-2 =the 4th root of x+2  i got no solutions to the real is that correct?
438101
Yes, in order to get a negative number after a positive number of multiplications or divisions, you must have the imaginary root of -1, or i. So if A x A x A x A equals a negative number, than A must be a real number with i.  In this case, it's  2 to the fourth power, or 16, and since it's -2, therefore it must be 16i.
498746
Square root of four has two solutions: 2 and -2.\n2 is a natural number. All natural numbers are integers. All integers are rational.\n-2 is NOT a natural number (no natural number is negative). But it is an integer, so it's a rational number.\nAs rational numbers are opposed to irrational numbers, none of the two solutions can be irrational.
1076805
y = -3 / (4x + 2)\n\nDomain: all acceptable x values:  We know that (4x + 2), or y would equal   -3/0, and that cannot be.  So actually set (4x + 2) =0 to find out what x CANNOT be\n\n4x + 2 = 0\n4x = -2\nx = -.5\n\nDomain (-infinity,-.5)U(-.5,+infinity)\n\nRange: All acceptable y values.  Looking at a graph of the function, you can see that y cannot equal zero, and looking at the equation,  there is nothing you can divide "-3" by and get zero, so \nRange (-infinity,0)U(0,+infinity)\n\nCheck this out to verify:\nhttp://www.coolmath.com/graphit/index.html
654517
x^2-2x+1\n(x-1)(x-1)\nso zero at x=1\n\nx^2-x=6\nx^2-x-6=0\n(x-3)(x+2)=0\nzero at x= 3, and x=-2\n\nx^2-3x-4\n(x-4)(x+1)\nzero at x=4  or x=-1\n\nx^2+8x+15\n(x+3)(x+5)\nzero at x=-5 or x=-3\n\nx^2-6x+5\n(x-5)(x-1)\nzero ar x=5, or x=1\n\nx^2+9x+20\n(x+4)(x+5)\nso zero at x=-4 or x=-5
747583
y= √(1/(x² + 1) – 2) \n\nWell 1/(x² + 1) – 2 ≥ 0\nie 1/(x² + 1) ≥ 2\nSince x² + 1 ≥ 0 for all x\nOn taking reciprocals\nx² + 1 ≤ ½\nx² ≤ -½ This is not possible as x² ≥ 0 for all x \nSo there is no domain\n----------------------- -----------------------\nWell that makes sense as  x² + 1 ≥ 1\nSo 1/(x² + 1) ≤ 1\nSo 1/(x² + 1) - 2 ≤ -1\n\nSo taking square roots gives no real solutions.
41284
Wait, all the answers are wrong\n\nx^2+4>4\nThen x^2-4x+4>0\nthen (x-2)^2>0\nBut hey, (x-2)^>0 is always true expect when x=2 which equals zero. An square is always positive or 0.\nSo the answer will be any value but 2 or R-{2}\nthese are the evidence:\nPut them in original non equation\nx=0 then 4>0 which is true\nx=-1 then 1+4>-4 so 5>-4 which is true\nx=3 then 9+4>12 so 13>12\n but at x=2 then 8>8 which is incorrect
878947
Sum of roots = 4, product = -7.\n\nNotice the first answer would give a positive product, so it's ruled out since the product is negative.  One of the others is ruled out too -- I'm sure you can figure out which.\n\nOf the two choices remaining, only one of them could give a positive sum, so it's the answer.\n\nPS  I'm sure you can do it from this, but if not, email h_chalker@yahoo.com.au\n\nPPS  I've just looked at Katya's answer.  She's right, except that it doesn't factorise, so you'd have to fall back on a graphing calculator, but that's unnecessary when you can use the simple logic I've outlined.
570705
Factoring out X gives:\nX(-X^2 + X + 6)=0\nUsing the quadratic formula to factor the terms inside the brackets gives:\n-X^2 + X + 6 = (X+2)(-X+3)\nSo we have:\nX(X+2)(-X+3)=0 and the zeros are X=0, X=-2, X=3
66783
(x-2)^2=x\nor x^2 - 4x + 4=x\nor x^2 - 5x + 4=0\nor x^2 - 4x - x + 4=0\nor x(x-4) - 1(x-4)=0\nor (x-4)(x-1)=0\n.'. x=4 or x=1;
296877
It has a solution, indeed. It actually has two solutions, -3 and 7. To find it out, set each pair of nomials, (e.g. (x + 3)) to zero. So it would be (x+3)=0. Then solve for x. Do the same for the other nomial, and there you have it.
982844
factor it -- notice that the relation of the x³ coefficient to the x² coefficient is the same as the relation of the x coefficient to the constant (1:25), therefore try x+4 as your first root and use long division to divide it evenly into your equation.\n\nx³ + 4x² + 25x + 100 = \n(x+4)(x²+25)\n\nSo your real root is x = -4\n\n(You have two other complex roots of x = -5i, x = 5i)
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
373728
Some folks seem to have forgotten this is a quadratic, and they have two (2) roots!\n\n(x - 7)^2 = 64\n\nx - 7 = ± 8 (take the square root of both sides)\n\nso \n\nx = 7 + 8 = 15 \n\nor \n\nx = 7 - 8 = -1\n\nyou could expand this out and use the quadratic equation to solve it as well:\n\nx^2 - 14x + 49 = 64\n\nx^2 -14x - 15 = 0  a = 1 b = -14  c = -15\n\nthe rest is left to the student as an exercise. :^)
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
1048224
This question is really only interesting if what you MEANT to type was:\n(4 - x^2) / (x + 2)\n\nSo you want to solve:\n\n. lim (4 - x^2) / (x + 2)\nx→ -2\n\nOf course the original expression is not defined at x = -2, because that would involve dividing by zero. However, for any real numbers x ≠ -2:\n(4 - x^2) / (x + 2) = (2 - x)(2 + x) / (x + 2)\n\nSince x ≠ -2 it is valid to divide top and bottom of the fraction by the common term (x + 2) which leaves:\n\n(4 - x^2) / (x + 2) = (2 - x), for all x ≠ -2.\n\nThus,\n. lim (4 - x^2)/(x + 2)\nx→ -2\n= lim (2 - x) = 2 - (-2) = 4 <<<<<<<<<<<\n. x→ -2
761784
Well, by inspection -2 is one root...\n\nP(-2) = (-2)^3 - 4(-2)^2 - 7(-2) + 10\nP(-2) = -8 - 16 + 14 + 10\nP(-2) = -24 + 24\nP(-2) = 0\n\nNow take your equation and divide by (x - (-2)).  In other words (x + 2):\n\n............ 1 -6 5\n1 2 ) 1 -4 -7 10\n....... 1 2\n....... ---------\n........... -6 -7\n........... -6 -12\n........... ----------\n............... 5 10\n............... 5 10\n............... ------\n.................... 0\n\nSo the result is x^2 - 6x + 5\n\nThis can be factored to (x - 5)(x - 1)\n\nThis results in (x + 2)(x - 5)(x - 1)\n\nAnd your roots are:\n-2, 5, 1\n\n(I don't know why I didn't notice 1 as a root... I usually check that first...)
1066820
If 'a' is a root of 2x^2=3x-4, then 2 a^2=3a - 4.\ni.e. a^2=(3a - 4)/2\n\nNow consider 4 a^3\n4 a^3\n= 4a(a^2)\n= 4a(3a - 4)/2\n= 2a(3a-4)\n= 6a^2 - 8a\n= 6(3a-4)/2 - 8a\n= a - 12\nQED