Trying to solve quadratic equations, have tried websites, but still don't get it.Can sumone IM me and explain?
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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-2 = 1/(1+x)-3/(x-1)\nmultiply both sides by (x^2-1)=(x+1)*(x-1):\n\n-2(x^2-1)= x-1 -3(x+1) \n\nrearrange\n2x^2 - 2x - 6 = 0\ndivide by 2\nx^2 - x -3 =0\napply formula for quadratic equations\nx=(1 +/- sqrt(1+4*3))/2\n = 0.5 +/- sqrt(13)/2\n\n"+/-" means plus or minus (there are two solutions)\n"^2" means squared
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a
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
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
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.
891393
are they all related? or are these just different equations??????\nnot clear at all..................................................\n \n\n3x-q=p: 3x=p+q, x=(p+q)/3\nx-2=r: x=2+r\ny-a=7: y= 7+a\nx-c=d: x= c+d\n4x-5c=3c: 4x= 8c, x= 2c\nbx=9b^2: x=9b
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
792339
A quadratic equation is a second-order polynomial equation in a single variable x\nax^2+bx+c==0,\n(1)\n\nwith a!=0. 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\nAmong his many other talents, Major General Stanley in Gilbert and Sullivan's operetta the Pirates of Penzance impresses the pirates with his knowledge of quadratic equations in "The Major General's Song" as follows: "I am the very model of a modern Major-General, I've information vegetable, animal, and mineral, I know the kings of England, and I quote the fights historical, From Marathon to Waterloo, in order categorical; I'm very well acquainted too with matters mathematical, I understand equations, both the simple and quadratical, About binomial theorem I'm teeming with a lot o' news-- With many cheerful facts about the square of the hypotenuse."\n\nThe roots x can be found by completing the square,\nx^2+b/ax==-c/a\n(2)\n(x+b/(2a))^2==-c/a+(b^2)/(4a^2)==(b^2-4ac)/(4a^2)\n(3)\nx+b/(2a)==(+/-sqrt(b^2-4ac))/(2a).\n(4)\n\nSolving for x then gives\nx==(-b+/-sqrt(b^2-4ac))/(2a).\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\nx^2+y^2=100\n(6)\ny=3/4x\n(7)\n\n(Smith 1953, p. 443). The Greeks were able to solve the quadratic equation by geometric methods, and Euclid's Eric Weisstein's World of Biography (ca. 325-270 BC) Data contains three problems involving quadratics. In his work Arithmetica, the Greek mathematician Diophantus Eric Weisstein's World of Biography (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 Eric Weisstein's World of Biography (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 Eric Weisstein's World of Biography (ca. 825) and Omar Khayyám Eric Weisstein's World of Biography (ca. 1100) also gave rules for finding the positive root.\n\nViète Eric Weisstein's World of Biography 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 x^2:\na+b/x+c/(x^2)==0\n(8)\nc(1/(x^2)+b/(cx))+a==0\n(9)\nc(1/x+b/(2c))^2==c(b/(2c))^2-a==(b^2)/(4c)-(4ac)/(4c)==(b^2-4ac)/(4c).\n(10)\n\nTherefore,\n1/x+b/(2c)==+/-(sqrt(b^2-4ac))/(2c)\n(11)\n1/x==(-b+/-sqrt(b^2-4ac))/(2c)\n(12)\nx==(2c)/(-b+/-sqrt(b^2-4ac)).
219130
I could not get your question. We all know that the quadratic equations are those equation which contain a square with the unknown (x). Quadratic equation will always have two roots. But, where did you get the "Four"? Would you explain a bit more.
797913
That's the quadratic formula. It is the solution to equations of the form\nax^2+bx+c=0
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.
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It is notable that the x-intercepts, or solution, of a quadratic equation are known as the ROOTS or the ZEROS of that particular equation.\n\nWhen you solve a quadratic equation, you are looking for the points that make the equation 0; recall that all problems have one side set to zero. For example:\n\nx^2 + 2x + 1 = 0, the original being y=x^2 + 2x + 1\n\nWhat you do when you solve the problem is look for the input values of x that give a y value of 0. It is precisely this that produces the effect of a solution being an x-intercept. Continuing:\n\n(x+1)(x+1) = 0, x = -1\n\nWhat you have done was assume that y=0 (and hence restricted yourself to the x-axis), then tried to find the x points where this \noccurs. By setting y=0, you are "searching" along the x-axis; the ssolution is given where the function CROSSES the axis.
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, ...
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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 -.
948294
Let's see:\nc² - 7c = 0\nUse the quadratic formula:\nc = [7 ± √49] / 2 = (7 ± 7) / 2 = 7/2 ± 7/2 = {0, 7}\n\n0 is one solution, but 7 is another.
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.