using the quadratic equation x^2-4x-5=0 solve by factoring?
399509
You are looking for factor of -5 such that when you add them together you get -4\nFactors of -5"\n-5 and 1 or 5 and -1\n-5+1=-4 and 5-1=4\nThus, we are looking for the -5 and 1\nx^2 - 4x - 5 = 0\n(x-5)(x+1) = 0\nwhich implies x-5 = 0 or x+1 = 0 which gives you\nx = 5 or x = -1\nCheck your answer by plugging these values back into the original equation.
301645
If you have an equation like 5x^2 + 7x - 6 = 0:\n\nGet the x^2 term by itself\nx^2 + 7x/5 - 6/5 = 0\n\nMove the constant to the other side\nx^2 + 7x/5 = 6/5\n\nTake half of the coefficient for the x term and square it: half of 7/5 is 7/10, squared is 49/100. Add this to both sides of the equation\n\nx^2 + 7x/5 + 49/100 = 6/5 + 49/100\nx^2 + 7x/5 + 49/100 = 169/100\n\nThe left side is a perfect square trinomial - factor it\n(x + 7/10)^2 = 169/100\n\nTake the square root of both sides and solve for x\nx + 7/10 = +/- 13/10\nx = -7/10 +/- 13/10\nx = 6/10 or -20/10\nx = 0.6 or -2
528350
(5x + 4)(6x + 4)\n30x^2 + 20x + 24x + 16\n30x^2 + 44x + 16\nANS : 16
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
952526
x^2-5x-2=0\n\nuse the quadratic formula\n\n[5  +/- sqrt(25-4*1*-2)]/2\n\n[5  +/-   sqrt(25+ 8)]/2\n\nI distributed the 1/2...easier to type\n\n5/2  +/-sqrt33/2
676772
If you must factor, this is how it is done...\n\n2x^4 - 16x^2 = 0\nx^4 - 8x^2 = 0\nx^2(x^2 - 8) = 0\nx^2 = 0 and x^2 = 8\nx = 0 and x = +/-sqrt(8)\n\nThere...
842037
We need to factor the equation down. This involves finding the roots of the equation. This will look like:\n\n(ax+b)(cx+d) = 12x^2-31x+20 = 0\n\nBecause the final term is positive and the middle term is negative, my first hunch is to have b = -4 and d = -5. By doing so, the equation looks like:\n\n(ax-4)(cx-5) = 12x^2-31x+20 = 0\n\nThe factors needed for 12x^2 could be 2 and 6 or 3 and 4. We should test each combination to see that they work out. The factoring ends up looking like this:\n\n(3x-4)(4x-5) = 0\n\nNow all that remains is to solve for x in each set of parentheses. If the first term is zero, then\n\n3x-4 = 0\nx = 4/3\n\nIf the second is zero, then\n\n4x-5 = 0\nx = 5/4\n\nThus, x = 5/4, 4/3
208014
Factorising gives\n\n(x-4)(x+1) = 0\n\nThe solutions are therefore x = +4 and -1
766816
This problem is best solved in a manner identical to how all other polynomials are solved. So x^2=x^3? This is the same as x^3-x^2=0, so factoring this yields x times x times (x-1)=0 so the only possible solutions are x=0 and x=1.
775457
multiplying by x\nx^2-5=4x\nx^2-4x-5=0\n(x-5)(x+1)=0\nx=5 or -1
1039684
for x^2+5x-9=0  u can't do this the same way.9 * 1 = 9 but 9 + or - 1 not = 5 and 3 * 3 = 9 but 3 + or - 3 not = 5.\nso there is a formula u got to use here.\nit is  x = {- b + - sqrt(b^2 - 4ac)} / { 2a}\nso from the given quadratic equation,\na = 1,b = 5 and c = 9\nand then u substitute , u will get 2 values of x and that's the answer.remeber we have to use this formula if we are unable to factorise the way u did for the previous one.
259003
Use the quadratic formula.  It works everytime!\nx = (-b plus or minus sprt(b^2 - 4ac))/2\nwhere\nax^2 + bx + c = 0\nSome of these will factor, however not all of them will.
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
796585
1.factor\n(2x+1)(x-4)=0\n2.set it to 0\n2x+1=0\nor\nx-4=0\n3.solve\n{x=-1/2 or x=4} <=answer
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.
815583
not sure what your question is, but here's a quick n' dirty to solving simple algebraic equations:\n\nwe'll use this simple equation 5 - x = 4x\n\n\nstep 1: get all your variables on one side of the = and constants on the other... think of the = as a balance point on a balance scale...  you can do anything to one side, and as long as you do it to the other, it will stay balanced!\n\nso to get the x from 5 - x to the other side, we need to add an x to both sides:  5 - x + x = 4x + x, 5 - 0 = 5x, 5 = 5x\n\nstep 2: get the variable down to one multiplier\n\nwe know what 5x is, but we want to know x... so how do we find out?  Well, since we know what 5 times x is, we can divide this by 5 to get x.  But, we want to stay "balanced" so we have to divide both sides by 5:\n\n5x / 5 = 5 / 5\nx = 1\n\nVoila!  Solved :)\n\nHope this helps.
670824
There are 2 ways to solve for this \n\n1. Separating the x^2\n\n2. Quadratic Formula\n   ax^2 + bx + c = 0\n x = (-b + or - sqrt(b^2 - 4ac))/2a\n http://www.1728.com/quadrtc.gif\n\n\nSolving it by separating x^2\n\na. First separate x^2\n(x + ?)(x +?)\n\nb. Now we have to figure out the 2 variables that would give us 4x and -12 when we multiply them out. notice, that 4 is positive so the larger number has to be Positive.\n\n-12 = 12 * -1   12x + -1x = 11x\n-12 = 6 * -2    6x + -2x = 4x\n-12 = 4 * -3    4x + -3x = x\n\ntherefore the 2 variables would be 6 and -2\n\n(x + 6)(x - 2) = 0\nSo x, would be either -6, or 2\n---------------------------------------------  \n2. SOLVING USING THE QUADRATIC FORMULA\nThis one is way easier\n\nax^2 + bx + c = 0\nx^2 + 4x - 12 = 0\n\n x = (-b + or - sqrt(b^2 - 4ac))/2a\n x = (-4 + or - sqrt(4^2 - 4*1*-12))/2*1\n\n x = (-4 + or - 8)/2\n x = (-4 + 8)/2 = 2\n\n x - (-4 - 8)/2 = -6