college algebra help?
958505
x={-(-4) (+/-)√[(-4)^2 - 4(1)(-5)]}/(2(1))\n x= 4(+-)6/2\n x=5,   x=-1\n\nCheck:\n with x=5,  5^2-4(5)-5=25-20-5=0  OK\n with x=-1   (-1)^2 -4(-1) -5 = 1+4-5=0  OK
104175
f=thanx\ng=for\nx=the\ny=two\nz=points\n\nTell your teacher you need some help with your algebra.
799321
Go to http://www.glencoe.com/sec/math/algebra/algebra2/pdf/alg2_20.pdf
631400
Factor out the y:\n\n(a-2) * y = c\n\nThen divide both sides by (a-2):\n\ny = c / (a-2)
747340
-2x-x > -6 + 4\n-3x > -2 \nx < 2/3
102673
If you can't get an answer here, try asking your Algebra teacher for extra help after school tomorrow.  It's sure to add to your grade that you asked for the help.
1046507
[2x/(x+3)(x-3)]*[(x-3)/4(x)]\ncancelling 2 and x-3 we will get\n=x/(x+3)2(x)\nsimplify we will get\n=x/(2x^2+6x)
896236
First expand the brackets, collect like terms and simplify.\n\n(-2x^2 + 5x - 1) - (3x^2 - 4x - 6)\n= -2x^2 + 5x - 1 - 3x^2 + 4x + 6\n= (-2 - 3)x^2 + (5 + 4)x + (-1 + 6)\n= -5x^2 + 9x + 5
283601
1) First we move all the factors with x to one side, and the ones without x to the other.  Result will be:\n\n4x/3 - 2x + x = 3/2 + 1/5\n\n2)  Then, we factor out an x from the left side and get:\n\nx(4/3 - 2 + 1) = 3/2 + 1/5\n\n3)  Calculate the actual numbers on both sides\n\nx(4/3 - 6/3 + 3/3) = 15/10 + 2/10\nx(1/3) = 17/10\n\n4)  Divide both sides by 1/3 and get a final result of...\n\nx = (17/10)/(1/3)\nx = (10/17)*(3/1)\nx = 30/17\n\n5)  Comform that I PWN n00bs.  And... I'm pretty sure I pwn n00bs.  Therefor, we are done!  ^-^
1181037
You will find the worksheets at the website of http://www.algebra.com. This website is a free site. You can also pose Maths. questions there and will get answers at the cost of $0 (zero dollar).\n\nI am a free tutor there and you may see my lessons and solutions to problems at\nhttp://www.algebra.com/tutors/aboutme.mpl?userid=psbhowmick\n\nYou will find the worksheets at\nhttp://www.edhelper.com/algebra.htm
1039099
This is the answer i got when i worked out the problem\n1. You have to distribute the (x + 2) & (x + 1)....you do the by using the FOIL method....Firsts Outsides Insides Last and you get: x + 2 = x^2 + 3x + 2\n\n2. Get rid of the 2 on the left side by subtracting it on the right side and you end up with : x = x^2 + 3x\n\n3. Now you can move the x on the left to the right side too so you have: 0 = x^2 + 2x\n\n4. Pull out and x leaving you with 0= x (x+2)\n\n5. x = 0 and x = -2\n\nMy algebra is rusty but I think this is how it's done.
58450
use the distributive property (poor u, u seem to be having a lot of trouble with your math homework tonight!! hope this helps:)\n\nthe property states that a(b+c)=ab+ac, basically you take the "a" and multiply it by the other 2, or distribute it.  say a=2, b=3, and c=1.  your equation would be 2(3+1) and, using the distributive property, make it become 6+2 which equals 8.  so its the same when your using x.  3(-4x+7)= -12x+21!!! yay!! hope this helped!
551906
Add each equation\nA + B = 3x^2 + 2x - 2\nA - B = -x^2 + 4x - 8\n2A = 2x^2 + 6x - 10\nDivide each side by 2\nA = x^2 + 3x - 5\nNow plug into one of the original equations to get B\nB = 3x^2 + 2x - 2 - (x^2 + 3x - 5)\nB = 2x^2 - x + 3
142621
x² + (a-3 )x – 2a² + 6a – 4 = 0\nx2 +ax-3x  = 2a2 -6a+4\n\nx= (-a +- sqrt(a2 +12))/2\nare roots of x\n(-a+sqrt(a2 + 12))/2  =  2 * (-a-sqrt(a2 + 12))/2\nas sqrt of any no is +ve\ncancel den. 2\n-a + sqrt(a2 + 12) = 2 * (-a-sqrt(a2+12))\na = -3sqrt(a2+12)\na2 = 9(a2 +12)\n8a2 = -12 * 9\na = sqrt(-27/2) = 3i * sqrt(0.5)
572480
breaking it apart, we have:\n\n(x+1)^3=(x+1)(x+1)(x+1)=\n(x^2 + 2x + 1)(x+1) =\n(x^3 + 3x^2 + 3x + 1) SO\n6(x+1)^3 = 6(x^3 + 3x^2 + 3x + 1) = 6x^3 + 18x^2 + 18x + 6\n\n(x+1)^2 = (x+1)(x+1)=\nx^2 + 2x + 1   SO\n-3(x+1)^2 = -3(x^2 + 2x + 1)=\n-3x^2 - 6x - 3\n\n4(x + 1) = 4x + 4\nand -9 stays the same.\n\nSo putting everything together, you have\n(6x^3 + 18x^2 + 18x + 6) + (-3x^2 - 6x - 3) + (4x + 4) - 9 \ncombine like terms and you get:\n6x^3 + (18x^2 -3x^2) + (18x - 6x + 4x) + (6 - 3 + 4 - 9)\n6x^3 + 15x^2 +  16x - 2\n\nWHEW!  That took me back to college pre-calc!
716170
since 2 is a root you take (x-2)*(x-2) and -3 is one also so you multiply (x+3) now you know it is a fifth degree so you need two more so you can take (4x^2+1) and multiply\n(x-2)(x-2)(x+3)(4X^2+1) you do the algebra
1085664
Good job!  You're halfway there!\n\nAs you are writing it out on your homework, underline the (4x+1) in each of the two terms.  That helps you see what you are factoring.\n\nFactoring out the (4x+1) from each term, you should get your final answer.\n\n=(5x + 4y)(4x+1)