how to simplify radicals.?
138792
To 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. \n\n \n\n If 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 \n \n\n\n2.  write 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.
780298
first u distribute the - to the stuff in the parenthesis so you have\n7x^2-3+x^2\nthen combine like terms\n8x^2-3
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
972961
Sites are there. But I would recommend to sole in this form\n\nthere is no x term.\n\nso collect x^2 and constants on 2 sides\n\n4x^2 = 9-189 = -180\nsimplify \n\nx^2 = -45\n\ntake square root of both sidex\n\nx = +/-3 sqrt(5) i\n\nIf yo still require a site you can visit
780960
I also have difficulty typing this thypes of math problems.  First, simplify the ax^2+bx+c parts in the first half to (x+c)(x+c) looking things first.  So...\n\n[(x-3)(x-1)]} is the first part.  With the bottom of the second half, divide by 3, so that you get 3(x+1).  Now line everything up like so...\n\n(x+1)(x+3)(x-3)\n_____________\n3(x-3)(x-1)(x+1)\n\nNow do the cancellations and you get the right answer...\n(x+3)\n____\n3\n\nSorry for not working to convention, but it was easier to type it that way.
49666
If you are saying simplify then:\nx = The square root of [5x] +2.5\nx = The square root of [5x] +2.5\nx = The square root of [12.5] +2.5\nx = The square root of [12.5] +2.5
531494
Ask your teacher.\n\nBut usually, you are supposed to simplify.
641733
[-b +/-  sqrt(b^2-4ac)]/(2a)
762985
(b-4)/(b-4)(b+4) simplifies to 1/(b+4)\nrestrict denominator...bcannot equal 4 or -4
879678
The first thing to do is to find the factors of the number you're taking the root of.\nFor example, teh factors of 24 include 2*12, 3*8, and 4*6.\nSince four is a perfect square (2^2), we'll use that set of factors.\nSo:\nsqrt(24)= sqrt(4*6)= sqrt(4)*sqrt(6)= 2sqrt(6)\n\nSame with sqrt(18):\n18=9*2\n9=3^2\nsqrt(18)= sqrt(9)*sqrt(2)= 3sqrt(2)\n\nsqrt(3) +sqrt(3) can only be reduced by realizing that it's\n2*sqrt(3), which could be the sqrt(12). THis is because you can take the two under the radical by squaring it. so  you get sqrt (2^2+3)= sqrt(12), but technically 2sqrt(3) is more reduced.\n\nsqrt(27)=3 sqrt(3)\nsqrt(12)= 2 sqrt(3)\nso 3sqrt(3) + 2sqrt(3) = 5sqrt(3)\nin this case, you're treating the sqrt(3) sort of like it was a variable. think of 3x +2x = 5x
648650
Sometimes dealing with radicals is easier if you can recognize square roots.  If not, it sometimes helps to factor out the number to the point where you have pairs of numbers under a radical.  If you have sqrt(16), that's really sqrt(4*4).  Since you have a pair of 4s under a radical, you can take them out and write one 4 on the outside.  So for 4sqrt(16) you'd end up with 4*4, or 16.\n\nIt's not always that clean though - sometimes you'll have something leftover under the radical.  Take sqrt(125) for example.  sqrt(125) = sqrt(5*5*5).  Since you have a pair of 5s, take two out and put one on the outside the radical.  So sqrt(125) = 5sqrt(5), since you still have a 5 under the radical and no pair to get it out.\n\nLook at something like sqrt(248).  That's really sqrt(2*2*2*31).  A pair of 2s come out, and you still have a 2*31 inside.  So that equals 2sqrt(62).\n\nIt may take some practice.  I used to teach it by factoring out and pulling pairs out, but eventually you'll be able to see square roots and see that sqrt(200) = 10sqrt(2) by noting that sqrt(200) = sqrt(100*2) = sqrt(100)*sqrt(2) = 10sqrt(2).
962705
8(x+4)² = 9\n(x+4)² = 9/8\nx+4 = ±√(9)/√8\nx+4 = ±3/√8\nx+4 = ±(3√2)/4\nx = -4 ± (3/4)√2\n\nso no, you're a little mixed up on simplifying the radical.\n\na/√b = (a/b)√b, or simply 1/√b = b/√b
618071
simplify to f(x)=(sqrt(x-2)(x-3))/(x-4)(x+5)\nthen sub in 1/x\nto get (sqrt(1/x -2)(1/x -3))/(1/x -4)(1/x +5)
1123203
You can multiply and divide them easier.\n\nfifth root of 2x times 2x^20\n           -or-\n2x^(1/5) times 2x^20
531417
(a+b)(b^6-ab^5+a^2b^4-a^3b^3+a^4b^2-a^5b+a^6)\nGood Night, and Good Luck!
204269
isolate one of the radicals, then square both sides.\nhere you get 2x+3=5x-6.......3x=9.......x=3\n\nworst case is sqrt(something) + something = sqrt(something)\n\nhere you'd isolate a radical, square both sides, clean up the equation, isolate the remaining radical, square both sides, and finish.  AND CHECK YOUR ANSWERS.  when you square both sides, "extraneous" solutions sneak in that might have you, back in the original equation, taking the square root of something negative.\n\nexample:  sqrt(3x+3) + 1 = sqrt(5x+6)\n....... (3x+3)+2*sqrt(3x+3)+1=(5x+6) ..... left side is a binomial\n.........2*sqrt(3x+3) = 2x+2\n...........sqrt(3x+3) = x+1\n...............(3x+3) = x^2+2x+1 ....... right side is a binomial\n....................0 = x^2-x-2 ......... and it factors!\n....................0 = (x-2)(x+1)\n....................0 = (x-2) so x=2\nor .................0 = (x+1) so x=-1\n\nnow check: sqrt(3*2+3)+1 =? sqrt(5*2+6)\n.............sqrt(9) + 1 =? sqrt(16)\n...................3 + 1 = 4.............yes.\n\n..........sqrt(3*-1+3)+1 = sqrt(5*-1+6)\n...............sqrt(0)+1 = sqrt(1)\n.......................1 = 1 ............yes\n\nsigh.  when I wrote it the equal signs lined up.
890645
Wow, there is a lot here!  I'll start with some tips for solving algebra problems.  The first thing you should always remember is to play the equal game.  Whenever you add, subtract, multiply, or divide, you should do the same to the other side.  The other tip is to use the order of operations when simplifying.  Remember the sentence: Please excuse my dear Aunt Sally.  The initial letters of each word stands for the order of operations.  Do the Parenthesis first, followed by the Exponents, then Multiplication or Division, and Addition and/or subtraction next.  Finally, you should always simplify first using the order of operations before you start shifting numbers from one side of the = to the other.  \n\nI'm not going to work all of these problems, but I'll show you a few to get you started.\n\n1)\n3x-7 = 2   Add 7 to both sides\n3x = 9   Divide by 3\nx = 3\n\n4)\n6x - 3 = 2x - (4-3)/3   Simplify the stuff inside the parenthesis first\n6x - 3 = 2x - 1/3  Add 3 to both sides\n6x = 2x - 5/3  Subtract 2x from both sides\n4x = -5/3  Multiply both sides by 3\n12x = -5  Divide both sides by 12\nx = -12/5\n\nI hope this helps.  Good luck!