How is the quadratic equation used in real life?
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One way is to predict how a body accelerates.  For example, a rock may be thrown upward with an initial velocity of 150 meters per second from a height of 2 meters above the ground.  When will the rock return to the ground.  From physics we have the equation\n\ny = -4.9x^2 + 150x + 2\n\nBy putting 0 in for y, we can use the quadratic to solve this equation.\n\nThere are other applications for the quadratic, this is perhaps the most often shown real life situation.\n\nBut quite frankly, there are lots of people on this planet who will never use the quadratic formula and will lead very successful lives.\n\nIt all depends on what you plan to do.  If you're going to be an engineer, scientist, or go into any field that requires math, you'll see it again.  \n\nIf you plan on going into a trade or sales, you will most likely never use it in the future.\n\nBut given all that, it's one of the first general formulas that math students learn and are expected to memorize.  The ability to substitute values into a formula is a skill that you will use even after the quadratic formula.
952054
To find the equation, just substitute the 2 roots to the "Quadramagic formula" (...)\nx² - (p + q)x + pq = 0\nwhere p and q are the roots.\n\nHence, p = 7 + i and q = 7 - i.  We substitute:\nx² - [(7 + i) + (7 - i)]x + (7 + i)(7 - i) = 0\n\nWe add and multiply\nx² - (14)x + (7² - i²) = 0\n\nSince i² = -1, then\nx² - 14x + 49 - (-1) = 0\n\nthen,\nx² - 14x + 49 + 1 = 0\n\nTherefore, the quadratic equation is\nx² - 14x + 50 = 0\n\nTherefore, the correct answer is 1).\n\n^_^
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 :)
1039684
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.
1049820
The quadratic formula:\nx= -b(+/-) the square root of (b squared - 4ac), all of this over 2a
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Ran into a little difficulty, I can't find my math editor.\n\nBut here goes:\nx = (-b +,- SQRT(b^2 - 4ac))/2a\na = 3, b = -5, c = 2;\nx = (-(-5) +,- SQRT(25 - 24))/ 6\nx = (5 + 1) / 6; x = (5 - 1) / 6\nx = (1, 2/3)\nThis would have looked a lot better if I had my math editor!!!\n\nClarification: I solved your quadratic equation using the quadratic formula.\n\nMashi below simplified your quadratic equation, which you don't need the quadratic formula to do. Substitute the values of x I obtained by solving into the simplified equation, they will equal 0
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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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.
781653
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
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(-b +/- SQRT(b^2-4*a*c)) / 2a\n\nSo plugging in the values\n\nx = (10 + SQRT(100-4*16))/2\nand\nx = (10 - SQRT(100-4*16))/2\n\nx = 8\nand\nx = 2\n\nYou can find it written better in wikipedia:\n\nhttp://en.wikipedia.org/wiki/Quadratic_function\n\nGood thing we didn't have to get into complex numbers.... ;-)
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The discriminant\n\n√b² - 4ac portion of the quadratic Formula\n\nx = - b ± √b² - 4ac/ 2a is negative , no real number solutions \n\n- - - - - -s-
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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 ...
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
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Note that the function is zero when x=1, this means that you can factor out an (x-1)\n\nx^4-x^2+4x-4 = (x-1)(x^3+x^2+4)\n\nNote that x^3+x^2+4 is zero when x=-2 so you can factor out an (x+2),\n\nx^3+x^2+4 = (x+2)(x^2-x+2)\n\nThe quadratic, x^2-x+2, does not factor with real numbers. (Try out the quadratic equation to confirm.) So the answer is\n\nx^4-x^2+4x-4 = (x-1)(x+2)(x^2-x+2)
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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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A quadratic equation is a second-order polynomial equation in a single variable  \n\n (1) \n\nwith . Because it is a second-order polynomial equation, the fundamental theorem of algebra guarantees that it has two solutions. These solutions may be both real, or both complex. \n\nThe roots  can be found by completing the square, \n\n (2) \n\n (3) \n\n (4) \n\nSolving for  then gives \n\n (5) \n\nThis equation is known as the quadratic formula. \n\nThe first known solution of a quadratic equation is the one given in the Berlin papyrus from the Middle Kingdom (ca. 2160-1700 BC) in Egypt. This problem reduces to solving \n\n   (6) \n   (7) \n\n(Smith 1953, p. 443). The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's  (ca. 325-270 BC) Data contains three problems involving quadratics. In his work Arithmetica, the Greek mathematician Diophantus  (ca. 210-290) solved the quadratic equation, but giving only one root, even when both roots were positive (Smith 1951, p. 134). \n\nA number of Indian mathematicians gave rules equivalent to the quadratic formula. It is possible that certain altar constructions dating from ca. 500 BC represent solutions of the equation, but even should this be the case, there is no record of the method of solution (Smith 1953, p. 444). The Hindu mathematician Aryabhata (475 or 476-550) gave a rule for the sum of a geometric series that shows knowledge of the quadratic equations with both solutions (Smith 1951, p. 159; Smith 1953, p. 444), while Brahmagupta  (ca. 628) appears to have considered only one of them (Smith 1951, p. 159; Smith 1953, pp. 444-445). Similarly, Mahavira (ca. 850) had substantially the modern rule for the positive root of a quadratic. Sridhara (ca. 1025) gave the positive root of the quadratic formula, as stated by Bhaskara (ca. 1150; Smith 1953, pp. 445-446). The Persian mathematicians al-Khwarizmi  (ca. 825) and Omar Khayyám  (ca. 1100) also gave rules for finding the positive root. \n\nViète  was among the first to replace geometric methods of solution with analytic ones, although he apparently did not grasp the idea of a general quadratic equation (Smith 1953, pp. 449-450). \n\nAn alternate form of the quadratic equation is given by dividing (◇) through by : \n\n (8) \n\n (9) \n\n (10) \n\nTherefore, \n\n (11) \n\n (12) \n\n (13) \n\nThis form is helpful if , where  denotes much greater, in which case the usual form of the quadratic formula can give inaccurate numerical results for one of the roots. This can be avoided by defining \n\n (14) \n\nso that  and the term under the square root sign always have the same sign. Now, if , then \n\n (15) \n\n   (16) \n   (17) \n\nso \n\n   (18) \n   (19) \n\nSimilarly, if , then \n\n (20) \n\n   (21) \n   (22) \n\nso \n\n   (23) \n   (24) \n\nTherefore, the roots are always given by  and . \n\nNow consider the equation expressed in the form \n\n (25) \n\nwith solutions  and . These solutions satisfy Vieta's formulas \n\n   (26) \n   (27) \n\nThe properties of the symmetric polynomials appearing in Vieta's formulas then give \n\n   (28) \n   (29) \n   (30) \n\n \nGiven a quadratic integer polynomial , consider the number such polynomials that are factorable over the integers for  and  taken from some set of integers . For example, for , there are four such polynomials, \n\n   (31) \n   (32) \n   (33) \n   (34) \n\nThe following table summarizes the counts of such factorable polynomials for simple  and small . Plots of the fractions of factorable polynomials for  (red),  (blue), and  (green) are also illustrated above. Amazingly, the sequence for  has the recurrence equation \n\n (35) \n\nwhere  is the number of divisors of  and  is the characteristic function of the square numbers. \n\n Sloane factorable over  for , 1, ... \n A067274 1, 4, 10, 16, 25, 31, 41, 47, 57, ... \n A091626 1, 2, 4, 6, 9, 11, 14, 16, 19, 22, ... \n A091627 0, 0, 1, 2, 4, 5, 7, 8, 10, 12, 14, ...
920068
The standard form of a quadratic:\n\nax^2 + bx +c\n\nIn your case a =1, and c=1. Since there is no x term, b=0