More Square root help please!?
928992
It is not a + sign underlined.It is plus and minus sign put together indicating that the result is either positive(+) or negative(-).In fact when when we find out a square root of any number say 16,the answer may be +4 or -4 because +4^2is 16 and -4^2 is also 16.Therefore square root of 16 is plus-minus 4 that is eithr plus 4 or minus 4.
737774
You figure the square root of each number separately...you've probably learned that in class alredy (how to do that)...then you take reach result, and multiply them together for your final answer.\n\nUse your textbook for reference...good luck!
460242
ur answer is [sqrt(3) - sqrt(7)].it can written as sqrt[x +/- 2*sqrt[y]] when x=a^2 + b^2 and y=a*b is equal to sqrt(a) +/- sqrt(b)
544837
I squared the 7 to start with and got 49.  I am also only showing the addition version, you can do the subtraction later yourself.\n\n(-7 + sqrt(49 - 4(9)(-1)))/2(9)\n(-7 + sqrt(49 - -36))/18\n(-7 + sqrt(49 + 36))/18\n(-7 + sqrt(85))/18\n\nand I think that that is it.  You cannot simplify the square root.
337853
The answer, before simplification, is sqrt(-8). There is no need to panic, as some of the other answerers did. This simplifies easily to\n\n2 * i * sqrt(2)
848127
When you added x to the right side, you should have gotten:\n\n4 = x^2\n\nThen take the square root of both sides to get\n\nx = 2, -2
84518
If the directions are to simplify, then what you have to do when multiplying square roots is to multiply the coefficients (numbers in front of the variables) and add the exponents of each variable.  Then if possible simplify the new coefficient (here, that won't be possible). Finally, if any exponent is 2 or higher, put one of that variable outside the square root for every 2 that the exponent has.  So if it is x^7 under the square root, x^3 goes outside and x is left inside.  For y^4, y^2 goes outside and no y is left inside.  I hope that makes sense and you can figure it out from here.
830100
'x= rcost\ny=rsint\n\nand r= sqrt(x^2+y^2)\nt=arctan(y/x)\nso\nif x=2, and y=2, then r= sqrqt(4+4)=sqrt(8)\nand t=arctan(2/2) = arctan(1) = pi/4
773471
√2 . √6 \n---- x ----- =\n. 3 .... 22\n\n√12\n------- =\n66\n\n\n√(2² x 3)\n------------- =\n66\n\n2√3\n------ =\n66\n\n√3\n-----\n33
714992
I'm 100% sure you will fail your tests if you dont at least try these yourself.  You should try them, get an answer, and then if YOU'RE not sure, ask the question here with your solutions and maybe someone will show you right or wrong.\nAs it stands now it sounds like you're just lazy and want someone else to do your homework.\nBesides, I see many many WRONG answers in this mathematics category.  Its like you'd just be copying your final exam answers from the class idiot.\n\nBY THE WAY...\nThere are two square roots of EVERY number.  So every answer which gives just one solution (unless the question specifically asks for just the negative square root, as in "the negative square root of of 169/4") is incomplete.  You dont have to specify "what is the negative and positive square root of (a number)...".  Asking for the square root of a number demands two solutions.
596727
Figure out what other number times itself equals your number.  That's its square root.
970218
Do you require mathematical signs, or are standard operations OK (like square root)?\n\n1: (1 + 1 + 1)!\n2: 2 + 2 + 2\n3: 3 * 3 - 3\n4: sqrt(4) + sqrt(4) + sqrt(4)\n5: 5 + 5/5\n6: 6 + 6 - 6\n7: 7 - 7/7\n8: 8 - sqrt(sqrt(8 + 8))\n9: 9/sqrt(9) + sqrt(9)\n\nAnd for a bonus: (0! + 0! + 0!)! = 6
297628
First of all, you look like a nice girl and I do not like to see you/people teary eyed, so if you promise me you will try to be happier and not cry anymore, I will make an effort to teach you a whole bunch of nice lil' math tricks (explained in the simplest possible language that I can) to help you with your homework, ok? \n\nIn the problem at hand, first try and remember what a square root means, for example, something very simple like sqrt of 4 or simply sqrt(4).    Every time you get a square root problem like that you should ask yourself: what is that number which multiplied by itself will give me the number inside the square root (or inside the box as you put it) and the answer to that question will be the answer to the square root as well.  In case of 4 - a most simple example - the answer is obviously 2, since 2 multiplied by itself will equal 4 (i.e. the number under the square root).  In other words:  sqrt(4) = 2, because 2x2=4 simple as that.   The difficulty is, not every sqrt expression is as simple to immediately resolve as sqrt(4) for instance, especially when they involve variables, etc.   So, what should one do in more complex situations where the answer is not immediately obvious?   The solution is simple: You must learn and apply the rules of square roots and others like factoring, etc., to solve the problem; that's why we study mathematical rules to begin with.   Ok, back to the question at hand, namely sqrt(18x^4y^3).  Having done so many similar problems and having applied the rules to so many analogous exercises in my life, I know immediately in my head what the answer in this case is, and I promise you in time and with lots of practice you will be able to do the same.  However, since you are a beginner at this let's try to apply the rules and solve the problem step by step.   First of all, notice that everything under the sqrt is being multiplied by one another.   This makes our situation much easier to deal with since we can breakup that one big sqrt into smaller ones that you can more easily deal with:\nsqrt(18x^4y^3) = sqrt(18)sqrt(x^4)sqrt(y^3) = sqrt (2*9)sqrt(x^4)sqrt(y*y^2) = sqrt(2)sqrt(9)sqrt(x^4)sqrt(y)sqrt(y^2)\n\nNow notice in the above expression:\n\nsqrt(2) is as simplified as it can be\nsqrt(9) = 3   (since 3 * 3 = 9)\nsqrt(x^4) = x^2   (since x^2 * x^2 = x^4)\nsqrt(y) is as simplified as it can be\nand finally,\nsqrt(y^2) = y (since y*y = y^2)\n\nSo now substituting these values into the above expression we will have:\n\nsqrt(2)*3*x^2*sqrt(y)*y  ….and now with simple rearrangement of the terms and knowing that sqrt(2)sqrt(y) = sqrt(2y) we will finally have:\n\n3x^2*y*sqrt(2y)   which is the answer we are looking for.\n\n\nI hope that was helpful.    Good Luck and Practice, Practice, Practice!\n\nP.S.  Please take the time to read my response to one of the older questions I’ve answered entitled: “What’s the key to Understand Mathematics?” as well.    I am sure it will help you with some of your math challenges.
773858
If you work out the square (of -3) and the multiplications and divisions to find the value that you need to take the square root of, you will find that the number is 121. And you should be able to calculate the square root of 121.\n\nIt would be helpful to you if you developed your ability to factor trinomials like this, so that you could solve problems like this by factoring. I used to use the quadratic formula all the time, and in real-world problems it's often the only practical way to get an answer. But for school problems, many, many times you can do it much faster by factoring.
1060817
Square Root:\n\nSQRT(1839472)=1356.271\n\nSQRT(75639)=275.0255\n\nSum:\n\nSQRT(1839472) + SQRT(75639)\n\n= 1631.2964 (ans)
1175507
square root of 2\n\nif 2 = x^(x^(x^...))\n\nthen 2 = x^ 2\n or x = √2\n\n[i hope you were truthful about the first answer gets the points....]\n\n\nI have an interesting follow up question however.  What if the quantity equals 4?  Then 4 = x^4 or x = 4^(1/4) = 2^(1/2) = √2 !!!  \n\nSo.. √2^(√2^(√2^(√2^...))) = 2 and √2^(√2^(√2^(√2^...))) = 4\nso...  2 = 4\nHow can this be??!!!!!\n(obviously its not.  where's the flaw?)
683937
Looks like someone is trying to get there math homework done.  Tsk tsk.  Square roots are easy - there is a button to do them on the calculator you should probably blow the dust off of.\n\nDid I say that?  Thanks for my 2 points and have fun cheating your way though high school math!\n\nAnd since it doesn't look like you're listening to your teacher, here's a math web site to get you started on the road to learning!