Interpretation of Quadratic Equation?
954948
a) Algebraically the x-coordinate of the vertex is x= -b/(2a)\n\nHere that is x = -10/(2 X 1) = -10/2 = -5\n\nPlug that in for x to get the y-coordinate of the vertex.\n\nb) To get the x-intercepts, replace y (here, g(x)) with 0 and solve.\n\n0 = x^2 + 10x + 21 = (x+7)(x+3)\n\nx = -7 or -3 so there are two x-intercepts: -7 and -3.\n\nc) to find the y-intercepts, replace x with zero. \nSo y = 0 + 0 + 21 = 21  \n21 is the y-intercept.
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^_^
1053962
I am assuming this is an "I hate math so I'm gonna complain about its lack of relevance" question and not a real question, but I'll answer it anyway.  I can't tell you every use of it, I suppose I've seen several, but here's one I see a lot.  If you have an object thrown in a trajectory, and you need to know when it will cross a certain height, or when it will be going a certain speed (I dunno, in case you're lobbing hand grenades or missiles someday when you're bored), the quadratic equation will come up in the solution, and it will give you BOTH of the answers, one for when the object is on the way up and one for when it's on the way down.
294522
COS_STUDYGROUP is COMPLETELY WRONG.  (2-7x)^2 == 4+49x^2, please don't get confused about that!\n\nWhile the first user above me is completely correct. I'd like to go a step further and explain how he arrived at that solution.  See that number in front of the x squared?  It should look familiar, if you take the square root of it you get an even 7.  That is how he got to 2 7x's.  The 4 when square rooted becomes a 2.  So now we have a 2 and a 7x in each set of parentheses.  Since the x squared is negative that means that the signs must mismatch so one + and one -.\n\nSo we arrive at the equation (-7x + 2)(7x + 2) or (2 - 7x)(2 + 7x) .... I was always taught the number with the x goes first.
54215
Your equation is a quadratic equation then it will have domain (-inf, +inf).\n\nLet f(x) = x^2 + 4x +1 \n\nThen f ‘(x) = 2x +4                {F ‘(x) = df(x)/dx \n\nPut f ‘(x) =0, U will find x = -2\n\nPut x = -2 in f(x), u will get f (-2) = -3\n\nNow, f “(x) = 2 i.e. f “(x) >0 for all x belongs to (-inf, +inf) {f “(x) = d^2 f(x) / dx^2, second derivative \n\nThen f(x) has a minima at x= -2 and the minimum value of f(x) on the domain (-inf, +inf) will be f (-2) = -3\n\n Let g(x) = ax + bx + c, compare f(x) with g(x), u will find a = 1, b= 4, c= 1 \n\nIf a >0 then parabola opens in the +x direction and if a< 0 then it opens in –x direction.\n\nIn our case a =1 >0 then the parabola f(x) = x^2 + 4x + 1 opens in +x direction. \n\nAs it’s minimum value is f(-2) = -3 then it is clear that the range of f(x) will belongs to [-3, +inf ).\nOther question can be done by the same way.
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
693908
Logs are just equivalent to power statements. \nlog_a (b) = c means a^c = b.\nSo that equation just means x^2 = 2x-3.\nNow thats just a quadratic;\nx^2 - 2x + 3 = 0\nHowever that has no real roots. So there are no such values of x. But hopefully you get the idea.
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 :)
737332
It's unclear what that equation says.
162293
On both equations, you have x^ but don't say what x was raised to. I'm assuming that it is x^2. Is this correct?
1072562
I'm reading the equation as \n\n4x^2 + 8x -1 = 0\n\nUsing quadratic formula x = (-8 +/- sqrt(64+16))/8 \n\nso x = -18/8 or 2/8 = -9/4 or 1/4\n\nThose are the roots.\n\nSubstitute each of those values as y into the second formula.
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.
373728
Some folks seem to have forgotten this is a quadratic, and they have two (2) roots!\n\n(x - 7)^2 = 64\n\nx - 7 = ± 8 (take the square root of both sides)\n\nso \n\nx = 7 + 8 = 15 \n\nor \n\nx = 7 - 8 = -1\n\nyou could expand this out and use the quadratic equation to solve it as well:\n\nx^2 - 14x + 49 = 64\n\nx^2 -14x - 15 = 0  a = 1 b = -14  c = -15\n\nthe rest is left to the student as an exercise. :^)
545305
Re-writing the equation: 9*Q^2-8*Q+(50-P)=0. Comparing with standard form of quadratic: A*X^2+B*X+C=0, A=9, B=-8, C=(50-P). Sridharacharya's formula for solving the standard equation is \nX= {-B+sqrt(B^2-4*A*C)}/(2*A) or {-B-sqrt(B^2-4*A*C)}/(2*A).\nWith this formula 'Q' can be expressed as a function of 'P' as:\n     Q = [2{2+sqrt(9*P-446)}]/9 or [2{2-sqrt(9*P-446)}]/9.
1172065
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)
913819
The "Hindu Solution" to the quadratic equation is the same method that we now call "completing the square."  This is a way to convert terms in the equation to the form of the square of a different polynomial, enabling it to be solved algebraically.  When applied to a quadratic equation in the general form, this method yields the quadratic formula.
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