what is the equation having 4-3i and 4+3i as roots ?
117698
so that x=4+ or - 3i\nthen transpose 4 to left\nx-4=+/- 3i square both sides\nthat's x^2-8x+16=-9\ntranspose -9 to left side\nx^2-8x+25=0
738757
3x - 4y + 17 = 0\n\n3x - 4y + 17 - 3x = 0 - 3x\n\n- 4y + 17 = -3x\n\n- 4y + 17 - 17 = - 3x - 17\n\n-4y = - 3x - 17\n\n-4y/-4 = - 3x/- 4 - 17/- 4\n\ny - 3x/4 + 17/4\n\n- - - - - - - - - - - - - -  \n\n4x + 3y + 6 = 0\n\n4x + 3y + 6 - 4x = 0 - 4x\n\n3y + 6 = - 4x\n\n3y + 6 - 6 = - 4x - 6\n\n3y = - 4x - 6\n\n3y/3 = - 4x/3 - 6/2\n\ny = -  4x/3 - 2\n\n- - - - - - - - - - - - - -\n\n3x + 4y - 41 = 0\n\n3x + 4y - 41 - 3x = 0 - 3x\n\n4y - 41 = - 3x\n\n4y - 41 + 41 = - 3x + 41\n\n4y = -3x + 41\n\n4y/4 = - 3x/4 + 41/4\n\ny = -3x/4 + 41/4\n\n- - - - - - - -s-
859715
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)
801068
-4(-1)^2 - 3(-1)\n-4(1) - -3\n-4 + 3\n= - 1
948364
Well, this is a standard square trinomial, which you can tell because it is written in the form a^2x^2 + 2abx + b^2, the square root of which is ax + b.  In this case, that means that your new equation is:\n(4y - 1)^2 = 0\nTake the square root of both sides to get:\n4y - 1 = 0\n4y = 1\ny = 1/4\n\nOne rational root
191203
(x-7i)(x+7i)=x^2 + (-i)^2 * 49 = x^2 + 49\n(which is your polinom having the roots -7i and +7i)
299363
Well, the square root is 27i, that makes it easy, cause 3X3X3 = 27, so 3 to the 6th is 729, and i X i is -1, so the sixth root of -729 is 3i.
819535
sqrt(x) - sqrt(3x + 3) = 1\nsqrt(x) = sqrt(3x + 3) + 1\n\nsquare both sides\n\nx = (sqrt(3x + 3) + 1)^2\nx = (sqrt(3x + 3) + 1)(sqrt(3x + 3) + 1)\nx = (3x + 3) + sqrt(3x + 3) + sqrt(3x + 3) + 1\nx = 3x + 3 + 2sqrt(3x + 3) + 1\nx = 3x + 4 + 2sqrt(3x + 3)\n\n-2x - 4 = 2sqrt(3x + 3)\n-x - 2 = sqrt(3x + 3)\n\nsquare both sides\n\n(-x - 2)^2 = 3x + 3\n(-x - 2)(-x - 2) = 3x + 3\nx^2 + 2x + 2x + 4 = 3x + 3\nx^2 + 4x + 4 = 3x + 3\nx^2 + x + 1 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/(2a)\n\nx = (-1 ± sqrt(1^2 - 4(1)(1)))/(2(1))\nx = (-1 ± sqrt(1 - 4))/2\nx = (-1 ± sqrt(-3))/2\n\nx = (1/2)(-1 ± isqrt(3))\n\nThere are no real solutions.
699632
Example:\n4^x = 8\n\nThe technique here is we change both sides into the same base.  Since we know that 4 = 2² and 8 = 2³, then we can change the equation into\n2^2x = 2^3\n\nThen, since the bases are equal, then the exponents must also be equal.  Thus,\n2x = 3\n\nTherefore,\nx = 3/2.\n\n----------------------------\nThe thing here is to change both sides into the same base.\nAnother example:\n81^[x(x + 2)] = 1/3^(9 - 20x)\n\nWe change both bases: 81 = 3^4, 1/3 = 3^-1.  Thus,\n3^[4x(x + 2)] = 3^[-(9 - 20x)]\n\nWe then equate the exponents:\n4x(x + 2) = (20x - 9)\n\nWe distribute:\n4x² + 8x = 20x - 9\n\nWe transpose everything to the left\n4x² - 12x + 9 = 0\n\nWe factor out\n(2x - 3)² = 0\n\nSquare root both sides\n2x - 3 = 0\n\nWe transpose -3\n2x = 3\n\nDivide 2.  Therefore,\nx = 3/2\n\n--------------------------\nHere's another example:\n\n2^x = 3\n\nI know that it is VERY DIFFICULT to change 3 into a power of 2.  Here, we use another tool: the logarithm.  If you are already ready, you can inform me so that I can teach you that.  Please feel free to contact me.\n\nkevin!\n^_^
757614
4 (x^2 + 3X -4)\n4(x+4)(X-1)
650764
Remember that i^2 is -1 so\n-3i+2(-1)-2+i=-4-2i
592356
-(-3) X 4\n= 3 X 4\n= 12
337374
i^2 = -1\nSo, i^4 = 1.\nSo, i^4n = 1.\n\nSo, i^335 = i^(4*83+3) = i^(4*83)*i^3 = 1*(-1)*i = -i
193529
if your equation is of the form\nax^2 +bx+c=0 then the general formula for the solution is\nthe two roots are:\n(-b+(b^2-4ac)^(1/2))/(2a)\n(-b-(b^2-4ac)^(1/2))/(2a)\nif b^2-4ac is less than zero then you will have imaginary roots, in such a case ur roots would be\n(-b+i(4ac-b^2)^(1/2))/(2a)\n(-b-i(4ac-b^2)^(1/2))/(2a)
872911
Let Z = 3 - 3i\n|Z| = √(3² + 3²)\n= 3√2\nThus Z = 3√2(1/√2 - 1/√2i)\nNow swwk an angle such that cosθ = 1/√2 and sinθ = -1/√2\nVlearly this is a 4th quadrabt angle\nTherefore θ = 2π - π/4\n= 7π/4\nThus Z = 3√2(cos7π/4 + i sin7π/4)\n[= 3√2e^(i7π/4)]
879323
x^4+4x^3+4x^2-4x-5 \n= (x^4 - x^2) + (4x^3 - 4x) + (5x^2 - 5)\n= x^2(x^2 -1) + 4x(x^2 -1) + 5(x^2 -1)\n= (x^2 -1)(x^2 + 4x + 5)\n= (x+1)(x-1)(x^2 + 4x + 5)\nSo, -1,1 and the roots of x^2 + 4x + 5 which are:\n\n-4/2 ± √(16 - 20) /2 = -2  ± √(-4) /2 = -2 ± 2i/2 = -2 ± i\n\nSo,\n1, -1, -2+i, -2-i
4519
(4/3)*pi*r^3\n\n[four thirds pi r cubed]\n\ngood luck!