Solving Quadratic equations using factoring?
828595
Let x and y be the two positive integers, then\nxy = 64 and x = 4y\nReplace x with 4y in the first equation.\n(4y)y = 64\n4y^2 = 64\nDivide both sides by 4\ny^2 = 16\nTake the square root of each side\ny = 4 (You can ignore y = -4 because you are looking for positive integers)\nIf y = 4, then x = 4(4) = 16\nAnswer: 4;16
1053403
Write it as:\n2x^2-4x+3x-6\n=2x[x-2]+3[x-2]\n=[2x+3][x-2]\nSimple middle-term factorisation!!!
893823
Use the quadratic equation \n\nt = -b + or - sqr(b^2 - 4ac)/2a\n\nyour equation is already in the form:\n\nat^2 + bt + c = 0\n\njust plug in the variables and solve for t twice
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a
939068
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~ ♥ ~
993104
take the equation x^2 - 7x + 12 and do the following  (x ) (x) now...see that there is a - and a + in the equation....you need to do this ( x -  ) ( x - ) 'cause you put the minus in the first factor in the second factor you take the - and the + and multiply them....So you get (x- ) (x -) the other 2 terms for each equation comes from the 7x and the 12.  What you do is to find 2 numbers that multiplied gives you 12 and added give you 7....The numbers are 3 and 4 ...Then put them into your factors (x-3)(x-4).\n\nFor the other 2 equations you do the same process\n\n(x^2 - 8x + 7) = (x-  ) (x-  ). In this one, you have to find 2 numbers that multiplied give you 7 and added give you 8. They are 1 and 7. Then (x^2 - 8x + 7) = (x- 1 ) (x- 7 ).\n\nw^2 - 8w + 15 = (w - ) (w- ). As you saw in one and two, three is the same....so the factors are 3 and 5. Then, w^2 - 8w + 15 = (w -3 ) (w-5 )\n\nFor the solve...\n1. x^2 - 6x = 0\n2. x^2 - 2x - 3 = 0\n\nTHe 1 is like this...you have to complete your quadratic equation then you do the following x^2 - 6x = 0 --> x^2 - 6x + 9 =0+ 9...The 9 in the right side was added in order to keep the two sides equal...IN the left side you got a perfect quadratic equation, hence it will be (x-3)^2= 9....Now you pass the 9 of right to the left ...then it will be [(x-3)^2]-9=0\n\n\nThe other equation x^2 - 2x - 3 = 0....This one is quite simple, it is the same as the part 1...The difference is that when you find the factors, they will be 2 numbers that multiplied give you 3 but substracted gives you 2.....The #s are 3 and 1....Here is the thing....when you break the equation into factors, you will have the following (x- ) (x+ )...Since you have a -2 that means that it is the -3 and the +1...Hence your factors will be the following (x-3)(x+1)
768295
Using an equation is the lazy, mechanical way to do it. If you want to understand what's going on, factor it if possible.\n\nIf you *really* want to know what's going on, derive the quadratic formula yourself. It isn't as hard as it looks, and it can be done on one page.
890098
I enjoy algebra, and this is easily my favorite type of problem.  So I hope you, too, one day enjoy them for their inner beauty.\n\nFor now, let's tackle this "factoring problem."  We may wish to use the method of "synthetic division" which is arguably easier than "long division of polynomials."\n\nThink back to grade school where you learned that 6 = 2 * 3.  This simple statement enabled you to 'deconstruct' the number 6 as either "two times three" or "three times two".  Even if you forgot one of the 'factors' you could conjure it up again quickly by asking yourself, "What is six divided by two?" for example.\n\nIn problem (a), you are essentially asking, "What are the coefficients of P(x) if the only factor we know is (x-1)?"\n\nLet's set up synthetic division (see link below), by placing all the coefficients of P(x), and using 1 (the zero implied by (x-1)) as the divisor:\n\n1 ||  a  b  c  d\n___________\n.....a\n\n1 ||  a  b  c  d\n_____a_____\n.....a (a+b)\n\n1 ||  a  b        c       d\n_____a___(a+b)____\n.....a (a+b) (a+b+c)\n\n1 ||  a  b        c                d\n_____a___(a+b)___(a+b+c)_\n.....a (a+b) (a+b+c) (a+b+c+d)\n\nNote that this last term must equal 0 since it is assumed that (x-1) is a factor ... thus, the "remainder" term must equal zero!\n\nThere's the result you were after:  a+b+c+d = 0\n\nHope this helps you tackle the others!
1113001
take squre root of both sides\nx-1=+ or - sqrt(7)\n\nthen \nthe first root \nx1=1-sqrt(7)\n\nthe second root\nx2=1+sqrt(7)\n\nBy factoring method:\n\ntake 7 to left\n(x-1)^2-(sqrt(7))^2=0\nfactor like a^2-b^2=(a-b)(a+b)\n\n[x-1-sqrt(7)] * [x-1+sqrt(7)]=0\n0*0=0\nx=1+sqrt(7)\nor\nx=1-sqrt(7)
826369
I have found sites like the following helpful:\n\nhttp://www.algebra.com/\nhttp://www.saab.org/mathdrills/factor.cgi\nhttp://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=algebra&s2=factor&s3=basic\n\nThe last one is a solution program, but it can be helpful to check your answers.
898442
You just factor\n(2x-1)(x-2)=0\nx=2, 1/2
371573
If i is a root, you may express it into x = i, a root of x^3 + i = 0.\nsince x = i, x - i = 0\nTherefore, x^3 + i has a root x - i. to get the other 2 factors, divide.\n(x^3 + i)/(x - i) = 0\nDivide it using long division, and you'll get\nx^2 + ix - 1 = 0\nActually, \nx^3 + i factors as (x - i)(x^2 + ix - 1) \ntherefore, \nx^2 + ix - 1 = 0\nuse the quadratic formula:\na = 1\nb = i\nc = -1\nx = [-b +or- sqrt(b^2 - 4ac)]/2a\nx = [-i +or- sqrt(i^2 - 4(1)(-1))]/2(1)\nx = [-i +or- sqrt(-1 + 4)/2\nx = (-i +or- sqrt -3)/2\nx = (-i +or- i sqrt3)/2\nTherefore, the other roots are\nx = (-i + i sqrt3)/2\nx = (-i - i sqrt3)/2\n\n^_^
1057648
You always can bring a quadratic equation\nax2 + bx + c = 0  in the form x2 + 2px + q = 0\nby dividing with a and p = b/(2a) while q = c/a\n\nx2 + 2px + q = 0\nx2 + 2px + p2 + q = p2\n(x + p)^2 = p2 - q\nx + p = +V(p2 - q) or x + p =  -V(p2 - q) \nx  = -p +V(p2 - q) or x = -p  - V(p2 - q)\n\nWhen you replace p and q you find the famous abc formula.\nSatisfied?
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
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
670366
Yes.  There are algorithms for factoring polynomials up to 4th degree.  These algorithms are guaranteed to work, however in the case of the 5th degree polynomials and above, there is no algorithm that is guaranteed to find the solution.  In part this is because the roots can be irrational numbers.  I was proved by Abel in 1824 that no such algorithm can even exist.\n\nThere are iterative methods to factor polynomials of 5th degree and above, but they provide approximations, they are not guaranteed to find the correct answer.  The iterative methods are quite complicated and are equivalent to finding eigenvalues of matrices.\n\nIf you are in high school or early college, all the problems they will assign you should be solvable with either the linear, quadratic, or cubic equations.
532215
Your question is to factor in the following form y=a(x-p)^2+q for the question y=x^2-8x+8. Okay, let's roll.\n\ny=x^2-8x+8\n = (x-4)^2 -16 +8   [using (a-b)^2 = a^2 -2ab +b^2]\n = (x-4)^2 -8\n\nTherefore, in the form  y=a(x-p)^2+q, a=1, p=4 and q= -8.