel you can bring up the calculator and enter the following equation Ch("[1]/results" )=Sqr(Ch("[1]/baseVals" )).
If you are doing this in script you can use the following.
Call ChnCalculate("Ch(""[1]/results"" )=Sqr(Ch(""[1]/baseVals"" ))" )
Hope this helps.
Message Edited by waynecj7 on 05-07-2009 08:51 PM -
Whether you use the Dashboard calculator widget or the Calculator that is built into the Apple OS as an application, there is no button for square root.
If you are using the Calculator application you can toggle the interface from Basic to Scientific to Programmer. Scientific includes buttons that square a number but no button for square root.
Why? Or am I missing a hidden keystroke or something?Remember Square Root is the same as x^(1/2)
If you have an x^y button (showing x raised to a power) enter the second number as .5, and you'll have a square root. And if you are really starving, there is Emu:
http://emu48mac.sourceforge.net/
Maybe you are looking for
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