Solve (x^2) <= 1- (y^2)?
831242
This is just the inside of the unit circle including its circumference.\nThe unit circle (circle at the origin with radius=1) is\nx^2 + y^2 = 1\nAnd you want x^2 + y^2 ≤ 1\nso its the circle plus its inside.\n( if it had been x^2 + y^2 ≥ 1 it would be the circle plus everything outside )\nIf you wanted x in terms of y, you just get |x| ≤ √(1-y^2) .
60538
if [ -7x + x^2 =0 ]\nthe you will get y = 7/x
207106
x^2-3=-2\nx^2=-2+3\nx^2= 1\nx=1\n\n1^2+-3+2=0
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
629654
x^2+y^2-2y-5=0\n\nx^2 + y^2-2y + 1 - 1 -5 =0\n(x-0)^2 + (y-1)^2 = 6\n\nCenter (0, 1) radius sqrt(6)
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
800348
b^2 - y^2 - 2yp - p^2\n=b^2-(y^2+2yp+p^2)\n=b^2-(y+p)^2\n=[b+y+p][b-(y+p)]\n=(b+y+p)(b-y-p)
676814
First equation: sqrt[x] + y = a\nor sqrt[x] = a - y\nSquaring both sides\nx = (a - y)^2__________(1)\n\nSecond equation: x + sqrt[y] = b\nor sqrt[y] = b - x\nSquaring both sides\ny = (b - x)^2___________(2)\n\nSubstituting 'x' from eqn. (1) in eqn. (2)\ny = [b - (a - y)^2]^2\n  = [b - a^2 + 2*a*y - y^2]^2\n\nSo it is clear that you have to solve a 4-th order equation.\nIt is really tough.
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
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
723313
Yes\n\n(1-2)^2 = (3-2)^2\n\n3 is not equal to one\n\na and c do not have to be equal since the terms are squared. On removing the squares, you have to account for both +ve and -ve roots.
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
902917
y=1/x^2-3 => y+3 = 1/x^2 => x= +/-sqrt(1/(y+3))
512448
x^2 + y^2 = 25\nx - y = 0\n\nx = y\n\nY^2 + Y^2 = 25\n2 Y^2 = 25\n\nY ^2 = 25/2\ny =  (25/2)^1/2 or\ny = Sqrt (12.5)
206555
is the question x=x^2?\nif so 1 is the only number that can always satisfy that
157812
[x^2/4][2(log(x) - 1) log(x) + 1]
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]