Two Algebra questions with quadratics!?
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1.  \n\n7x + 5i = 7 - 6yi\n\nIsolate x:  \n\n7x = 7 - 6yi - 5i\n\nSolve the right hand side in the proper form of complex numbers (a + bi)\n\n7x = 7 + i(-6y - 5)\nx = 1 + ( (-6y - 5)/7 ) i\n\nYour goal is to get some value for a (the number without the i) and some value for b (the number next to the i).  That's why I factored out the i so that we get the complex number proper form.\n\n2.\n\nThe quadratic formula for ax^2 + bx + c = 0 goes as follows:\n\nx = (-b +/- sqrt(b^2 - 4ac))/2a\n\nIn the above question, a = 2, b = 7, and c = 6.  All you need to do is substitute.  Note that "+/-" means "plus or minus", meaning you'll get two solutions for x.
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Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
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the 1st one is simple\n(x+y) dy/dx + y =0\nor xdy/dx + ydy/dx + y =0\nor (xdy/dx + y)  + ydy/dx = 0\nxdy/dx + y = d/dx(xy)\n\nand d(y^2/2)/dx \n\nso d/dx(xy+y^2/2) = 0\n\nor xy+y^2/2 = c\nor 2xy+y^2 = 2c or C\n\nproved\nfor the second one\n\ndy/dx -(1/x)y = xy^2\n\nlet y/x = t\ny = xt\ndy/dx -y/x = t+ x dt/dx -t = x dt/dx\nso\nx dt/dx = - x (xt)^2\nor dt/dx = x^2t^2\ndt/t^2 = x^2 dx\nyou should be able to proceed
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This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
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i think u typed the question wrong, can u please correct it and if u have 2 unknowns, please give 2 equations, else u cant get a definite solution
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No prob dude... take a deep breath, grab a cold drink, and concentrate on this... ready? ok....\n\nOften, the simplest way to solve "ax2 + bx + c = 0" for the value of x is to factor the quadratic, set each factor equal to zero, and then solve each factor. But sometimes the quadratic is too messy, or it doesn't factor at all, or you just don't feel like factoring. So, while factoring may not always be successful, the Quadratic Formula can always find the solution. \nThe Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers. The Formula is derived from the process of completing the square, and is formally stated as:\n\n For ax2 + bx + c = 0, the value of x is given by:\n   \nNote that, for the Formula to work, you must have "(quadratic) = 0". Note also that the "2a" at the bottom of the Formula is underneath everything above, not just the square root. And don't forget that it's a "2a" under there, not just a "2"! And make sure that you are careful not to drop the square root or the "plus/minus" in the middle of your calculations, or I can guarantee that you will forget to "put them back" on your test, and you'll mess yourself up. And remember that "b2" means "the square of ALL of b, including the sign", so don't leave b2 being negative, even if b is negative, because the square of a negative is a positive. In other words, don't be sloppy and don't try to take shortcuts, because it will only hurt you in the long run. Trust me on this!
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Is this one problem?  If so, you can solve it like this:\n\n(a/b) / (c/d) = (a/b) * (d/c)\n\n~ ~ ~ ~ \n\n( (x² - 4) / x² ) / ( (x²+5x+6) / x ) = \n\n( (x² - 4) / x² ) * ( x / (x²+5x+6) ) = \n\n( (x² - 4) * x) ) / (x² * (x²+5x+6) ) = \n\n( (x² - 4)  / (x * (x²+5x+6) ) = \n\n{{ now notice you can factor your quadratic equations, to get the following }}\n\n(x+2)(x-2)  / (x * (x+2)(x+3) ) = \n\n(x-2) / (x(x+3))\n\nHope that helped!\n\n~ ♥ ~
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Let's use the variables a and b to represent the two positive integers:\n\na=b+8\na*b=84\n\nPlugin the first equation into the second (substituting the a):\n\n(b+8)*b=84\n(b^2)+8b=84\n(b^2)+8b-84=0\n\nUse the quadratic formula to find the a value.\na=14\n\nPlugin the b value into one of the original equations to get the b value.\n\na=b+8\n14=b+8\nb=6
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1)   f(x) = x+1/x =(x^2+1)/x  \n      g(x) = x-2/x =(x^2-2)/x\n(f+g)(x) = (x^2+x^2+1-2)/x = (2*x^2-1)/x = 2*x -1/x\n\n2)  f(x) = -2/x\n     g(x) = (x+1) / (x-3)\n(f+g)(x) = -2/x + (x+1)/(x-3) = {-2*(x-3)+x*(x+1)}/x*(x-3)\n              = (-2x+6+x^2-x)/(x^2-3x) =(x^2-x+6)/x^2-3x
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There's lots of help online. Try these web sites:\n\nhttp://www.algebrahelp.com/\nhttp://www.purplemath.com/\nhttp://www.sosmath.com/algebra/algebra.html\nhttp://www.helpalgebra.com/\n\nThere's even more. So start with these, and good luck!
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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
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(x^2+4x+1)/(x-2) = x+2+3/(x-2)\n\nx  : x^2+4x+1-x^2-2x =2x+1\n2  : 2x+1 -2x-4 =3
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Let's call :\n(a) : x + 2y < 3\n(b) : 2x - 3y < 6\n\nWith the 1st one:\nx + 2y < 3\n<=> 2x + 4y < 6 \n\nThe 2nd one is 2x - 3y < 6\nThat means : \n(a) - (b)\n<=> 4y - (-3y) < 0 \nor 7y < 0 \nor y < 0\n\nWe have also: 2x - 3y < 6 <=> (2/3).2x - (2/3).3y < (2/3).6\n<=> (4/3)x -  2y < 4\nThat means:\n(a) + (b)\n<=> x + (4/3)x < 7\n<=> (7/3)x < 7\n<=> x < 3\n\nSo : x < 3 and y < 0
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first one is x^2=361/25.....the answer to the second one is 9^x=1
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Answer 1:\n(4y + 2)/2 = 21\n4y + 2 = 21*2\n4y + 2 = 42\n4y = 42 - 2\n4y = 40\ny = 40/4\ny = 10\n\nAnswer 2:\na + 4 = 3a - 81\na = 3a - 81 - 4\na = 3a - 85\na - 3a = -85\n-2a = -85\n2a = 85 (The minus signs cancel each other out)\na = 85/2 (or) 42.5\n\nAnswer 3:\n146 = (36y + 2)/4 = 236.5?\nThis means that 146 = 236.5 which is never possible\nThis implies that the equation has no solution.
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b1=(-b+ (b^2-4ac)^1/2 )/2a=(-8+(64-48)^1/2)/6=(-8+4)/6\n    =-4/6=-2/3\nb2=(-b- (b^2-4ac)^1/2 )/2a=(-8-4)/6=-12/6=-2
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1) 2x^2 + 3 = x\n2x^2 - x + 3\n\nQuadratic formula:\n[1 +/- sqrt(1-4*2*3)]/(2*2)\n[1 +/- sqrt(-23)]/4\n\nx= 1/4 + sqrt(23)/4 * i\nor\nx= 1/4 - sqrt(23)/4 * i\n\n2) (2m - 7)^2 - 2(2m - 7) - 24 = 0\nSay X = 2m-7\nSo: x^2 - 2x - 24 = 0\n(x-6)(x+4)=0\nx=6,-4\n\n2m-7=6\nm=13/2\n\nor\n\n2m-7=-4\nm=3/2