Can u tell me and explain to me what the quadratic equation is?
146612
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, ...
848564
Substitute 2x-1 for y in the middle part, you get:\nx^2 + x -6(2x-1) = 2x-1\nThis you can reduce to a quadratic equation for x.
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
576236
I'm in my final year of university and I still cannot remember the quadratic formula. And my degree in in math. haha becoz it can be easily derived, so why waste your brain cell remembering it?\n\n\nanyway, when sqrt(b^2-4ac)<0, meaning it has no REAL root**, it means the quadratic equation has no REAL solution.\n\nWhen I say REAL, I don't mean real or fake. I mean, REAL NUMBER solution. In other words, it does not cross over at the x-axis.\n\n\n\n**It does have a solution in the complex number space though. As u might know, when u sqrt(-1) u get i, the imaginary number. but u don't need to know that yet.
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.
857303
Hmm. It could just be something like forgetting a parenthesis. Let's go through it step by step:\n\nThe quadratic formula is:\n\n[ -b +/- sqrt(b^2-4ac) ] / 2a\n\nBecause of the +/-, we will have two equations:\n\n[ -b + sqrt(b^2-4ac) ] / 2a\nand\n[ -b - sqrt(b^2-4ac) ] / 2a\n\nThe equation is x^2 + 3x - 84 = 0\na = 1\nb = 3\nc = (-84)\n\nSo we have:\n[ -3 + sqrt(3^2-4(1)(-84)) ] / 2(1)\nand\n[ -3 - sqrt(3^2-4(1)(-84)) ] / 2(1)\n\n[ -3 + sqrt(345) ] / 2\nand\n[ -3 - sqrt(345) ] / 2\n\n7.78709\nand\n-10.7871
736851
I got a different answer than you.  Let's do this together:\n\nRecall that for an equation ax² + bx + c = 0, that the quadratic formula is:\n\nx = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\nSo we want to find the zeros of x² + 10x + 23\n\nSo, x = [ -b ± √ (b² - 4ac) ] ÷ 2a\n\n= [ -10 ± √ (10² - 4*1*23) ] ÷ 2*1\n\n= [ -10 ± √ (100 - 92) ] ÷ 2\n\n= [ -10 ± √8 ] ÷ 2\n\n= [ -10 ± 2√2 ] ÷ 2\n\n= -5 ±√2\n\nHope that helped.\n\n~ ♥ ~
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
530596
Given\nax^2 + bx + c = 0\nDivide all terms by a \nx^2 + (b / a) x + c / a = 0 \nSubtract c / a from both sides \nx^2 + (b / a) x + c / a - c / a = - c / a \nand simplify \nx^2 + (b / a) x = - c / a \nAdd (b / 2a)^2 to both sides \nx^2 + (b / a) x + (b / 2a)^2 = - c / a + (b / 2a)^2 \nto complete the square \n[ x + (b / 2a) ]^2 = - c / a + (b / 2a)^2 \nGroup the two terms on the right side of the equation \n[ x + (b / a) ]^2 = [ b^2 - 4a c ] / ( 4a^2 ) \nSolve by taking the square root \nx + (b / a) = ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nSolve for x to obtain two solutions \nx = - b / a ± sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } \nThe term sqrt { [ b^2 - 4a c ] / ( 4a^2 ) } may be written \nsqrt { [ b^2 - 4a c ] / ( 4a^2 ) } = sqrt(b^2 - 4a c) / 2 | a | \nSince 2 | a | = 2a when a > 0 and 2 | a | = -2a when a < 0, the two solutions to the quadratic equation may be written \nx = [ -b + sqrt( b^2 - 4a c ) ] / 2 a \nx = [ -b - sqrt( b^2 - 4a c ) ] / 2 a \nThe term b^2 - 4a c which is under the square root in both solutions is called the discriminant of the quadratic equation. It can be used to determine the number and nature of the solutions of the quadratic equation. 3 cases are possible \ncase 1: If b 2 - 4a c > 0 , the equation has 2 solutions. \ncase 2: If b 2 - 4a c = 0 , the equation has one solutions of mutliplicity 2. \ncase 3: If b 2 - 4a c < 0 , the equation has 2 imaginary solutions\nps. I did this as quick as possible, hope everything copied correctly.
894370
It just stands for the coefficients (integers) in front of your terms.\n\nSo if you have an equation x² - 3x + 8,\n\n.   a is the coefficient of your x² term -- in this case a = 1\n\n.   b is the coefficient of your x term -- in this case b = -3\n\n.   c is the coefficient of the non-x term -- in this case c = 8\n\nThe quadratic formula assumes your equation is in the format of a² + bx + c; that's where the letters come from and that's how you determine what they stand for.\n\nHope that helped!\n\n~ ♥ ~
595518
1 + 4.9t^2 = 6t\n4.9t^2 - 6t + 1 = 0\n\nusing the quadratic formula\n\nt = (-b ± sqrt(b^2 - 4ac))/2a\nt = (-(-6) ± sqrt((-6)^2 - 4(4.9)(1)))/2(4.9)\nt = (6 ± sqrt(36 - 19.6))/9.8\nt = (6 ± sqrt(16.4))/9.8\nt = (6 ± sqrt(4 * 4.1))/9.8\nt = (6 ± 2sqrt(4.1))/9.8\nt = (3 ± sqrt(4.1))/4.9\nt = about 1.02548 or .199011\n\n-----------------------------------\n\n1 - 4.9t^2 = 6t\n4.9t^2 + 6t - 1 = 0\n\nt = (-6 ± sqrt(36 + 4(4.9)))/9.8\nt = (-6 ± sqrt(36 + 16.4))/9.8\nt = (-6 ± sqrt(52.4))/9.8\nt = (-6 ± sqrt(4 * 13.1))/9.8\nt = (-6 ± 2sqrt(13.1))/9.8\nt = (-3 ± sqrt(13.1))/4.9\nt = about -1.37312 or .148627
309372
If you have a quadratic equation of the form ax squared + bx + c = 0, the discriminant is\nb squared minus 4 times a times c, or b^2 - 4ac\n\nIt's the part under the square root in the Quadratic Formula.  It tells if the roots  are real, rational, or imaginary.\n\nSo in your first example, it's 9 - 4X1X-4, or 9 - -16, or 25.
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.
44408
5x2 = 9 − x\n\n5x2 + x − 9 = 0\n\nIt is a quadratic equation with\na = 5, b = 1, c = -9
36555
There are few ways of solving for the unknown in a quadratic equation.\n\nIn general, a quadratic equation is of the form:\n   a*x^2 + b*x + c = 0, a != 0    --- (1)\nDividing (1) by "a" through out, we get,\n   x^2 + (b/a)*x + (c/a) = 0  --- (2)\n\nOne way of solving is to really guess the solution. It works like this:\nAssume we have (somehow) factorised (2) into a form like:\n   (x - p)(x - q) = 0   --- (3)\n(3) can also be written as (by mutiplying term by term),\n   x^2 - (p+q)*x + p*q = 0   --- (4)\nComparing (2) and (4),\n   p + q = -b/a\n   p*q   = c/a\nSo, if the equation is written like in (2), the sum of roots is negative of x co-efficient, and the product is the constant. A "root" is a value of x that satisfies the given equation (1).\n\nFor e.g, given,\n   x^2 - 6*x + 5 = 0,\nthe sum of roots must be 6 and the product must be 5. So (we guess), the roots must be 1 and 5. Easy!\n\nBut "guessing" is not always possible :(\n\nLook at this equation:\n   x^2 - 6*x + 10 = 0\nThe roots of this are 3 + i and 3 - i. Not easy to guess right?\n\nA generic way of solving is called "completing the square". And it is a very clever piece of work! It works like this:\nEquation (2) again:\n   x^2 + (b/a)*x + (c/a) = 0\nThe aim is to get the equation into a form that looks like:\n   (x + something)^2 = somethingelse\nThen we can solve for x easily.\n\nLet us first get rid of the constant term to the other side:\n   x^2 + (b/a)*x = -(c/a)\nLet us re-write this again like:\n   x^2 + 2*(b/2a)*x = -(c/a) --- (6)\nLook at the left hand side (LHS) of the equation.. it is looking like: x^2 + 2*something*x. If only we have something^2, then we can put it in the form we want. So we'll add that term:\n   x^2 + 2*(b/2a)*x + (b/2a)^2 = -(c/a) + (b/2a)^2\nWe've got it! We now have,\n   ( x + (b/2a) )^2 = -(c/a) + (b/2a)^2\n\nWe can solve for x quite easily from here (left as exercise to the reader).\n\nThe "magic" equation (-b plus or minus sqare-root of etc...) is really derived by completing the square.
745544
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
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. :^)