How can a radical equation be used in our daily life?
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well, you could be an architect and be using a 30-60-90 triangle and you needed to know a particular side of it. you can use x times the square root of 3, x being the shorter leg, to find the length of the longer leg.
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New Radicals - You get what you give
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One big advantage is that rational exponents are easier to combine when more than one appears in an expression.\n\nI would say that something like the following would be easier to solve with rational exponents:\nThe square root of x to the 5th power, all divided by the cube root of x to the fourth power.\nWhen I see this equation, the way to simplify is to change the radicals to rational exponents--x to the 5/2 power divided by x to the 4/3 power--which can be further reduced by subtracting exponents.\n\nSince my first step in that problem is to change the radicals to rational exponents, that seems to "prove" (in some sense) that rational exponents are easier to work with in this type of problem.
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Usually, when people talk about solvable equations, they are talking about taking a polynomial and setting it to zero & then finding all of the roots of the equation.\n\nA linear equation (ax+b = 0) is easy to solve.\n\nA quadratic equation (ax^2+bx+c = 0) is not so hard.  You probably learned the quadratic equation in high school.  That formula has been around since the ancient Greeks.\n\nWhat about cubic equations?  (ax^3+bx^2+cx+d = 0)  It turns out that there is a formula that you can use to find the three roots of the equation.  It was discovered in the 1300s by Tartaglia.  It is a very complicated formula.\n\nThe obvious next choice is to see if we can solve fourth degree equations using a formula.  Legendre discovered the formula for this in the late 18th century.\n\nThen a few interesting things happened.  Gauss had already proven that every polynomial with integral coefficients has a solution in the complex numbers.  But in the early 1800s, Nicholas Abel proved that there is no general formula for solving equations of degree five or higher.  Sure -- we know that some can be solved (x^5-32=0 has five roots that can be found, including 2).  But there is no general formula.\n\nA few years later, a young Frenchman named Evariste Galois created a beautiful system that allows us to take any polynomial and through a series of steps tells us exactly which polynomial equations can be solved and which ones cannot be solved.  It took another 50 years for people to find out about his work, because he was killed in a duel before he turned 21.\n\nSo -- if you want to know if an equation is unsolvable, just use Galois' method.
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you must apply oxidation numbers to the equation
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THE POSSIBILITIES:\n\n1.1:1→1*1=1\n2.1:2→1*2=2\n3.1:3→1*3=3\n4.1:4→1*4=4\n5.1:5→1*5=5 √\n6.1:6→1*6=6\n\n7.2:1→2*1=2\n8.2:2→2*2=2\n9.2:3→2*3=5\n10.2:4→2*4=8\n11.2:5→2*5=10 √\n12.2:6→2*6=12\n\n13.3:1→3*1=3\n14.3:2→3*2=6\n15.3:3→3*3=9\n16.3:4→3*4=12\n17.3:5→3*5=15 √\n18.3:6→3*6=18\n\n19.4:1→4*1=4\n20.4:2→4*2=8\n21.4:3→4*3=12\n22.4:4→4*4=16\n23.4:5→4*5=20 √\n24.4:6→4*6=24\n\n25.5:1→5*1=5    √\n26.5:2→5*2=10   √ \n27.5:3→5*3=15   √\n28.5:4→5*4=20   √\n29.5:5→5*5=25   √\n30.5:6→5*6=30   √\n\n31.6:1→6*1=6\n32.6:2→6*2=12\n33.6:3→6*3=18\n34.6:4→6*4=24\n35.6:5→6*5=30   √\n36.6:6→6*6=36\n\nProbability = No. of required possibilities/No. of total possible events =11/36\nSo the right answer is D
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Find next closest square to 80 that is also a factor: 16\n\nSince 4*4 is 16, you can bring the 4 outside the radical and leave the product of the other factors underneath the radical: 4rt(5)
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if the multiplication involves the same radical, (squarel root, cube root, etc.) then that is equal to the product of the radicals\nie sqrt(5) * sqrt (2) = sqrt(10).\n\nif they are different radicals, then it is more difficult. the only way is to change the radical to the same radical through manipulation of the number/radical.\n\nfor example 4 * sqrt (2) = sqrt (4^2) * sqrt (2) = sqrt (16) * sqrt (2) = sqrt(32).
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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
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Do you mean real AND imaginary? Or just all imaginary?  Well, either way, since we \nhave a polynomial with four factors, we will factor by grouping to simplify.\n\nf(x)=(x^3)-(9x^2)+26x-24\n\nf(x)=[(x^3)-(9x^2)+26x]-24\n\nf(x)=x[(x^2)-9x-26)]-24\n\nf(x)=(x-24)[(x^2)-9x+26]\n\nWe have one real solution. Our linear factor is (x-24), so our Real solution is x=24.\n\nNow we need to use the quadratic formula:\n\n{-b(+or-)[The Square Root of:(b-4ac)]}ALL over 2a.\n\n[(x^2)-9x+26] a=1, b=-9, c=26\n\n-(-9) ± √(-9-(4*1*26))ALL divided by 2\n\n9 ± √(-9-104)ALL divided by 2\n\n9 ± √(-113)ALL divided by 2\n\nSo our two imaginary solutions are:\n\n [9+√(-113)]/2] and [9-√(-113)]/2]\n\nRecall that i=√(-1)\n\nSo, our first solution [9+√(-113)]/2]=9+i√(113)or 9+√(113)i\n\nand our second solution [9-√(-113)]/2]=9-i√(113)or 9-√(113)i\n\nTherefore, All Solutions, Real and Imaginary, for \n\nf(x)=(x^3)-(9x^2)+26x-24 are:\n\nx=24, x=9+√(113)i, and x=9-√(113)i\n\n:)
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Free radicals, during our process of intake and metaboism some of the oxygen are emerge in our body. These are the main free radical which is highly potensive to destroy our body. \n\nAs you can see, oxygen radicals are very powerful and that is the same one which leads for the corrosion of iron.\n\nIn the same way accumulation of free radicals are very much danger to the body. Almost all diseases in our body are due to the act of this highly reactive free radicals.\n\nThe anti-oxidants are the only one to fight against the free radicals. We can get all nececessary anit oxidants from the colored fruits. \n\nThe another method to come out of these free radicals is detoxify our body.\n\nHave a great day.
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Steps:\n1) Divide by 2\nx^2=32\n2)Find square root\nx=sq.rt.32\n3)Simplify Radical\nx=sq.rt.(2)(16)\nANSWER:\nx=4+sq.rt.2\n\n*=times
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You have to solve the equation so that you can eliminate the sqrt by raising it to the second power .for instance,\n\nsqrt(2x+10)=5     you square it and get\n\n2x+10=25   easy, isnt it?\n\nbe careful if you have something like this\n\nsqrt(x+3)+sqrt(x-2)= 5\n\nhere you cannot do that you have to distribute one sqrt on each side of the equation.\n\nsqrt(x+3)=5-sqrt(x-2)\n\nnow you can square them
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You use math everyday
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http://mathworld.wolfram.com/Radical.html\nhttp://mathworld.wolfram.com/RationalNumber.html\nhttp://mathworld.wolfram.com/Exponent.html\nhttp://mathworld.wolfram.com/Notation.html
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Rational exponents are something like x^(2/3)\n\nwhich can be rewritten with radical sign as something like\n ____ \n/ 2^3\n\n\nThere is really no advantage except that most high school math curriculums de-emphasized the rational exponents in favor of the radical sign.  Even the radical is de-emphasized since it is easier to use the calculator.  The reason for the de-emphasis is that most high school graduate would never need to see a radical or rational exponents, except know why some calculators have a square root button.  If a student needs it in higher education, then that is left up to the higher educators.
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I'm going to guess that you mean for the entire expression (4x+1) to be under the radical sign.  (I'm basing that guess purely on what sort of problems I typically see in math textbooks.)\n\nSo, start by subtracting the 3 from both sides:\n\n√(4x+1) = -3\n\nSquare both sides:\n\n4x+1 = 9\n\nSubtract 1 from both sides:\n\n4x = 8\n\nDivide both sides by 4:\n\nx = 2.\n\nNow, whenever you solve an equation with a square root in it, you've got to test your answers, because you can do everything right and still generate false solutions.  And, in fact, that's what happens here:\n\n√(4*2 + 1) + 3 = √9 + 3 = 3 + 3 = 6 which does not equal 0.\n\nThat means there are no solutions to this problem, and that's as it should be: since we always assume that a square root is positive, there's no way that we can add √(4x+1), a positive number, to 3, another positive number, and get zero as an answer.\n\nAs it happens, if the problem had been\n\n√(4x+1) - 3 = 0,\n\nwe would have gotten the exact same answer, x = 2, and it would have worked.  But the problem as stated has no solution.  (Assuming, that is, that my initial assumption about it was correct.)\n\nHope that helps!