How many solutions sqrt(x-4) – sqrt(x+8)=2?
345141
0 solutions!\n\nMove sqrt(x+8) on the other side and then raise by the power of 2. You will get\n\nx-4=4+4*sqrt(x+8)+x+8    (btw if there is a solution it has to be greater or equal to 4)\n\nthis is equivallent to \n\nsqrt(x+8)=-4\n\nWhich is impossible by the definition of sqrt(). It always has to be greater or equal to 0.
919399
6(x+4)-2x=-8\n6x+24-2x=-8\n6x-2x+24+8=0\n4x+32=0\n4x=-32\nx=-32/4\nx=-8
302992
2 x 2 = 4\nHe asked for square, not square root.
556363
(8/x+2) + (8/2) = 5\n\n(8/x+2) + 4 = 5\n\nSubtract 4\n\n(8/x+2)=1\n\nMultiply x+2\n\n8= x+2\n\nSubtract 2\n\nx=6
896553
4x^2+8x-x-2\n(4x-1)(x+2)
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
805274
1) 3 * sqrt(-8)\n= 3 * sqrt(4) * sqrt(-2)\n= 6 * sqrt(2) * i\n\n2) sqrt(1/6)\n= sqrt(6/36)\n= sqrt(6) / sqrt(36)\n= sqrt(6) / 6
320716
8(a^4)(b^4) - 28(a^3)(b^3) + 4(a^2)(b^2)\nRemember that (a^N)(b^N) = (ab)^N and let c=ab, then\n8c^4 - 28c^3 + 4c^2 = c^2(8c^2 - 28c +4) or\nc^2(2c^2-7c + 1)/4\n\nAt this point, use the quadratic equation to find the 2 roots from within the parentheses and then make the appropriate substitions for c = ab.
636051
x=4\n\nplease solve...\n\n8-(1) :-) ;-(
752254
Well, let's check if y=4.\nJust replace y with 4\n4(4) + 2(4) = 24\n16 + 8 = 24\n24 = 24 \nThis is equation is true, so you are correct.
713553
you mean simplify squared root of 8?\n8= squared root of 2x2x2\n8= 2 radical 2
1056888
2(x^2-4x)-1=\n2(x^2-4x+4) -8-1=\n2(x-2)^2-9
134165
On top..\n4 plus or minus the square root of 8\non bottom..\nall over 4\n=(4 +or- sqrt8)/4\nIf you can find any factor of the number inside the radical sign which is a perfect square, then you may rewrite that number into the product of that perfect square and the 'other". Let's say 45. 9 is a factor of 45 and a perfect square, so 45 = 9 * 5.\nIn your given, 8 = 4 * 2. (4 is a perfect square) You may substitute this to the original equation:\n=(4 +or- sqrt(4*2))/4\nyou may then get the square root of that perfect square and place it out of the radical sign, leaving the 'other' number. Since 2 is the sqrt of 4, then\n=(4 +or- 2 (sqrt 2))/4\nThe numerator has a common monomial factor of 2 (One technique is to find if all terms are even). you can factor this out and use the cancellation method.\n=2(2 +or- sqrt 2)/4\nCancelling 2 from both numerator and denominator,\n=(2 +or- sqrt 2)/2\nYou have a +or- sign, so there are to answers:\n=(2 + sqrt 2)/2 or\n=(2 - sqrt 2)/2\nThese answers are accepted, but if you want, you can simplify sqrt 2 into 1.4142. So you will get:\n=1.7071 or\n=0.2929\nHope this helps ;-)
769101
= 6 + root(8/2)\n= 6 + root(4)\n= 6 + 2\n= 8
243121
first square two sides\nyou will get 3x+1 =9+6√(x-4)+x-4\n= (x-2)=6√(x-4)  square it \nyou will get x^2 -4x +4=9(x-4)\n=x^2 -13x + 40=0\nfactor  (x-5)(x-8)\nso x=5 or x=8
117585
4 and 64\nSQRT4 = 2\nSQRT64 = 8\n8 + 2 = 10\n64 - 4 = 60
799076
i will use = for is congruent to:\n\n1) x^2 + 14x = 35 (m 337)\n\nx^2+14x+49 = 84 (m 337)\n\n(x+7)^2 = 84  (m 337)\n\nNow 84 is a quadratic residue (m 337), so this equation has two solutions.\n\n2) x^2-64x+1024 = 81 (m 3011)\n\n(x-32)^2 = 81 (m 3011)\n\nAgain, 81 is a quadratic residue (m 3011), so this equation has two solutions.\n\nSteve