How do you simplify a square root of a fraction?
953882
The square root of a fraction is the square root of the numerator over the square root of the denominator.\n\nSo if you have sqrt(9/25) that is equal to sqrt(9) / sqrt(25) = 3/5.\n\nThat's the easy case where everything works out cleanly.\n\nLet's try another example of sqrt(7/100).  This is equal to sqrt(7)/sqrt(100).  The bottom works out nicely to give you sqrt(7)/10.\n\nThat's about as far as you can go to simplify it, so your answer is\nsqrt(7)/10.\n\nNow the other case is where you get a sqrt in the denominator.\n\nsqrt(4/7) = sqrt(4)/sqrt(7) = 2/sqrt(7).\n\nHere we don't like to leave sqrt(7) in the denominator, so multiply top and bottom by another sqrt(7).  Notice how this is equal to 1 so you aren't changing the value.\n\n2/sqrt(7) = 2sqrt(7) / sqrt(7)sqrt(7).  And what happens if you multiply sqrt(7) * sqrt(7)?  You just need to remember that sqrt(a) * sqrt(a) = a.  In our case we have sqrt(7) * sqrt(7) = sqrt(49) which is simply 7.  So the final answer is:\n2 sqrt(7) / 7.\n\nFrom the website:\nsqrt(1/2) = sqrt(1) / sqrt(2) = 1 / sqrt(2).\n\nMultiply top and bottom by sqrt(2):\nsqrt(2) / sqrt(2) * sqrt(2)\n= sqrt(2) / 2
875144
you have to take the square root of each side, then just simplyfy\n\nx = 1.2472191289247
152941
Make the whole number a fraction. ex: 3 = 3/1
243752
Well, the definition of a square root has been explained pretty well.  The square root of 4 is 2.  sqrt(9)=3.  sqrt(2)=sqrt(2).  This is in simplist form becuase you cannot write sqrt(2) in a finite decimal expansion.\nHow to Simplify:\n\nConsider sqrt(99).\n99=11*9.\nso, sqrt(99)=sqrt(11*9)=sqrt(11)*sqrt(9)=sqrt(11)*3\nThe steps to follow are as described:\n1) try to break down the number being considered into its prime factors.  In this case: 11,3,3\n2)  look at those factors.  Are any of them perfect squares?  In our case, 9 is 3*3.\n3)  Use this property: sqrt(a*b)=sqrt(a)*sqrt(b) to break thye original number up into an irreducible number and a perfect square if possible.\n\nSome notes:\nThis is not possible if the number does not reduced to perfect squares.  For example: 143=11*13 neither of which are perfect squares so we are stuck here and sqrt(143) is in simplist form.  Hope this helps.
978895
Can you use a square root symbol?  If so, sqrt(9) + 9 - 9
372531
first simplify the square root of 1/25.\nthe square root of 1 is one, and the square root of 25 is 5.  that leaves it to 1/5.  then multiply 5 to it - 5 X 1/5 = 1.\ngood luck!
801012
inverse square root of a number say"x" is\nx^[-(1/2)]
416099
The simplest, but most tedious, way to calculate the decimal digits of a square root is to figure out each digit in turn, as follows. \nFirst of all, find how many digits to the left of the decimal point your square root will have. \n\nA number with one digit to the left of the decimal point will be somewhere from 1 up to, but not including, 10; therefore, its square will be somewhere from 1 up to, but not including, 100, and hence will have either one or two digits to the left of the decimal point. \n\nA number with two digits to the left of the decimal point will be somewhere from 10 up to, but not including, 100; therefore, its square will be somewhere from 100 up to, but not including, 10000, and hence will have either three or four digits to the left of the decimal point. \n\nAnd so on. \n\nTherefore, if the number you're trying to find the square root of has one or two digits to the left of the decimal point, its square root will have one; if it has three or four, its square root will have two; and so on. \n\nNow, having found how many digits the square root has, just start from the left and figure out each digit in turn. \n\nFor example, to find the square root of 1234 to 3 decimal places: \n\nThe square root has two digits to the left of the decimal point, so we're looking for a number of the form AB.CDE. . . (and we want to find A through E). \n\nNow, 10^2=100, 20^2 = 400, 30^2=900, all too small, but 40^2=1600, too big. This tells us the first digit A is 3. \n\nNext, 31^2 = 961, 32^2 = 1024, 33^2 = 1089, 34^2=1156, 35^2 = 1225, all too small; 36^2 = 1296, too big, so the second digit B is 5. \n\nThirdly, 35.1^2 = 1232.01, too small, but 35.2^2 = 1239.04, too big, so the third digit C is 1. \n\nAnd so on; continuing in this way, you see that . \n\nThis method is horribly slow and inefficient but that's no problem in an era of calculators. All the necessary ideas are there. \n\nYou can apply lots of little shortcuts to this procedure to drastically reduce the number of computations you need to do. Then you come up with a procedure similar to the long-division algorithm. Its complicated to remember, though; I don't remember it offhand. It is seldom taught today because there is little need for it. \n\nTo actually calculate a square root to a large number of digits quickly, one uses a different method. For example, saying that x^2 = a is the same as saying x = (1/2)(x + a/x) and you can use this as the basis of an iteration. Start with one value of x, for example, x=1. Then get a second value y = (1/2)(x + a/x). Then get a third value z = (1/2)(y + a/y). And so on. It turns out that these numbers x, y, z, . . . converge quite quickly to the desired square root, especially if you start with a number that is already close. \n\nDoing this calculation in the above example, for a=1234, and starting with the number a=40 which is obviously just a little bit bigger than the desired square root, you get the numbers 40, 35.425, 35.129578. . . , 35.128336. . . , 35.128336. . . and you see that just four calculations are enough to get the square root accurate to six decimal places.
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
759262
The rules for simplifying integer exponents are pretty simple:\n\nhttp://en.wikipedia.org/wiki/Exponentiation
752173
Sqrt(3x/2) = Sqrt(3x/2) * SQRT(2/2) = SQRT(6x/4) = 1/2 * SQRT(6x)\n\n6*SQRT(7) - SQR(28) = 6*SQRT(7) - SQR(4*7) = 6*SQRT(7) - 2*SQR(7) = 4 * SQRT(7)
197500
Factor the numerator and denominator of the fraction in the radical - then take the square root of any perfect squares that result.  E.g.  for 76. 28/36 = (4*7)/36).  The square root of 4/36 is 2/6 or 1/3.  So the answer is sqrt(7)/3.  Sometimes you'll end up with a radical in the denominator.  Most math teachers want you to get rid of that by multiplying the expression by 1 in the form (denominator radical)/(denominator radical) to get a whole number in the denominator.  Then divide out any common factors that result.\n\nFor the velocity problem, take the derivative of the equation wrt t to get an expression for velocity:\n   dH/dt = V = -32t+v.  Substitute the initial velocity for v and solve for t when V = 0 (at the top of its arc).  Stick that value of t in the original equation and calculate the height.
964858
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
923840
Here's how I would do it:\n\nTo remove the sqrt, you square both sides:\n\nsqrt(p^2 - 3p + 16) = (p+1)(p+1)\np^2 - 3p + 16 = p^2 + 2p + 1\n\nThen it's basic algebra. The p^2 are out of there, and you are left with\n\n-3p + 16 = 2p + 1\n16-1=2p+3p\n15=5p\np=3\n\nSo, let's check:\nsqrt(3^2 - 3(3) + 16) = 3+1\nsqrt(9-9+16)=4\nsqrt 16 =4\n\nThe person who put -5 as one of the answers... it doesn't work:\n\nsqrt ((-5)^2 -3(-5) + 16) = -5+1\nsqrt (25+15+16)= -4\nsqrt 56= -4 This is false, so -5 is not an answer.
258864
there is a method of getting the square root of any number!!\n\nfirst, group the digits in two's, starting from the decimal pt.\n\n1'23'45'43'21.00'00\n\nThen, do the ff:\n1.Get the sq.root or the nearest sq.root of the first group of digits (sqrt3=1, sqrt5=2), that is the first digit of your square root.\n2. Square that number and subtract it from the first group of digits. Bring down the first digit of the next pair of digits.\n3. Multiply the digits of your sq.root by 2 and use this as the divisor of the numbers you brought down. this will be the next digit of your sq.root. Put it next to the previous digit.\n4. then, multiply, and subtract, just like in division. Bring down the next digit of the pair.\n5. Square the so far last digit of the sq.root, and subtract this from the brought-down number.\n6. Repeat the steps until the right decimal pt.\n\nThus,\n\nsqrt1'23'45'43'21\n\nsqrt1 = 1 (1st digit)\n1x1 = 1\n1 - 1 = 0\nBring down 2\n\n1x2 = 2\n2/2 = 1 (2nd digit)\n1x2 = 2\n2-2 = 0\nbring down 3\n1² = 1\n3-1 = 2\nBring down 4\n\n11x2 = 22\n24/22 = 1 (3rd digit)\n1x22 = 22\n24-22 = 2\nbring down 5\n1²=1\n25-1=24\nBring down 4\n\n111x2=222\n244/222=1 (4th digit)\n1x222=222\n244-222=22\nBring down 3\n1²=1\n223-1=222\nBring down 2\n\n1111x2=2222\n2222/2222=1 (5th digit)\n1x2222=2222\n2222-2222=0\nBring down 1\n1²=1\n1-1=0 STOP..\n\nThe answer is 11111.^_^
752580
It just takes foiling-\n\n(5*2)+(5*5i)+(-3i*2) +(-3i*5i)\n(10)+(25i)+(-6i)+(-15*i^2)\ni^2=-1\nso...\n25+19i\n\nTo those who have said that you can't simplify it past 10 + 19i - 15i^2, that isn't correct considering that since i is the square root of -1, i^2 is -1. That's why you can simplify -15i^2 to 15, thus making it 10+19i+15, or 25+19i
1185125
√3 (√3+2) is of the form a(b+c)\na(b+c) = (ab+ac)\nso we simplify to \n√3√3+√3*2\n√a√a = √a*a = √a² = a\nso it's 3+2√3.\nThe rules for roots are exactly the same as the rules for exponents, since a (square) root is merely a exponent of (1/2).  A cube root is an exponent of (1/3).\nⁿ√a root is the same as a^(1/n).