For each equation of the form (ax mod n = b), solve for x in the range [0,n-1]19x mod 26=1?
109196
For range 0 - Answer = 0\n\nFor range n-1\n\nA= 19(n-1) = 19N-19\nN= 26\nB = 1\n\nTherefore A = 19(26) - 19 = 475
686655
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!
720847
Use logarithms. If you haven't learned them yet, it just helps to solve equations with powers.\na = b^x\nlna = ln(b^x) = xlnb\nx = lna/lnb\nx = ln(10)/ln(32.6087)\nx = 0.6607928852\nYou can use your calculator to check. I guarentee that it works.
652953
I got:\n1. 6x^2 - 36x +48\n2. 6x + 22
924424
The equations:\n(a+b)^2=a^2+2ab+b^2\nA^2-B^2=(A+B)(A-B)\nJames Chan's solutions were right.
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
798499
Given ax² + bx + c = 0, the quadratic formula tells us that:\n\nx = [ -b ± √(b²-4ac) ] ÷ 2a\n\nIn your problem, a=3, b=12, c=1  --  simply plug these values into the quadratic formula to solve for x.
304206
the clue would be to simplify the equation  to \nlim {x +[-(x+1)+1]}/(x+1)\n x-->0 \nlim [x + 1 -x]/(x+1)\nx-->\nlim (1/(x+1)) =1\nx-->0
1177677
shouldn't it be x^2??? if it IS x-squared, then sure, you could use the factorization method. \n\nit goes like this:  (6x + 1) (x - 1) = 0\n\ntherefore, x = -1/6 , 1
657080
(x + 1)^2 = 25\nx + 1 = -5 or 5\nx = -6 or 4\n\n---------------------\n\nx^2 + 4x + 4 = 16\nx^2 + 4x - 12 = 0\n(x + 6)(x - 2) = 0\nx = -6 or 2\n\n----------------------\n\nx^2 - 14x + 49 = 13\nx^2 - 14x + 36 = 0\n\nx = (-b ± sqrt(b^2 - 4ac))/2a\nx = (-(-14) ± sqrt((-14)^2 - 4(1)(36)))/2(1)\nx = (14 ± sqrt(196 - 144))/2\nx = (14 ± sqrt(52))/2\nx = (14 ± sqrt(4 * 13))/2\nx = (14 ± 2sqrt(13))/2\nx = 7 ± sqrt(13)
402617
No solutions.\nFrom the first equation you have\ny=x^2+2x+9/2\n\nUsing the second:\nx^2+2x+3=x^2+2x+9/2\n3=9/2 , so there is no solution
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
681319
-18=x-9\n+9   +9\n-9=x\n\n27=x-11\n+11   +11\nx=38\n\n15=x-16\n+16  +16\nx=31
754366
You need to use the quadratic equation as it is not factorable into rational expressions.\nThe form of the equation is:\nax^2 + bx + c = 0 \nSo in your case,\na = -2, b = 6, c = -3\nand the solution is:\nx = [ -b ± √(b^2 - 4ac) ] / (2a)\n\nso substitute and reduce.\n\nx = [ -6 ± √(6^2 - 4(-2)(-3)) ] / (2(-2))\n\nx = [ -6 ± √(36 - 24) ] / (-4)\n\n= [ -6 ± √12 ] / (-4)\n= [ -6 ± 2√3 ] / (-4) \n= (3 ± √3)/2
121228
here is a simple program algo\n\nprogram quad\nbegin\nvar\n     declare variables;\nget variable values of a,b,c for equation ax2+bx+c=0\nd=)b*b)-4*a*c;\nif(d<0)\nbegin\n\n   root are imaginary;\nend\nelse\n\n    root1=b(sqrt(d))/2a;\n    root2=-b(sqrt(d))/2a;\n ( note : formula i am not sure make changes as per required)\nprint the value of root1 and root2;\nend
626693
ax - b = -c\nax = b - c\nx = (b - c) / a.
282131
Note that although x1=x2 satisfy the equations but some distinct values for x1 and x2 can satisfy it too: \n\nsuppose x1 and x2 are distinct and so x1-x2 is not zero\nthen eq1 is:\n\n4(x1-x2)(x1+x2)-4(x1-x2)=0\n\n---> 4(x1+x2)-4=0\n---> x1+x2=1\n\nand eq2 is:\n\n-3(x1-x2)(x1+x2)(x1^2+x2^2)+2(x1-x2)(x1+x2)=0\n\nomit (x1-x2)\nand substitute x1 by 1-x2 to get:\n(1-x2)^2+x2^2=2/3\n\n---> x2=(3+-sqrt(3))/6\n\nso we have two values for x1 and x2 which satisfy the equation and x1 and x2 are distinct.