^3 sqr x^2=4, and is sqrx^2=x an identity, for solve for all values of x that satisfies equation x-sqr x=0?
441224
x - sqrt(x) = 0\nAdd sqrt(x) to each side\nx = sqrt(x)\nSquare each side\nx^2 = x \nSubtract x from each side\nx^2 - x = 0\nFactor x\nx(x-1) = 0\nx = 0 or x = 1\nCheck in orginal problem: x- sqrt(x) = 0\nFor x = 0\n0 - sqrt(0) = 0\n0 - 0 = 0 \n0 = 0\nFor x = 1\n1 - sqrt(1) = 0\n1 - 1 = 0\n0 = 0\nThus, the answers are \n0 and 1\n\nAlso, sqrt(x^2) = x is only tru if x is positive.\nsqrt(x^2) = |x| is true for any number.\nThat is, the square root of x^2 is equal to the absolute value of x.\n\nx^2 = 4 means x = -2 or x = 2
156166
that's easy.\nD=b^2-4ac\nx1= [-b-sqrt(D)]/2a\nx2= [-b+sqrt(D)]/2a\n-------------------\na,b,c are the coeficients (ax^2+bx+c=0) so a=1, b=3, c=-5;\nsqrt is square root
945218
[(x^2)^2]^2\n\n(x^2 -4x +4)^2\n\nmultiply      x^2  (x^2 -4x +4)^  = x^4 -4x^3 +4x^2\n\nmultiply     -4x (x^2 -4x +4)^    = -4x^3 +16x^2 -16x\n\nmultiply     4  (x^2 -4x +4)^      =  4x^2 -16x +16\n__________________________________________\n\nthe answer should be                 x^4 -8x^3 +24x^2 -32x +16\n\ni hope that helped
265248
the general form of a quadratic equation ax^2 + bx + c =0\n\ntherefore there is two solution\nx=[-b ± sqrt (b^2-4ac)]/(2a)\n\nx^2+2x+2=0\n\na=1 b=2 c=2\n\nx1=[-2+sqrt(2^2-4*1*2)]/2 =(-2+2i)/2 = -1+i\nx2=[-2-sqrt(2^2-4*1*2)]/2 =(-2-2i)/2 = -1-i\n\nwhere i is the square root of -1 and it is imaginary\n\nthe solution is X={-1+i , -1-i}\n\nThose solution means that this equation in the from y=x^2+2x+2 has no intersection with y=0 since no real solution for x^2+2x+2=0
749124
x^2 + 4x + 4=7  (first put it in the form ax^2 + bx + c = 0, by subtracting 7 from both sides of the equation)\n\nx^2 + 4x - 3 = 0   (this is called a quadratic equation)\n\nNow substitute a =1, b = 4, c = -3 in the following formula\n\nx equals -b plus or minus the square root of (b^2 - 4ac) all over 2a\n\nBecause you are solving for x^2 you expect to get 2 answers.  The plus or minus in the middle of the formula gets you 2 answers as you solve the formula for both plus and minus the square root of (b^2 minus 4ac).\n\n[If you're going to have to do this in an exam, it's probably worth learning the formula off by heart.  It's boring to have to do so but you'll probably have more confidence in the exam if you see an quadratic equation and already know the formula.]\n\nGood luck for the working out - you can do this bit!
469289
3x^2 = 12\n\nFirst, divide by three.\n\nx^2 = 4\n\nTake the square root.\n\nx = +2, -2
857154
sure.\n\nx ^ 4 - a ^ 4 = (x ^ 2 - a ^ 2) * (x ^ 2 + a ^ 2)\n\nthen for x no equals to a or - a :\n\n(x^2-a^2) / (x^4-a^4) = 1 / (x ^ 2 + a ^2) -------> 1 / a ^ 4
1047363
I'm going to guess that you mean for the entire expression (4x+1) to be under the radical sign.  (I'm basing that guess purely on what sort of problems I typically see in math textbooks.)\n\nSo, start by subtracting the 3 from both sides:\n\n√(4x+1) = -3\n\nSquare both sides:\n\n4x+1 = 9\n\nSubtract 1 from both sides:\n\n4x = 8\n\nDivide both sides by 4:\n\nx = 2.\n\nNow, whenever you solve an equation with a square root in it, you've got to test your answers, because you can do everything right and still generate false solutions.  And, in fact, that's what happens here:\n\n√(4*2 + 1) + 3 = √9 + 3 = 3 + 3 = 6 which does not equal 0.\n\nThat means there are no solutions to this problem, and that's as it should be: since we always assume that a square root is positive, there's no way that we can add √(4x+1), a positive number, to 3, another positive number, and get zero as an answer.\n\nAs it happens, if the problem had been\n\n√(4x+1) - 3 = 0,\n\nwe would have gotten the exact same answer, x = 2, and it would have worked.  But the problem as stated has no solution.  (Assuming, that is, that my initial assumption about it was correct.)\n\nHope that helps!
164954
x^3 – 3x^2 + 4x – 12\n= x^2(x – 3) + 4(x – 3)\n= (x-3)(x^2 +4)
764942
√(4x + 1) +3 = 0\nSo √(4x + 1) = -3 (which means that 4x + 1 cannot be real as √N  ≥ 0)\nSo no solution\n\nIf you want to solve: \n\nFnd an x such that one of the square roots of (4x + 1) is - 3,\n\nthen 4x + 1 = 9\nx = 2
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
839343
x^2 + 4x + 4 = 7 (subtrat 7 from both sides)\n\nx^2 + 4x -3 = 0 (use the quadratic equation formula)\n\n-b + or - the square root of (b^2 - 4ac) / a^2\n\n-4 + or - the square root of 16 - (4) (1) (-3) / 2\n\n-4 + or - the square root of 28 / 2\n\n-4 + or - (the square root of 4) (the square root of 7) / 2\n\n-4 + or - 2 (the square root of 7) / 2 (simplify by dividing 2 into  the -4 and 2 on the top to get rid of the 2 on the bottom)\n\nthat gives you x = - 2 + or - the square root of 7.
580733
First equation:\n(x+a+b)^2 = x^2 + (x+a)^2 - (x+b)^2\nor x^2 + 2x(a+b) + (a+b)^2 = x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2\nor x^2 + 2ax + 2bx + a^2 + 2ab + b^2 - [x^2 + x^2 + 2ax + a^2 - x^2 - 2bx - b^2] = 0\nor 4bx + 2ab + 2b^2 = 0\nor 2b(2x+a+b) = 0\nor  2x + (a + b) = 0\nor 2x = -(a+b)\nTherefore x = -(a+b)/2\n\nSecond equation:\n(x+a)^2=(x+b)^2\nor (x+a)^2 - (x+b)^2 = 0\nor (x+a+x+b)(x+a-x-b) = 0\nor (2x+a+b)(a-b) = 0\nIf a = b then the equation becomes identity. So a is not equal to b.\nHence 2x + a + b = 0 \nor x = -(a+b)/2\n\nThus from both equations we have x = -(a+b)/2
342699
Completing the square involves creating a perfect square trinomial.  Consider\n\nx^2 + 4x - 1 = 0\n\nCompleting the square would involve trying to make one side a perfect square: (x + 2)^2 in this case.  Step by step:\n\n1.  Get all your x terms on one side, and constants on the other.\n\nx^2 + 4x = 1\n\n2.  Take half of the coefficient of the x term and square it.  Half of 4 is 2, and 2 squared is 4.\n\n3.  Add this to both sides of the equation.\n\nx^2 + 4x +4 = 1 + 4\n\n4.  Now factor the left side.  It works out to be a perfect square.\n\n(x + 2)^2 = 5\n\n5.  Take the square root of both sides and solve for x.\n\nx + 2 = +/- sqrt(5)\n\nx = +/- sqrt(5) - 2\n\nThe quadratic equation is pretty easy:\n\nx = [-b +/- sqrt (b^2 - 4ac)] / 2a\n\nIn the above example, a = 1, b = 4, and c = -1 (coefficients of the x^2 and x terms are a and b, and the constant is c).  So plugging in to the quadratic formula gives you\n\nx = [-4 +/- sqrt(4^2 - 4x1x(-1))] / 2x1\n  = [-4 +/- sqrt(16 - (-4))] / 2\n  = [-4 +/- sqrt(20)] / 2\n  = [-4 +/- 2sqrt(5)] / 2\n  = -2 +/- sqrt(5)\n\nwhich is the same as we got via completing the square.
501220
Are you sure it is x^3 and not x^2? If you're sure, then there is one relatively easy way to solve the equation:\n\n(x^3)-4x=24\n\nNow, this equation is of the form: x^3+px=q (where p=-4, q=24). The solution to this equation was given by the Italian mathematician "Cardan" to be:\n\nx=(sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3) - (sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3)\n\n\nFor your case, if you plug in the values for p,q you will get:\n\nx = (√(-64/27+144)+12)^(1/3) - (√(-64/27+144)-12)^(1/3)\n\nx = 2.888 - (-0.4629) = 3.3434\n\n\nThus, x=3.3434 is the real solution to the equation.
319669
did you mean the square root of x-1 = 3?\nin that case it will be 2:\nadd positive 1 to both sides:\n   x-1+1= 3+1\nthen solve the remaining equation:\n   x=4\nget the square root of 4:\n   x=2
730483
Note that the function is zero when x=1, this means that you can factor out an (x-1)\n\nx^4-x^2+4x-4 = (x-1)(x^3+x^2+4)\n\nNote that x^3+x^2+4 is zero when x=-2 so you can factor out an (x+2),\n\nx^3+x^2+4 = (x+2)(x^2-x+2)\n\nThe quadratic, x^2-x+2, does not factor with real numbers. (Try out the quadratic equation to confirm.) So the answer is\n\nx^4-x^2+4x-4 = (x-1)(x+2)(x^2-x+2)