How do you use quadratic equations in the real world?
826926
playing a simple game of catch invovles the quadratic equation. Im positive nasa uses the quadratic equation for determining flight angles and all that stuff.
825052
The first step would be to try to factor the equation, so you'd want to get all of the terms on the left side of the equation:\nx^2 + 4x -3 = 0\n\nSince this won't easily factor, you need to use a different method.  I'm sure someone will suggest the quadratic equation, which is fine _if you can remember it_.  I much prefer an alternate method, "completing the square," because it's intuitive enough that you can remember how to work out the process even if you forget the steps.\n\n1. Start by moving the constant to the right side of the equation:\nx^2 + 4x = 3\n\n2. Then divide the coefficient of x (in this case, 4) by two and square it, adding it to both sides of the equation:\nx^2 + 4x + 4 = 7\n\n3. Factor the left side of the equation to (x + a)^2, where 'a' will be the number you just squared:\n(x + 2)^2 = 7\n\n4. Take the square root of both sides, remembering that you'll have both a positive and a negative root:\nx + 2 = +/- (sqrt)7\n\n5. Move the constant on the left to the right, and you have your two solutions:\nx = -2 +/- (sqrt)7
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
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 :)
282293
The techniques used to solve systems of linear equations (substitution and elimination) tend to do a pretty good job of solving quadratic systems.\n\nThis page does a pretty good job of demonstrating how:
291823
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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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.
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 -.
264348
The best way to start is to create a function based on the equation. Do this by setting the equation equal to 0. For instance, the first step to solving the equation x^3=5 would be to subtract 5 from both sides to get x^3-5=0. This creates the function f(x)=x^3-5.\n\nNow, look at the graph of the function. Are there any obvious places it appears to cross the x-axis? If the function appears to cross at x = 2 then perform synthetic division, dividing the cubic equation by x-2 (in other words, put a +2 in the synthetic division "box"). If you get a remainder, then it wasn't a root. If there's no remainder, you've factored out the root, and you'll get (x-2)(ax^2+bx+c), where a, b, and c are the numbers that come out of the synthetic division you just completed.\n\nNow you can factor ax^2+bx+c (if it's factorable) or use the quadratic formula (if it's not) to get the other two solutions. There will be 3, even if some of them are imaginary (have negatives inside radicals).\n\nIf none of the x-intercepts are rational (none of the synthetic division worked), pick an x-value close to each x-intercept on the graph and apply Newton's method to each one to estimate each root as accurate as you'd like.\n\nClick on the link below to see how I apply Newton's Method to solve f(x) = x^3-5. It's a big graphic, and your browser may smoosh it to make it fit in the screen, which makes it unreadable. Usually, clicking on the image or holding your mouse over the image until a resize box appears in the lower right corner will let you expand it to the real size. Either that or save it to your computer and double-click it there.\n\nBest of luck.
708644
Use the quadratic formula.\nx1 = 1\nx2 = -1/2
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Eliminate the y to get\n2-x^2=2x+4\nx^2+2x+2=0\nusing quadratic formula\n(-2+-sqrt(4-4*2))/2\nwe see there are no real roots: the two functions do not intersect\nx= -1+-i
756992
Don't you just love the quadratic equation?\n\nThe first thing you got to do is get the given equation into the correct form, and in order to do that you have to combine all like terms. Since there are no like terms yet you have to do some math first.\n\n-2.0^2 = 4.0because neg * neg = pos\n\nNow you have 2 like terms 4.0 and 0.5 combine to get 4.5. Your eqution should look like this:\n\nf(t)=10t + 4.5 since you need at^2 + bt^2 + c just insert t^2. Now your equation will look like this:\n\nf(t)=t^2 + 10t + 4.5\n\nAnd just in case you need the quadratic equation:\n\nt= -b + or - sqr(b^2 - 4ac) / 2a\n\noh yeah and a = 1
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for x^2+5x-9=0  u can't do this the same way.9 * 1 = 9 but 9 + or - 1 not = 5 and 3 * 3 = 9 but 3 + or - 3 not = 5.\nso there is a formula u got to use here.\nit is  x = {- b + - sqrt(b^2 - 4ac)} / { 2a}\nso from the given quadratic equation,\na = 1,b = 5 and c = 9\nand then u substitute , u will get 2 values of x and that's the answer.remeber we have to use this formula if we are unable to factorise the way u did for the previous one.
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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~ ♥ ~
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On a graph with x and y-axes , if you plot out the values of y calculated from an equation involving x squared you draw out a parabola.  A quadratic equation is one in the form y = a.x-squared + b.x + c \n a can be positive or negative (not zero or you don't involve x-squared). If negative, the parabola is upside down (open end downward).\nb and c can be positive negative or zero.\nWhen you are asked to "solve" a quadratic, you are finding where the parabola cuts the x-axis, that is to say the values of x where the equation calculates out to y = 0
285238
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
705433
Derivation of the quadratic formula by completing the square.\n\nax^2 + bx + c = 0\nx^2 + (b/a)x + (c/a) = 0\nx^2 + (b/a)x = -c/a\nx^2 + (b/a)x + b^2/(4a^2) =  b^2/(4a^2) - c/a \n(x + b/2a)^2 =  b^2/(4a^2) - 4ac/(4a^2) \nx + b/2a = ±√[ b^2/(4a^2) - 4ac/(4a^2) ]\n\nx = -b/2a ± √[ b^2/(4a^2) - 4ac/(4a^2) ]\n= -b/2a ± √[ b^2- 4ac] / 2a\n\nx = {-b ± √[ b^2- 4ac] } / 2a