MATH!! HELP!!  Quadratic function and the rule!!?
913768
y = ax² + bx + c\n\nWhen x = 2 y = 30 so 4a + 2b + c = 30 ...... (1)\nWhen x = 3 y = 15 so 9a + 3b + c = 15 ...... (2)\nWhen x = 4 y = 6 so 16a + 4b + c = 6 ........ (3)\n\n(2) - (1) 5a + b = -15 ....... (4)\n(3) - (2) 7a + b = -9 ......... (5)\n(5) - (4) 2a = 6\na = 3 so b = - 30 and c = 78\n\nie y = 3x² - 30x + 78
925669
You need to use the quadratic equation...\n\nhttp://mathworld.wolfram.com/QuadraticEquation.html\n\nOR\n\nx (x - 5) = 0
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
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.
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.
783117
Without getting into the graphing explanation, the simplest reason I have is that you can find X using the quadratic equation if the equation is in the form ax^2 + bx + c = 0. You set the equation equal to zero so it matches the form and then you can use the quadratic equation to find X.
932849
The quadratic formula goes: (-b ± √b² - 4ac)/ 2a\nSo x²=a, -6x=b, -7=c\nNow plug it in:\n [-(-6) ± √(6)² -4(1)(-7)]/ 2(1) \nsimplify: (6± √36 + 28)/ 2\nsimplify further: (6±√64)/2\nwhich equals: (6±8)/2\nand if you separate it out: (6-8)/2 = -1      and   (6+8)/2 = 7
826907
ill write down the formula for finding the root of a quadratic equation :  ax^2 + bx + c = 0\nsolution: x = {-b +- sqrt(b^2 - 4ac) } / 2a\n\nequation x^2=5x + 2 can be written as :\n\nx^2 - 5x - 2 = 0\nhere a = 1 ; b = -5 ; c = -2 ;\nsimply apply the formula to find the value of x .\n\nx ={ -(-5)+-sqrt(25 - 4*-2) } /  2\nx ={ 5 +- sqrt(24+8) } / 2\nx = { 5 +- sqrt(33) } / 2\n\nThe second equation :\n\n4x^2 + 2x - 3 = 0\nhere a = 4 ; b = 2 ; c = -3\n\nx = {-2 +- sqrt( 4 - 4*4*-3) } / 2*4\nx = {-2 +- sqrt(4+48) } / 8\nx = { -2 +- sqrt(52) } /8\n\nsimplify the term inside the square root ,  52 = 13*4\nso 4 gets square rooted and\n\nx = { -2 +- 2 sqrt(13) } /8\n\ntake 2 common from the numerator and denominator and cancel them.\n\nx = 2{ -1 +- sqrt(13) } / 4*2\nx = { -1 +- sqrt(13) } / 4\n\nthats your answer.\n\nhave a nice day :)
817524
For a quadratic equation, the maximum or minimum will be the vertex. Since the first term of the quadratic is positive, the parabola goes up, so you're looking for the minimum.\n\nThere are 2 methods:\nCalculus (using derivatives):\nf(x) = x2 - 2x - 4\nf'(x) = 2x - 2\nthe maximum/minimum will be where f'(x) = 0.\n2x - 2 = 0\nx = 1\ny = 1^2 - 2*1 - 4 = -5\n\nAlgebra:\nThe vertex can be found using the formula:\n(-b/2a , -(b^2 - 4ac)/4a)\n(-(-2)/2, -(4 - (4*1*-4))/4)\n(1,-5)\n\nCheck: x = 1, y = 1^2 - 2*1 - 4 = -5
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.
892157
wait 1 sec.\nOk this is reasonably difficult, you have to use the chain rule and the quotient rule.\nUsing the chain rule:\nderivative = [ (1/2) * ( (2x + 3) / (7x+4) ) ^ (-1/2) ]  *  [d/dx (2x+3) / (7x+4) ]\n\nSo now you to use the quotient rule to find:\nd/dx (2x+3) / (7x+4) = (14x+8 - 14x + 21) / (7x+4)^2\n= 29 / (7x+4)^2\n\nSo the derivative is [1/2 * ((2x+3)/(7x+4))^(-1/2)]*[29 / (7x+4)^2]\n\n\nBy the way i got the thumbs down before I wrote all this up, so it doesn't mean it's wrong.
891271
first into the second:\n\nx^2+(x-1)^2=25\n\nexpand:\n\nx^2+x^2-2x+1=25\n\nget 0 on the right side:\n\n2*x^2 - 2*x -24 = 0\n\na=2, b=-2, c=-24\n\nuse the root finding formula for quadratoic equation:\n\nx=(-b+/-sqrt(b*b-4*a*c))/(2*a)=(2+/-sqrt(4+4*2*24))/4=0.5 +/- 3.5\n\nyour solutions are: x = -3 and 4\n\ncheck:\n(-3)^2+(-4)^2 = 9 + 15 = 24, correct\n(4)^2+(3)^2 = 15 + 9 = 24, correct\n\nsubstitute to the first eqn: y = -4 for the 1st. x and 3 for the 2nd.
833237
so far, so good\n\nn!/(n-4)! = 56 [n!/(n-2)!] \n\nmultiply both sides by (n-2)!/n!\n\n[n!(n-2)!] / [n!(n-4)!] = 56\n\n(n-2)! / (n-4)! = 56\n\n(n-2)(n-3) = 56\n\nYou could multiply this out and solve the quadratic, in which case you will get one positive and one negative root, or if you remember that 56 = 8*7, then that quickly leads to n = 10
754733
Let y = u/v   where u=some function and v =some funtion\n\nthen dy/dx = (v*du/dx - u*dv/dx)/ (v^2)     **Quotient Rule**\n\nIn this case u = 6x^2-1 and v = 2*sqrt(x)\n\nSo you just differentiate and plug into the equation for dy/dx
852286
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
922434
There is only one step. Take the quadratic formula and plug in the numbers.\n\nIf ax^2 + bx + c = 0,\nx = (-b +/- sqrt(b^2-4ac))/2a\n\nFor the equation you gave, a=1, b=19, c=88. Plug those numbers in and see what you get.\n\nIf you're confused by the +/-, it's shorthand for writing two things in one. There are usually two values of x that work, one with + and one with -.
390615
For #1, try fully factoring the numerator and the denominator. You will find that some of the terms cancel. Start by factoring x from each one. Your resulting term in the numerator is a difference of squares. Your resulting term in the denominator is a factorable quadratic.\n\nFor #2 the expression will be undefined for any value that makes the denominator equal to zero. Again factoring so you can solve for the zeroes is the way to go.\n\nI don't like to work out problems for people, so I hope this is helpful. Likely enough someone else will give you the answers, but I wanted to make sure you knew the process.