how do you do this problem? Algebra 2?
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www.bbc.co.uk/revisewise\nwww.murderousmaths.co.uk\nwww.a729.piczo.com\nwww.bbc.co.uk/bitesize
658148
given 4a^2-12ab+9b^2\n    =(2a)^2-2(2a)(3b)+(3a)^2\n    =(2a-3b)^2
1160760
what you have is something like\na(b+c)\n\nthe solution is \nab+ac\n\n(x^3)(2x^-2)+(x^3)(4x)\nnow to multiply x^a * x^b you get x^(a+b)\n\n2x^((3+(-2))+4x^(3+1)\n\n2x+4x^4
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
5554
Solving the problem stepwise -\n(x - 8 ) ( x - 2 )\n= x(x-2) - 8(x-2)\n=x^2-2x-8x+2\n=x^2-10x+2\n\nThese two sites will help you for the concepts.
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take out 4x first\n4x(x2-6x+8)\n4x(x-4)(x-2)
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Multiply each term in the first () by each term in the second ():\n\n2x times 2x = 4x^2\n2x times 1 = 2x\n-3 times 2x = -6x\n-3 times 1 = -3\n\nThen combine 2x and -6x\nAnswer: 4x^2 - 4x - 3
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2(x - 3) = 2\n\n2(x - 3) / 2 = 8 / 2\n\n(x - 3)² = 4\n\n√(x - 3)² = √4\n\nx - 3 = ± 2\n\nx - 3 + 3 = 3 ± 2\n\nx = 3 ± 2\n\n- - - - - - - - - \n\nSolving for ( + ) \n\nx = 3 + 2 \n\nX = 5\n\n- - - - - - - - - \n\nSolving for ( - )\n\nx = 3 - 2 \n\nx = 1\n\n- - - - - - - - -s-
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x2+5x-2x-10\n\n*answer*\n\nx2+3x-10
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12 simply place the 1 where the x is\n\n(1+3)(1+2)\n4*3\n12
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Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
1108240
2(x-5)^2 = 3\ndivide by 2:  (x - 5)^2 = 3/2\nsquare root: (x - 5) = +/- root(6)/2\nadd 5: x = 5 +/- root(6)/2
665603
you put (x+2) in for x in the equation first\n\n((x+2)^3)+(2(x+2)^2)-4\n\nthen simplify \n\nx^3 + 8x^2 + 20x + 12\n\nand multiply the entire thing by 3\n\n3(x^3 + 8x^2 + 20x + 12)\n\n3x^3 + 24x^2 + 60x + 36
708943
You have to complete the square.\n\ny = a(x^2 + b/a x) + c\ny = a(x^2 +b/a x + b^2/4a^2) + c - b^2/4a\ny = a(x + b/2a)^2 + (4ac - b^2)/4a\n\n[4ay - (4ac - b^2)]/4a = a(x + b/2a)^2\n[4ay - 4ac + b^2]/(4a^2) = (x + b/2a)^2\n\n+/- sqrt(4ay - 4ac + b^2)/2a = x + b/2a\n\nx = (-b +/- sqrt(4ay - 4ac + b^2))/2a
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2x^2 - 2y^2 = 2(x^2 -y^2) = 2(x-y)(x+y)\n-2x^2 + 2xy = -2x(x-y)\nx^2 + xy = x(x+y)\n2(x-y)(x+y)/[-2x(x -y)]/[x(x+y)]\n= 2(x-y)(x+y)/[-2x(x-y)(x)(x+y)]\n= -1/x^2
873100
Here you go!\n\nhttp://www.math.com/homeworkhelp/Algebra.html
651993
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest