Use the quadratic formula to solve each of the following quadratic equations.?
952497
Ok, I assume you know the quadratic equation, but just for convenience I'm going to write it out:  x = (-b +/- root(b^2 - 4ac))/2a.  \n\n1) x = (5 +/- root(25 + 24))/4.  (Remember, you have to subtract the 3 from each side to get the equation to equal 0, so c is -3.)  That simplifies to x = (5 +/- 7)/4--->x = 12/4 or -2/4--->x = 3 or -1/2.\n\n2) x = (2 +/- root(4 - 12))/6.  Are you sure that wasn't supposed to be a negative 3x^2 or minus 1?  If it were, the term inside the square root would be 16, a positive perfect square.  As it is, it's -8, whose square root is 2root(2)i.  That seems pretty complex (no pun intended) compared to the other one.  \n\nI'm going to solve that one assuming that it should have ended in -1 rather than 1.  Then x = (2 +/- root(4 + 12))/6--->x = (2 +/- 4)/6--->x = -2/6 or 6/6--->x = -1/3 or 1.
1141591
1. 2\n\n2. 2a^2b\n\n3. 6xyz^2(2z-3y)\n\n4. 3a(4a-b+3ab^2)\n\n5.(d+2)(d+3)\n\n6.(x-9)(x-2)\n\n7. (4x-1)(x+3)\n\n8. (4b-9)(4b+9)\n\n9.(y+7)^2\n\n10. (2m+7n)^2
530596
Given\nax^2 + bx + c = 0\nDivide all terms by a \nx^2 + (b / a) x + c / a = 0 \nSubtract c / a from both sides \nx^2 + (b / a) x + c / a - c / a = - c / a \nand simplify \nx^2 + (b / a) x = - c / a \nAdd (b / 2a)^2 to both sides \nx^2 + (b / a) x + (b / 2a)^2 = - c / a + (b / 2a)^2 \nto complete the square \n[ x + (b / 2a) ]^2 = - c / a + (b / 2a)^2 \nGroup the two terms on the right side of the equation \n[ x + (b / a) ]^2 = [ b^2 - 4a c ] / ( 4a^2 ) \nSolve by taking the square root \nx + (b / a) = ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nSolve for x to obtain two solutions \nx = - b / a ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nThe term sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } may be written \nsqrt { [ b^2 - 4a c ] / ( 4a^2 ) } = sqrt(b^2 - 4a c) / 2 | a | \nSince 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written \nx = [ -b + sqrt( b^2 - 4a c ) ] / 2 a \nx = [ -b - sqrt( b^2 - 4a c ) ] / 2 a \nThe term b^2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible \ncase 1: If b 2 - 4a c > 0 , the equation has 2 solutions. \ncase 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. \ncase 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions\nps. I did this as quick as possible, hope everything copied correctly.
883463
Quadratic formula:\nX=(-b+Sqrt(b^2-4ac))/2a\nOR\n(-b-Sqrt(b^2-4ac))/2a\n\nIn case you don't know what a,b,&c stand for, they are the coefficients of the quadratic equation (besides c.)... OK, that was kind of condusing, so I'll just write out the quadratic equation, using a,b,&c where the numbers should be:\nax^2-bx+c\n\nIn your case the equation is:\nx^2 -2x +1=5, which is equal to:\n\n(1)x^6 + (-2)x + (-4)= 0, \n(the correct form for a quadratic equation)\nTherefore, a=1,\n                  b=-2\n                  c=-4\nTherefore,\nX=(-b+Sqrt(b^2-4ac))/2a\nOR\n(-b-Sqrt(b^2-4ac))/2a (The Quadratic Equation)\nbecomes:\n\nX=(2+sqrt(4+16))/2 = (2+sqrt(20))/2 = (2+2sqrt(5))/2 = 1+sqrt(5)\nOR \n1-sqrt(5)
972476
........._______________\nx - 5 ) x³ - 3x² - 13x + 15 ( x²\n......... x³ - 5x²\n......... -------- ....↓\n............... 2x² - 13x ....... ( 2x\n............... 2x² - 10x\n............... ----------- ....↓ \n............... ....... -3x + 15. (-3\n............... ....... -3x + 15\n............... ....... -----------\n............... ....... -----------\n\nSo P(x) = x³ - 3x² - 13x + 15\n= (x - 5)(x² + 2x - 3)
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?
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.
541499
2x(x-5)-3(2x+1)\n2x^2-10x-6x-3\n2x^2-16x-3
534396
quadratic equations of the form \n\nax^2+bx+c=0\n\ncan be used with the quadratic formula, see\n\nhttp://mathworld.wolfram.com/QuadraticFormula.html\n\nso just transform all your equations into the form\n\nax^2+bx+c = 0 (x can be called differently, of course).\n\nEdit: for example, let's take \n\n5t^2 + 20t - 60 = 0\n\naccording to the quadratic formula, the solutions are\n\nt_1 = (-20+sqrt(20*20-4*5*(-60))) / (2*5) = (-20+sqrt(1600))/10 = 2\n\nand\n\nt_2 = (-20-sqrt(20*20-4*5*(-60))) / (2*5) = (-20-sqrt(1600))/10 = -6\n\nNote: if the term below the root (sqrt) is negative, it is not a valid solution - sometimes the quadratic equation has only one solution, or none.
1088212
if solving for  t\nmutipying both sides by t gives \nct = -2t^2 + 4 which is equal to 2t^2 + ct - 4 = 0\nusing the quadratic formula we get t = (-c +/- sqrt(c^2 + 32)) / 4
778763
1) (a - b)(x - y)\n2) (x^9 + y^3)(x^9 - y^3)\n3) (6x + 5)(x + 8)\n4) (x - 9)(x - 7)
651993
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest
879937
ax^2 + bx+c = 0\nax^2 + bx = -c\nx^2+(b/a)x+ ... = -c/a + ...\nx^2+(b/a)x+ (b/2a)^2 = -c/a + (b/2a)^2\n{x - (b/2a)}^2 = -4ac/4(a^2) + b^2/4(a^2)\nx - (b/2a)}^2 = + or - root{(-4ac + b^2)/4(a^2)}\nx= {-b + or - sqrt (b^2-4ac)}/2a\n\nTh
870960
Do you mean that the roots of x^2+pX+r are α  and β and you need to find quadratics in terms of p and q?\nIf so you can use the following:\nIn any quadratic ax^2 + bx + c = 0, the sum of the roots = -b/a and the product of the roots = c/a.\nUsing this,  -(α + β) = p, and αβ= r.\n\nAnswers are for terms b and c in the quadratic x^2 + bx + c\ni)\nb = -(1/α + 1/β) = -(α + β)/αβ = p/r\nc = 1/αβ = 1/r\n\nii) \nb = -(1/α^2 + 1/β^2) =  -(α^2 + β^2)/(αβ)^2 = -[(α + β)^2 - 2αβ])/(αβ)^2 = -(p^2 - 2r)/r^2\nc = 1/(αβ)^2 = 1/r^2\n\niii) \nb = (α^2β + αβ^2) = αβ(α + β) = -pr\nc = -α^2β * -αβ^2 = (αβ)^3 = r^3
82007
Just type in quadratic equation and maybe roots in a serch engine. You will have your pick of dozens of sites.
681236
Multiply each term in the first () by each term in the second ():\n\n2x times 2x = 4x^2\n2x times 1 = 2x\n-3 times 2x = -6x\n-3 times 1 = -3\n\nThen combine 2x and -6x\nAnswer: 4x^2 - 4x - 3
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