What is x squared plus x?
599364
x^2 + x = x(x+1)\n\nIt can be simplified to x(x+1).
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
12044
By using the FOIL method, the final answer will be \n\n3x squared - x - 4
540085
(a-2)(a+2)  \n(difference of squares)
17048
= x^2 - 7x + 4x -28\n= x^2 - 3x - 28
712870
You only can say\nRoot[x^2 - 25x + 40]\nType for x sq only x^2.\n\nTh
338855
You are looking for values of x such that\nx^3 > 4x^2  Solve for x by bringing everything to one side.\nSo subtract 4x^2 from both sides\nx^3 - 4x^2 > 0\nFactor x^2 out\nx^2(x-4) > 0\nx^2(x-4) = 0 when x = 0 or x = 4\nSince you're looking for positive numbers, you only have to test a number between 0 and 4, and then another number greater than 4\nAny number between 0 and 4 will result in a negative value.\nAny nummber greater than 4 will result in a positive value.  You are looking for the positive values because  the expression is greater than zero.\nThus, any number that is greater than 4 will give you the result that the cube of a number is greater than 4 times its square.
953913
It's a difference of squares...\nIf you have (a² - b²) that is equal to (a + b)(a - b)\n\nSo in your example:\n16 - 9x²\n\n16 = 4² = a²\na = 4\n\n9x² = (3x)² = b²\nb = 3x\n\n(a + b)(a - b)\n= (4 + 3x)(4 - 3x)
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
700473
(x^5 + y^5) =  (x+y)(x^4 - x^3*y + x^2*y^2 - x*y^3 + y^4).
848127
When you added x to the right side, you should have gotten:\n\n4 = x^2\n\nThen take the square root of both sides to get\n\nx = 2, -2
914435
x > 5\n\n- - - - - -s-
170337
numbers 3 and 4\n\nx(squared) is still an integer and xy/x=y, which is an integer
614151
Your equation is pretty much simplified.\n\nif 11 had "x" as denominator you could sum them all. But since 11 is just a numeric number, adding them would result in: three 'x' squared plus four 'x' squared minus eleven 'x' everything divided by 'x'. Because you must find a common denominator for all three and that would be 'x'
471125
Hi!\nI'm suprised nobody used the easiest way that i will demonstrate now.\n\nLets assume further that x is positive real.\n\nWe start with a simple fact that ANY squares are nonnegative. The idea is we "choose" a square intelligently to suit our purpose. Also not that x does NOT equal zero.\n\nBy experience, \nI choose the square ( x^(1/2) - 1 / [ x^(1/2)] )^2.\nIt must be nonnegative for whatever positive real values of x.\nThus,\n\n    ( x^(1/2) - 1 / [ x^(1/2)] )^2 >= 0\n\nExpanding and regrouping, we get,\n\n    x - 2 + 1/x >= 0 \n      \nor equivalently,\n\n      x + 1/x >= 2\n\nThus, minimum is 2. \nWith equality when x^(1/2) = 1 / [ x^(1/2)], or\n x = 1.\n\nDone.\n\nAlternative Method.\n\nOne can use two variable AM - GM,\nFor any two positive reals m and n, we have\n\n ( m + n )/ 2 >= (mn)^(1/2)\n\nWith equality if and only if m = n.\n\nJust set m = x and n = 1/x, to get,\n\n( x + 1/x )/ 2 >= 1\n\nor  x + 1/x >= 2\n\nAgain, the minimum values is 2 with equality if and only if x = 1 .\n\nRegards
408554
yes. take out an x \nyou answer should be x(x+3)
811666
cross multiplying\n(x-4)(x+4)<x(x-5)\nx^2-16<x^2-5x\nadding 5x-x^2\n5x-16<0\nadding 16\n5x<16\ndividing by 5\nx<16/5\nx<3 1/5