More Quadratic Equations?
487271
suppose the size of the square need to be cut is x^2, (it means that the length of each side is x).\nwe will have an equation:\n   80= (14-2x)(11-2x)\nthen you will get the answer.\n\ni am from china, i not sure i fully expressed my solution:)
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.
236212
i hope u know c++ well.\n\n/* Roots Of The Quaditic Equation*/\n\n#include<iostream.h>\n#include<conio.h>\n#include<math.h>\nvoid main()\n{\nint a, b, c;\nfloat root1, root2 ;\nclrscr() ;\ncout << "Let the Quadratic Equation be aX2 + bX + c" ;\ncout << "&#92;n&#92;nEnter the value of a : " ;\ncin >> a ;\ncout << "&#92;nEnter the value of b : " ;\ncin >> b ;\ncout << "&#92;nEnter the value of c : " ;\ncin >> c ;\n\n//Finding The Roots Of The Quadratic Equation.\nroot1 = ( -b + sqrt( abs(b*b - 4*a*c) ) ) / 2*a ;\nroot2 = ( -b - sqrt( abs(b*b - 4*a*c) ) ) / 2*a ;\ncout <<"&#92;n&#92;nThe roots of the Given Equation are : "\n        << root1 << " , " << root2 ;\ngetch();\n}\n\nHeres the program.\nany ways i will explain u wht it is\n\nfirstly accept there values a,b,c bcoze quadratic equation is ax2+bx+c now if want 2 calculate d roots then d above program will work fine. because for finding square root we cannot have -ve values i used abs() its a funtion which converts -ve values into +ve values hope that ur qustion is answered bye
612720
-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
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
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.
919864
1.  ax + xb\n2.  21x^3 + 24x*2y\n3.  24x^2 + 31x + 10\n4.  24x^2 - 8x - 6\n5.  40x^3 + 66x^2 + 38x + 6\n6.  x^2 - 1\n7.  x^3 - 8
891271
first into the second:\n\nx^2+(x-1)^2=25\n\nexpand:\n\nx^2+x^2-2x+1=25\n\nget 0 on the right side:\n\n2*x^2 - 2*x -24 = 0\n\na=2, b=-2, c=-24\n\nuse the root finding formula for quadratoic equation:\n\nx=(-b+/-sqrt(b*b-4*a*c))/(2*a)=(2+/-sqrt(4+4*2*24))/4=0.5 +/- 3.5\n\nyour solutions are: x = -3 and 4\n\ncheck:\n(-3)^2+(-4)^2 = 9 + 15 = 24, correct\n(4)^2+(3)^2 = 15 + 9 = 24, correct\n\nsubstitute to the first eqn: y = -4 for the 1st. x and 3 for the 2nd.
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)).
986257
If the equation is X^4 - 12x^2 -64 =0, there is.\n\nUse u = x^2, now your equation will look like this:\n\nu^2 - 12u - 64 =0\nFind u1 and u2 using the quadratic formula.\n\nThe solutions are u1 = 16, u2 = -4\n\nNow you go back to x, first deal with u1, then u2.  Because u = x^2 each u solution will give you two x solutions.\n\n\nu = x^2 = 16\nx = +/- SQRT(16)\nx1 = 4\nx2 = -4\n\nx2 = -4.  This is not possible with real numbers. so it will give you two imaginary numbers. Do the same as before\n\nx = +/- SQRT(-4) = +/- 2*SQRT(-1) = +/- 2i\n\nx3 = i\nx4 = -i
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 :)
739088
Consider a = 1, b = 0, c = 1.  Clearly, b^2-4ac = -4 < 0.  This corresponds to the equation x^2 + 1 = 0.  But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.\n\nIf you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.\n\nYou clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."\n\nRational equations do indeed give complex solutions, many of them nonreal.  The example I gave above (x^2 + 1 = 0) is an example.  The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers.  If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.
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.
651993
Here's the best I can do to help with the first one.\n   __x_+9___\nx-2)X^2+7x-2\n    x^2-2x\n    ------\n        9x-2\n        9x-18\n        -----\n           16\nFollow the example for the rest
534396
quadratic equations of the form \n\nax^2+bx+c=0\n\ncan be used with the quadratic formula, see\n\nhttp://mathworld.wolfram.com/QuadraticFormula.html\n\nso just transform all your equations into the form\n\nax^2+bx+c = 0 (x can be called differently, of course).\n\nEdit: for example, let's take \n\n5t^2 + 20t - 60 = 0\n\naccording to the quadratic formula, the solutions are\n\nt_1 = (-20+sqrt(20*20-4*5*(-60))) / (2*5) = (-20+sqrt(1600))/10 = 2\n\nand\n\nt_2 = (-20-sqrt(20*20-4*5*(-60))) / (2*5) = (-20-sqrt(1600))/10 = -6\n\nNote: if the term below the root (sqrt) is negative, it is not a valid solution - sometimes the quadratic equation has only one solution, or none.
939068
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~ ♥ ~
374428
The most strait forward way is to use the quadratic formula. You just do simple adding, powering, rooting, and you get the answer. It works the 100% of the time. There is also the filling-in-the-square, but that's too much work compaired to the quadratic formula.