can someone please help me answer x+3 over 2 times x squared-x over x+3 = 21 [show steps please]?
368125
if this is the problem:\n((x+3)/2)*((x^2-x)/(x+3))=21\nthen you solve by\ncancelling out (x+3)\n(1/2)*(x^2-x)=21\nmultiplying by 2\nx^2-x=42\nmaking it quadratic\nx^2-x-42=0\nand using x=(-b(+-)srt(b^2-4ac))/2a\nx=(1(+-)srt(1-4*1*(-42)))/2\nx=(1(+-)srt(169))/2\nx=(1+13)/2  or x=(1-13)/2\nx=7   or x= -6
819535
sqrt(x) - sqrt(3x + 3) = 1\nsqrt(x) = sqrt(3x + 3) + 1\n\nsquare both sides\n\nx = (sqrt(3x + 3) + 1)^2\nx = (sqrt(3x + 3) + 1)(sqrt(3x + 3) + 1)\nx = (3x + 3) + sqrt(3x + 3) + sqrt(3x + 3) + 1\nx = 3x + 3 + 2sqrt(3x + 3) + 1\nx = 3x + 4 + 2sqrt(3x + 3)\n\n-2x - 4 = 2sqrt(3x + 3)\n-x - 2 = sqrt(3x + 3)\n\nsquare both sides\n\n(-x - 2)^2 = 3x + 3\n(-x - 2)(-x - 2) = 3x + 3\nx^2 + 2x + 2x + 4 = 3x + 3\nx^2 + 4x + 4 = 3x + 3\nx^2 + x + 1 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/(2a)\n\nx = (-1 ± sqrt(1^2 - 4(1)(1)))/(2(1))\nx = (-1 ± sqrt(1 - 4))/2\nx = (-1 ± sqrt(-3))/2\n\nx = (1/2)(-1 ± isqrt(3))\n\nThere are no real solutions.
79707
16 - 3(8-3)squared / 2\n\n1. 16-3x(8-3)²÷2\n2. 16-3x(5)²÷2\n3. 16-3x25÷2\n4. 16-37.5\n5. -21.5
817835
x^3-x^2+3x-1\nx+4&#92;x^4+3x^3-x^2+11x-4\n      -(x^4+4x^3)\n              -x^3-x^2\n            -(-x^3-4x^2)\n                       3x^2+11x\n                     -(3x^2+12x)\n                                   -x-4\n                                 -(-x-4)\n                                         0\nx^4+3x^3-x^2+11x-4=(x+4)(x^3-x^2+3x-1)
872966
1.) 3 &#92;/ x+2 - 4 = 3\nif you mean\n3 sqrt( x+2) -4 =3,\nthen\n3 sqrt(x+2) =7\nsqrt(x+2) = 7/3\nx+2 = 49/9\nx=49/9 -2\n\n2.) &#92;/ x+3 - &#92;/2x-8 = 1\nsqrt(x+3) -sqrt(2x-8) =1\nsqrt(x+3) = sqrt(2x-8) +1\nx+3 = 2x-8 +2 sqrt(2x-8) +1\n-x + 10 = 2 sqrt(2x-8)\nx^2 -20x + 100 = 4 ( 2x-8)\nx^2 - 28x  +132 =0\n'
965748
|x - x²| \n= x - x² for x - x² ≥ 0\n= x² - x for x - x² < 0\n\nx - x² = x(1 - x) = 0 when x = 0 and x = 1\n\nSo x - x² ≥ 0 when 0 ≤ x ≤ 1\nand x - x² < 0 when x < 0 and x > 1\n\nSo  dx under the interval [-1,3]\n\n= ∫(x² - x) dx under the interval [-1, 0] +\n∫(x - x²) dx under the interval [0, 1] +\n∫(x² - x) dx under the interval [1, 3]\n\n=  [⅓x³ - ½x² ] [-1, 0] + [½x² - ⅓x³] [0, 1] + [⅓x³ - ½x² ] [1, 3]\n\n= [(⅓0³ - ½0²) - (⅓ (-1)³ - ½ (-1)²) ] + [(½(1)² - ⅓(1)³) - (½0² - ⅓0³)] + [(⅓ 3³ - ½ 3²) - (⅓1³ - ½1²) ]\n\n= [5/6] + [1/6] + [9/2 + 1/6]\n\n= 5 2/3
83153
There are a couple of ways to work this out, I will take the easiest to explain.\n\n(x-3)(x^3+3x^2+9x+27) Can be rewritten as:\n\n(x)(x^3+3x^2+9x+27)+(-3)(x^3+3x^2+9x+27) Now, multiply those out.\n\nx^4+3x^3+9x^2+27x-3x^3-9x^2-27x-81 Now, sort by like powers of x\n\nx^4+3x^3-3x^3+9x^2-9x^2+27x-27x-81 Simplify\n\nx^4-81
942253
the answer is -3x square root of 2
123892
3(x – 2)^2 – 3(x – 2) – 6\n3((x-2)^2 - (x-2) -2)\n3(x^2-4x+4-x+2-2)\n3(x^2-5x+4)\n\n3(x-4)(x-1)\n\nYay!!
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
598233
just a little correction to taylor's answer.\nsqrt(x-2)=1, square it on both sides \n=> x-2 => square(1)=1\n=> x=3
1012196
3(x+12)=x+2\n3x+36 =x+2\n2x+36=2\n2x=34\nx= 17
1182929
7x+27 over (x+9)(x-9)
194716
x^2 + 13x + 52 + 160/(x-3)
605879
(x+3)(x+12)(x+2)(x+8)= [(x+3)(x+8)][(x+12)(x+2)]\n                     = (x^2+11x+24)*(x^2+14x+24)\nlet t=x^2+24+(25X)&#92;2 . \nSo we have: x^2+11x+24=t-(3x)&#92;2 and x^2+14x+24=t+(3x)&#92;2                                 \nso the equation becomes: [t-(3x)&#92;2][t+(3x)&#92;2] = 4x^2\n                    <-->t^2-(9x^2)&#92;4=4x^2  \n                    <-->t^2=(25x^2)&#92;4\n                    <-->t=(5x)&#92;2 or t=-(5x)&#92;2\n                    <-->x^2+10x+24=0 or x^2+15x+24=0\ncontinue by yourself
408554
yes. take out an x \nyou answer should be x(x+3)
1059114
3x^2 - 9x + 6 - 3(x - 1)(x - 2)\n3x^2 - 9x + 6 - 3(x^2 - 2x - x + 2)\n3x^2 - 9x + 6 - 3(x^2 - 3x + 2)\n3x^2 - 9x + 6 - 3x^2 + 9x - 6\n\nEverything gets cancelled out, so it doesn't matter if i put 10 in for x or not.\n\n3(10)^2 - 9(10) + 6\n3(100) - 90 + 6\n300 - 90 + 6\n210 + 6\n216