what is the problem of this proof?  (-1)=(-1)^1=(-1)^(2/2)= square root of(-1)^2=/-1/=1?
926410
the fallacy is square root of (-1)^2=-1 and not mod -1
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...)
166642
First subsitute -1 wherever you see x\n\nf(-1) = 4(-1) - 3\n      = -4 - 3\n      = -7
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
466400
0 can't be in the domain.\nAlso, -1/2 cannot be in the domain because\n(-1/2)^(-1/2) = 1/sqrt(-1/2) which is not a real number\nI'm not sure exactly how to right this:\nThe domain is any real number except 0 and any number of the form (p/q), where p and q do not have any common factors and q is even and (p/q) is negative.\nIf (p/q) is negative and reduced and q is even, then you would be taking an even root of a negative number(which is never a real number).
220918
2*x^2 - 3*x - 2 = 0 <=> x = -1/2 V x = 2 \n\n==> 2*x^2 - 3*x - 2 <=0 <=> x = [-1/2 , 2]
153289
The answer is: f(-x)=1-(-x-1)^1/2.\n\nIf f(x)=f(x)=1-(x-1)^1/2, then it means that f(y)=1-(y-1)^1/2. Let y=(-x), then we get f(-x)=1-(-x-1)^1/2. Note that -x must be less than -1 in order to get the answer in real number. Otherwise, it will be a complex number.
915401
this means\n\nx^3 = -1\nor x^3+1 = 0\n\nf(x) = x^3+1\n\nf(-1) = 0\nso one root is -1\n\nfor the other root by division\n(x+1)(x^2-x+1)\nx^2-x+1 = 0 => x= (1+/-sqrt(3)i)/2
772527
x^2 + 3x  = -1\nx^2 + 3x +(3/2)^2 = -1 + (3/2)^2\n(x +(3/2))^2 = (5/4)\nx + 3/2 = sqrt(5)/2 OR x + 3/2 = - sqrt(5)/2\n\n(sqrt means the positive square root) \n\nx = -(3/2) + (1/2)sqrt(5)\nx = -(3/2) - (1/2)sqrt(5)\n\n.......................................................................\n\n2x^2 +4x + 1 = 0\nx^2 + 2x = -(1/2)\nx^2 + 2x + 1 = (1/2)\n(x+1)^2 = (1/2)\n\n(x + 1) = sqrt(1/2) OR (x + 1) = -sqrt(1/2)\n\nx = -1 + sqrt(1/2) OR\nx = -1 - sqrt(1/2)\n....................................\nLW = 783\n\n\n\n(W+2)W = 783\nW^2 + 2W = 783\nW^2 + 2W  + 1 = 784\nW + 1 = sqrt(784)\n\nW = srqt(784) -1\nW + 2 = sqrt(784) + 1\n\n2(W + L) = 2(2) sqrt(784)\n                = 4* 28\n                 =   112
325592
You need to first find 2 numbers that multiplies to give you a -4 and adds to give you a negative 3.  -4 and 1 work.\n(p^2 -4) (p^2+1) = 0\n\nThen, set them both = to 0.\np^2-4 = 0\np^2 = 4\np = 2\n\nand P^2 +1 = o\nP^2 = -1\nP = the square root of -1, also known as i\n\nFINAL ANSWER: P = 2, and i
574042
we need to multiply something by (x+1) so that the result would be x^2-1. but we know x^2-1 = (x-1)(x+1).\nso we have to multiply (x-1) with (x+1) to get x^2-1.
19513
This seems to be a popular question. There is no real number that is the square root of -1.  However, it turns out that it is useful to develop as set of numbers that allows the square root of -1.    This is called "i" as is known as an imaginary number.  The set of numbers that includes i is called the set of complex numbers.
200519
+/- 1\n\nThe answer is not 1, because -1 * -1 = 1 also.
846663
(x+3)(x-1)\n\n x= -3\nx=1
812806
The error comes from getting the square root of:\n(3 - 5/2)(3 - 5/2) = (2 - 5/2)(2 - 5/2)\n\nLet's give a similar example.  We are given:\n(1)² = (-1)²\n\nBut here we cannot get the square root of both sides, since we know that\n1 ≠ -1\n\nThe same thing goes with your example.\n^_^
140009
Note that it has two answers:\n\na^2=-4 ---> a=2i or a=-2i\n\n(2i)^2=2^2*i^2=4*(-1)=4\n\nand\n\n(-2i)^2=(-2)^2*i^2=4*(-1)=4
401494
From,\n(3-5/2)^ 2 = (2-5/2)^ 2\n\nThe mistake is in \n"Taking positive square root on both sides"\n\nLHS = Left Hand Side\nRHS = Right Hand Side\n\nFor we cannot simply take square roots. \nLook at LHS (3-5/2) is the POSITIVE root. \nOn the RHS (2-5/2) is the NEGATIVE root.\nYou made them "equal" to get the "wrong proof".\n\nActually, to take the roots note that \n(2-5/2)^ 2 = (5/2 - 2)^ 2\nSo, If you wanted to take "positive root" then,\nits obviously (5/2 - 2).\n\nThus comparing positive roots,\n3 - 5/2 = 5/2 - 2 which is a identity.\n\nRegards