is sqr x^2=x an identity ( true for all values of x)?
338782
sqrt x^2 = x is  not an identity.Its true only 4 positive values of x.\nThe reason is explained as follows:-\n\nConsider x > 0\nthen x^2 > 0 (of course,yes)\nthen sqrt x^2 = ± x.\nsince x > 0(as considered above),\nhence sqrt x^2 = x (for all x > 0).\n\nNow,Consider x < 0\nthen x^2 > 0(of course,yes)\nthen sqrt x^2 = ± x.\nsince x < 0(as considered above),\nhence sqrt x^2 = - x (for all x < 0).\n\nSo in this case,after taking sqrt,we have 2 insert a negative sign 2 balance the equation.\n\nHence,sqrt x^2 = x is not an identity(coz its true only 4 positive values of x).
523409
say you had x^2 + 4x + 3 = 0\nto complete the square you want the right side to be "squarable." i.e. in the form (x+a)^2 = b\nin this case the form is\n(x+2)^2 = x^2 + 4x + 4\n\nso you add one to both sides and yet\nx^2 + 4x + 4 = 1\n\nthis equals\n(x+2)^2 = 1\nx+2 = sqrt(1) = +/-1\nx therefore = -3, or -1
166521
Yes it is; and let us see how:\n\n(a+b)square = (a+b) x (a+b)\n= a x a + a x b + b x a + b x b\n= a x a + a x b + a x b + b x b\n= a square + 2ab + b square\n\nsimple, isnt it?
955603
x+0=x is always true. This is the definition of the identity property of addition.
46351
Sy   = |x| - x^2 + 2   \n     = (x^2/2) - (X^3/3) + 2X
803771
sq.rt of x*y = Sq.rt X * sq.rt. Y\n\nIt can be proven that:\n √(x*y) = √x * √y\n\nConsider the first part → √(x*y).\nTwo number values, one is x and the other is y, they are multiplied together.  You then take the square root of this result.\n\nConsider the second part → √x * √y.\nYou take the square root of x and y and then you multiply the results together.\nIt can be shown that the results of the first part is the same as the result of the second part.
1147786
x - sqrt(x) = 0\nx = sqrt(x)\nx^2 = x\nx(x-1) = 0\nx = 0,1\n\nBoth check.
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
984623
It will be 1/(x^2)\n\nFor example\n\n(1/2)^2 = (1^2)/(2^2) = 1/ 4\n(1/3)^2 = (1^2)/(3^2) = 1/ 9\n(1/4)^2 = (1^2)/(4^2) = 1/16\n\nSo the result actually gets smaller even though you are squaring it
115194
3 + 2sqrt(2) = (1 + sqrt(2))^2 so sqrt(3+2sqrt(2))=1+sqrt(2).\n\nBugzpodder, (x^2-3)^2 - 8 = x^4 - 6 x^2 + 1 is reducible.  Namely, it is x^4 - 2 x^2 + 1 - 4 x^2 = (x^2-1)^2 - (2x)^2\n   = (x^2 - 2x - 1)(x^2 + 2x - 1)
704948
Taking x common from (x^2 - ax)/x , we get x(x - a) /x which gives us (x-a). Now, (x + a)(x - a) gives us (x^2 - a^2)
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
586893
If its differentiate x * sqrt 1 + x^2 , the answer is\n  1 + 2x ( since sqrt 1 is a constant and d x^2/dx = 2x)\n\n \nIf your question is sqrt1 + x^2 answer is 0 + 2x = 2x
321878
The statement is true for all values of x.\n\nBasically, from negative infinity to positive infinity.\n\nIf you work out the equation:\n\nx+5-7 = x-7+5\nx-2 = x-2\nx = x
67086
in the above answer, it wouldn't be d^2, it would be d. So the answer would be\n \nd =sqrt[(x2-x1)^2+(y2-y1)^2+\n(z2-z1)^2] \n\nor\n\nd^2=(x2-x1)^2+ (y2-y1)^2+(z2-z1)^2
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
600230
Do you mean (x-h)^2? If so, yes, this complies with the general equation of a parabola y = ax^2+c. \nThe value of y is proportional to the square on x, that has to be a parabola.