wut is the rationalized denominator for 5 over the square root of 3  PLEZ HELP!!!!!! i already asked my parents, they cant help. so anyone pleez help. im in the 8th grade and in algebra
132128
You have to multiply top and bottom by sqrt 3. This makes the top 5 sqrt 3 and the bottom sqrt 9, which is just 3.\nSo the answer is 5 sqrt 3 over 3. The denominator is now rational; the irrational sq rt is now in the numerator.
459427
This cannot be simplified without guessing because the problem is ambiguos for the following reasons:\n\n1) You have an odd number of parentheses. You have one open parenthesis and two close parentheses.\n\n2) It is unclear what the exponents are. Following the u^  you should always enclose in parentheses what you intend the exponent to be. Exception is when it is totally clear as in u^5.\n\nOne cannot tell what part of SQRT 63ux^2 is the intended exponent of u. \n\nMy guess is that the expression should be (8u^(sqrt63))(ux^2) -\nx*sqrt(28u^5). If this guess is accurate, then:\n\n8u^(3*sqrt(7))(ux^2)-x*2u^2*sqrt(7) =\n2ux[4u^(3sqrt(7))x-2sqrt(7)]\n\nBut my guess is probably wrong. Hope I was not rude.
459186
When working with the simplification of radicals you must remember some basic information about perfect square numbers. \n\nYou need to remember:  \n\nPerfect Squares \n4 = 2 x 2 \n9 = 3 x 3 \n16 = 4 x 4 \n25 = 5 x 5 \n36 = 6 x 6 \n49 = 7 x 7 \n64 = 8 x 8  \n81 = 9 x 9  \n100 = 10 x 10 \n \n Radicals (square roots) \n = 2 \n = 3 \n = 4 \n = 5 \n = 6 \n = 7 \n = 8 \n = 9 \n = 10 \n \n \n\nWhile there are certainly many more perfect squares, the ones appearing in the charts above are the ones most commonly used. \n\n \n\nTo simplify means to find another expression with the same value.  It does not mean to find a decimal approximation.\n \n\n \n\nTo simplify (or reduce) a radical: \n\n1.  find the largest perfect square which will divide evenly into the number under your radical sign.  This means that when you divide, you get no remainders, no decimals, no fractions.\n\nReduce:  the largest perfect square that divides evenly into 48 is 16. \nIf the number under your radical cannot be divided evenly by any of the perfect squares, your radical is already in simplest form and cannot be reduced further.\n\nwrite the number appearing under your radical as the product (multiplication) of the perfect square and your answer from dividing.\n\n\n \n\n3.  give each number in the product its own radical sign.\n\n\n \n\n4.  reduce the "perfect" radical which you have now created.\n\n\n\n5.  you now have your answer.
420190
1)  √10 · ³√100\n= 10^(1/2) · 10^(2/3)\n= 10^(7/6)\n\n2)  (6x)^(1/2) · (36x²)^(1/4)\n= (6x)^(1/2) · (6x)²^(1/4)\n= (6x)^(1/2) · (6x)^(2/4)\n= 6x\n\n...and so on.  Change the radicands to rational exponents with common bases, and follow the rules from there.  Good luck!
510313
the roots of the numerator are  5  and -1.5. So, the numerator is divisible by  a-5 . if you divide  the numerator, you find 2a+3\n\nthe answer 2a+3
157411
16^(3/4) means take a fourth root of 16, and then cube it.  Anytime you take an even root (square root, fourth root, etc.), you always have +/- answers.\n\nConsider that the square root of 4 is +/- 2.  That makes sense because if you square 2, it is 4, and if you square -2, it's 4 as well.\n\nNow consider the fourth root of 16.  That gives you +/- 2.  If you reverse it, you'll see it makes sense: 2^4 = 2*2*2*2 = 16, and (-2)^4 = (-2)(-2)(-2)(-2) = 16, since all the negatives end up cancelling out.\n\nSo for your original problem,\n\n(3x-1)^(4/3) = 16\n3x - 1 = 16^(3/4)\n3x - 1 = +/- 8\n3x = 1 +/- 8\nx = (1 +/- 8)/3\n= 3 or -7/3\n\nAnd if you plug in either value into your original equation, it works out.
254431
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.
539135
That just means that keep it in the square root instead of simplifying it.\nEx: Keep it as square root of 2 instead of 1.414\n\nSo if you're finding the square root of 27, for example.  You can split up 27 into 9*3.  Square root of 9 is 3, but 3 is not a perfect square.  So your answer would be : 3 square root of 3.  Instead of 5.196
417325
You're not doing anything wrong! Your answer and their answer are actually the same.\nWhat do you get when you square sqrt(6)/3? You get 6/(3*3) = 6/9 = 2/3. So sqrt(6)/3 = sqrt(2/3).\nI would prefer the answer you came up with, but theres really no difference.
289094
Here's how to work it out:\n\n(Sorry, because I can't use special symbols, I'm using _/ for the square root symbol and <= for the less than or equal to symbol.)\n\n_/3x + 6 + 4 <= 7\n_/3x + 10 <= 7\n3x + 10 <= 7 x 7\n3x + 10 <= 49\n3x + 10 - 10 <= 49 - 10\n3x <= 39\n3x/3 <= 39/3\nx <= 13
487767
Why not?  If you divide top and bottom of this expression by x^7 you get\n(3 + 4/(x^5))/(9 - 1/(x^4) + 2/(x^7))\n\nEvery term except the 3 and the 9 now has a power of x as denominator, and therefore approaches 0 as x goes to infinity,\ntherefore it approaches 3/9.
499062
Multiply the top and bottom by: 4 - i sqrt(3)\n(4 - i sqrt(3))(4 - i sqrt(3))\n---------------------------------\n(4 + i sqrt(3))(4 - i sqrt(3))\n\nThe denominator becomes a difference of squares:\n(4 - i sqrt(3))²\n-----------------\n4² - (i sqrt(3))²\n\nExpand the denominator:\n(4 - i sqrt(3))²\n-----------------\n16 - (-1)(3)\n\n(4 - i sqrt(3))²\n-----------------\n19\n\nExpand the numerator:\n4² - 8i sqrt(3) - (i sqrt(3))²\n---------------------------------\n19\n\nSimplify:\n16 - 8i sqrt(3) - (-1)(3)\n----------------------------\n19\n\n19 - 8i sqrt(3)\n-----------------\n19\n\nAnswer:\n1 - i (8/19)*sqrt(3)
354138
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.
397130
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.
242193
ok, lets say that Sqrt() means square root \n\nSqrt(2x-5)=13   , elevate both sides of the equation to square\n (Sqrt(2x-5))^2 = 13^2  , the square root and the square power cancell each other, so\n\n(2x-5) = 169\n2x = 164, x = 82\n\nThe second one is the same, only that you have to leave the square root alone in the left side of the equation\n\nsqrt(3x-5)+7 = 3 , \nsqrt(3x-5) = -4\n\nand do the same as the first one\n\n3x-5 = 16\nx = 7\n\nthe last one is a bit harder cause there's an 'x' outside of the square root, and you do it like this\n\nsqrt(x+2) = x - 4    , elevate both sides to the square\n\n(x+2) = (x-4)^2    , and do the special product \n(x-4)^2 = x^2 -8x + 16 , then you get\n\n(x+2) = x^2 - 8x + 16 ,  this is a quadratic equation, take it to the general form\n\nx^2 -9x + 14 = 0    and factorize\n\n(x-7)(x-2) = 0  then you get\n\nx1 = 7 and x2 = 2 \n\nthat's it, if you have doubts contact me
529948
Ok, I assume you know the quadratic equation, but just for convenience I'm going to write it out:  x = (-b +/- root(b^2 - 4ac))/2a.  \n\n1) x = (5 +/- root(25 + 24))/4.  (Remember, you have to subtract the 3 from each side to get the equation to equal 0, so c is -3.)  That simplifies to x = (5 +/- 7)/4--->x = 12/4 or -2/4--->x = 3 or -1/2.\n\n2) x = (2 +/- root(4 - 12))/6.  Are you sure that wasn't supposed to be a negative 3x^2 or minus 1?  If it were, the term inside the square root would be 16, a positive perfect square.  As it is, it's -8, whose square root is 2root(2)i.  That seems pretty complex (no pun intended) compared to the other one.  \n\nI'm going to solve that one assuming that it should have ended in -1 rather than 1.  Then x = (2 +/- root(4 + 12))/6--->x = (2 +/- 4)/6--->x = -2/6 or 6/6--->x = -1/3 or 1.
171790
For #1, try fully factoring the numerator and the denominator. You will find that some of the terms cancel. Start by factoring x from each one. Your resulting term in the numerator is a difference of squares. Your resulting term in the denominator is a factorable quadratic.\n\nFor #2 the expression will be undefined for any value that makes the denominator equal to zero. Again factoring so you can solve for the zeroes is the way to go.\n\nI don't like to work out problems for people, so I hope this is helpful. Likely enough someone else will give you the answers, but I wanted to make sure you knew the process.