Solve the quadratic equation 3x2+6x+3=0 by factoring?
919554
Factor out 3.\n3(x^2+2x+1)=0\n3(x+1)^2=0\nThe only answer then is x=-1 to make the expression equal 0.
825419
x^3 - 6x^2 - 3x + 18 \n= x^2 (x -6) - 3(x-6) \n= (x^2 - 3)(x - 6)
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
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
959079
For the y-intercept, set x to zero and solve:\n\nf(0)=-3(0)^2+6(0)-2\nf(0)=-2\n\nFor the x-intercepts, use the quadratic formula since this equation can't be factored. The x-intercepts come out to be 0.42264973 and 1.5773503
607601
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
727026
The equation to start with is : x^3 = 27\nor, x^3 = 3^3\nor, x^3 - 3^3 = 0\n\nThis is a difference of two cubes,\nso it can be factored :\n\n(x - 3)(x^2 + 3x + 9) = 0\n\nOne solution is seen to be x = 3.\n\nThe other solutions are obtained by\napplying the quadratic formula to :\n\nx^2 + 3x + 9 = 0\n\nThis gives :\n\nx = {-3 ± sqrt[3^2 - 4(1)(9)]} / (2 * 1)\n\n= [-3 ± sqrt(-27)] / 2\n\n= [-3 ± i*3*sqrt(3)] / 2\n\n= (-3 / 2) ± [3*sqrt(3) / 2] * i
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~ ♥ ~
731670
(-4x3 + 3x2 - 5) + (x2 - 5x + 6) =\n-4x3 + 3x2 - 5 + x2 - 5x + 6 =\n-4x3 + 4x2 - 5x + 1
734679
5a^2 + 3a - 2 = 0\n\ncan be factored as (5a - 2) (a + 1) = 0\nThus a = 2/5 and a = -1
846617
You would set your two equations equal to each other to solve:\n\ny = 2x-6\ny = x² + 5x + 6\n\nSo, 2x-6 = x² + 5x + 6\n\nSubtract 2x from both sides:\n\n-6 = x² + 3x + 6\n\nAdd 6 to both sides:\n\n0 = x² + 3x + 12\n\nNow you can factor or use quadratic formula to solve for x.  (I don't see any way that this can be factored, unfortunately, so we'll have to use the ol' quadratic formula.)\n\nx = [ -3 ± √ (3*3 - 4*1*12) ] ÷ 2*1 = \n\n[ -3 ± √ (-39) ] ÷ 2\n\nThis equation has complex roots, but no real roots.  Therefore in your real plane these two graphs do not intersect.\n\n.
254834
if you use quadratic formula you get the factors of \n\nx= ( ( 3 +/- square rt.(5) ) / -6 )
695176
6x^2-3x-4x+2\n3x(2x-1)-2(2x-1)\n(2x-1)(3x-2)
36762
Factor by grouping:\nx^2 (3x - 5) - 3 (3x - 5)\nAnswer: (3x - 5)(x^2 - 3)
945902
We have the equation\nx² - 6x + c = 0\n\nNow, we let p and q be the roots of the equation.  If we are given the quadratic equation ax² + bx + c = 0, then the sum of the roots is -b/a.  Thus,\np + q = -b/a = -(-6)/(1) = 6\np + q = 6\n\nNow, since they differ by 2√3, then\np - q = 2√3\n\nThus, we now have 2 equations:\np + q = 6\np - q = 2√3\n\nWe now add them\n2p = 6 + 2√3\n\nThus,\np = 3 + √3\n\nWe can solve for q:\np + q = 6\nq = 6 - p\nq = 6 - (3 + √3)\nq = 3 - √3\n\nTherefore, the roots of the equation are:\n3 + √3 and 3 - √3\n\nNow, given the quadratic ax² + bx + c = 0, the product of the roots is c/a.  Thus,\npq = c/a = c/1 = c\n\nNow,\npq = c\n\nWe can substitute the value of p and q\n(3 + √3)(3 - √3) = c\n\nthus, using the special multiplication (a + b)(a - b) = a² - b²\nc = 3² - √3²\n\nThus,\nc = 9 - 3\n\ntherefore,\nc = 6\n\nNow, here are the answers to your question:\na) What are the roots of the equation?\n3 + √3 and 3 - √3\nB) what is the value of c?\nc = 6\n\nHope this helped!\n\n^_^
905417
By dividing the equation by 2:\n\nx^2 + 3x +1 = 0\n\nThe quadratic formula is = [-b +/- sqrt(b^2 - 4*a*c)]/2a\n\nWhere a =1;     b = 3;       c = 1.\n\nNow, we plug in this values in the quadratic formula, and we'll obtain a value of x, using the minus sign, and other, using the plus sign, as follows:\n\nx1= [ - 3 + sqrt( 3^2  - 4*1*1)]/(2*1)\nx1= [ - 3 + sqrt(9 - 4)]/2\nx1= [ - 3 + sqrt(5)]/2\nx1= [ -3 + 2.236]/2\nx1= -0.382\n\nx2= [ - 3 - sqrt( 3^2  - 4*1*1)]/(2*1)\nx2= [ - 3 - sqrt(9 - 4)]/2\nx2= [ - 3 - sqrt(5)]/2\nx2= [ -3 - 2.236]/2\nx2= -2.618
494989
It seems that many people here don't realize that your equation is 3x^2 - 6x + 9.\n\n3 is the basic factor to get 3*(x^2 - 2x + 3)\n\nIf you are allowed to use imaginary numbers (where i = sqrt(-1)) then,\n\n3*[x - 1 + i*sqrt(2)][x - 1 - i*sqrt(2)]