What is the quadratic equation Alice in Wonderland?
545190
The part where she meets the Mad Hatter
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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.
778230
y = ax^2 + bx + c, a not equal to 0\n\nvertex = [(-b/2a), f(-b/2a)]\n\n-b/2a = -4/2 = -2\n\n-2^2 + (4 x -2) + 5 = 4 - 2 + 5 = 7\n\nthe vertex of the graph of the equation is -2, 7\n\nTherefore the axis of symmetry is x = -2
857762
Not sure if it's the shortest or nicest way, but I was able to get to a solution using substitution u=x^2, and v=y^2, then adding 1/4 to the v side to make it a quadratic equation, solving for v, then re-substituting, and solving for y.\n\nI got y = sqrt ( sqrt (x^2-x^4+1/4) + 1/2 )\n\nYou should be able to plot that ...
439883
Use the following four steps to solve a quadratic equation, in this order:\n\nStep 1: Guess the answer\n\nIf x^2=0, can you guess the answer?\nIf x^2=1, you can guess, but remember that there may be two answers. (1 and -1)\n\nYou can tell if you are guessing right by plugging back into the original equation. If it worked, you're done.\n\nIf you can't guess in 30 seconds, you are done with Step 1. Go to Step 2.\n\n\nStep 2: Try factoring\n\nx^2 + 4x + 4 = (x+2)(x+2)\n\nRemember that the constant term of your factors has to multiply to the constant term of the quadratic. \n\nFor example, if you have x squared plus something plus 6, remember that your constant terms must be multiples of 6. \n\nTry:\n\n(x + 1)(x + 6)\n(x - 1)(x - 6)\n(x + 2)(x + 3)\n(x - 2)(x - 3)\n\nYou just ran out of factors of 6. If this didn't work, go to step 3.\n\n\nStep 3 -- Complete the Square\n\nOne important thing to remember about completing the square is that it does exactly the same thing as the quadratic equation. Pick either Step 3 or Step 4 as your last step. Guessing and factoring don't always work because you can't always guess and you can't always factor. But completing the square and using the quadratic formula will always find the answer if there is one.\n\nSuppose you have \n\nx^2 - 6x - 16 = 0\n\nHere's how to complete the square:\n\n#1. Move the constant (whatever it is) to the right side\n\nx^2 - 6x = 16\n\n#2. Find half of the coeficient of the x term and add the square of that number to both sides of the equation:\n\nThe x term here is -6x\nThe coefficient is -6\nHalf of -6 is -3 (Remember this number -3)\n-3 sqared is 9. Add 9 to both sides of the equation:\n\nx^2 - 6x + 9 = 16 + 9\n\n#3. If you did #2 correctly, then the left side is a perfect square and the constant term is -3 (remember -3?)\n\n(x - 3)^2 = 25\n\n(If you're not sure, multiply it out and see if you get what you had before.)\n\n#4. Take the square root of both sides. Remember, it could be negative or positive.\n\nx - 3 = 5\nx - 3 = -5\n\n#5. Solve your two equations\n\nx = 8\nx = -2\n\n#6. Put your answers back into the original equation and see if you got the right answer. \n\n\nStep 4. \n\nIf you did Step 3, you already have the answer, and you're already done. If you didn't do Step 3, use the quadratic formula to get the answer. You can find the quadratic formula here:\n\nhttp://en.wikipedia.org/wiki/Quadratic_formula\n\nMemorize the quadratic formula and never forget it.  Write it on your forhead until you have it memorized.
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the quadratic formula is a useful tool in solving an equation involving a variable and its square without breaking it down into smaller pieces. It can only be used when it is in the\nax^2+bx+c=0  format. (a, b, and c can be negative.)\n\nIn this problem, a=2, b=-2, and c=-1. The equation for the quadratic is:\n\nx= (-b+/- sq.rt (b^2-4ac))/2a\n\nSolving for the equation given, you get:\n\nx= (2 +/- sq.rt.(4-4*2*-1))/4\n   = (2+/- sq.rt 12)/4\n   = 1.366, 0.366
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if it x - 4x/(x+1) = 4/(x+1)\nthen x(x+1) - 4x = 4\nor x^2-3x + 4 = 0\n\nthis is quadratic and so 2 solutions so Megan is right\n\nyou do not have tro evaluate
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Parametric equations are as follow:\n\nx = X1 + A1t\ny = Y1 + A2t\nz = Z1 + A3t
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the sides are 9 and 12 units long.  Here is the algebra:\narea: a=xy=108\nperimiter:p=2x+2y=42\n                     -2y   -2y\n               =(2x=42-2y)/2\n               =x=(-y+21)\nsubstitute that back into the area formula:\n(-y+21)(y)=108\n0=-y^2+21y-108\nuse the quadratic formulat to get the roots 9,12
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1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
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you can take 9v^2 out and then it becomes a simple quadratic equation\ni can give u d ans \nit will be\n\n9v^2(v+2)(v+1)\ni hope u understood
443949
Use quadratic formula and do your own homework.
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a^2 + 321 a - 51 = ( a - 0.158799954)*(a + 321.158799954)\nUsing the general law for solving a quadratic equation.
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if it is a quadratic (ax^2 + bx + c) then put it into the quadratic equation. that automatically will give you 0, 1, or 2 real answers. the equation is: x = [-b +/- (square root of b^2 - 4*a*c)] / 2a\n\nif (b^2 - 4*a*c) is negative, you have no real solutions. if it is zero, you have one real solution. if it is positive, you have 2 real solutions.\n\nthen, if for example you get  x=4 and x= -2, your factors are (x-4) and (x+2)
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8(16) = 128\n\nx^29squared) + 10x\n\nmove all terms to same side x^2 + 10x -128 = 0\n\nuse quadratic equation: -b +/- (plus or minus) the square root of b^2 - 4ac.\n\nThis would make x equal to 7.3693168.... it is an irrational number- i had to take the square root of 612. but if you round x is equal to about 7 or -7
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the quadratic formula is a derivation of the equation \n\nax^2+bx+c=0 \n\nit is: (-b+/-sqrt(b^2-4ac))/2a\n\nit gets the answer out of any and all quadratic trinomials equal to zero...\n\nhere is a site that shows the derivation process:\nhttp://www.purplemath.com/modules/sqrquad.htm#formula
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Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.