how does the zero product relate to the quadratic equation?
96998
The zero product rule says that if two numbers multiply out to be zero, then one of the numbers must be zero.  It's used in quadratic equations (not to be confused with the quadratic formula) when they are factored by a method often called "FOIL".\n\nExample:  Solve x^2 - 8x + 15 = 0\n\nFactored, this is (x-3)(x-5)=0\n\nBecause of the zero property you know that either x-3=0 or x-5=0.\n\nSo you solve each small equation to get the possible values of x (3 or 5).
797453
you set the bottom of the equation equal to 0. So x^2 - x = 0, which leads to two cases. The first one is x = 0. The second one is x-1 = 0, or x = 1.
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.
279743
in this case, the cross-product uses the sine of the angle between the two vectors to multiply the product of the magnitudes of the two vectors.   the angle between any vector and itself is zero.\n\nthe sine of zero is zero.\n\nthen, i x i = |i|*|i|*sin(0) = 1*1*0 = 0
115065
after you factor out the y   you are going to set both parts = to zero because you are multiplying either the y term,  or everything within the parathenses has to equal zero.  You set everything in the parathenses to zero,  and solve for y.  and in this problem you will have two soloutions for y
234829
With factorisation you turn the equation x^2 + px + q = 0 into two smaller equations in brackets like this:\n\n(x+a)(x+b) = 0\n\nSo that a+b = p\nand a*b = q\n\nEG. Your first example you have p = 4 and q = 3\n\nSo you've got a+b = 4 so a could be 2 and b 2 or a 1 and b 3 etc\n\nand a*b = 3 so a could be 1 and b 3... Yes it matches.\n\n(x+3)(x+1) = 0\n\nThe solution is just x = -a and x = -b\n\n-> x = -1 or x = -3
103674
2x^2 - x = 0  { by adding 1 to both sides }\nx^2 - (1/2)x = 0  { by dividing both sides by 2 }\nx (x - 1/2) = 0 { by factoring }\n\nsolves when either x = 0 or x = 1/2 . QED
769595
Assuming your equation is:\n\nmx" + bx' + kx = 0\n\nYou can solve this equation using Euler transformation:\n\nx = e^(nt)\n\nThis means our equation will become:\n\nm(e^nt)" + b(e^nt)' + k(e^nt) = 0\nm(n^2)(e^nt) + b(n)(e^nt) + k(e^nt) = 0\n\nCancelling out the common factor (e^nt), we get:\n\nm(n^2) + bn + k = 0\n\nNow we have a quadratic in 'n'. The roots of this equation are:\n\nn = [(-b) +/- sqrt{(b^2) - (4*k*m)}] / 2\n\nThe discriminant is the one within the square root:\n\n{(b^2) - (4*k*m)}\n\nIf the discriminant is zero or positive, then the root 'n' will be a real number, but if negative then 'n' will be a complex number. This means our initial assumption 'x=e^(nt)' will give us a real value of 'x' for the former case & a complex value of 'x' for the latter. \n\nIn reality, this equation resembles a physical experiment such as a weight hanging from a spring, or a LCR electric circuit. The term 'x' resembles the position or charge in these cases. But by common sense, we can say that the position or charge of an object can only be numerical, real values & not complex values. \n\nTherefore, if our equation resembled a physical situation, then we are forced to say that 'x' has to be a real number, and subsequently 'n' has to be real - which implies our discriminant has to be zero or positive.\n\nThus, we say:\n\n[(b^2) - (4*k*m)] >= 0\n\nFor CRITICAL STATE, the discriminant should equal zero. Because if it is greater than zero, our solution is real & if less than zero the solution is complex. So, at zero it is like a border or boundary - where it is still real - this border value is the critical state:\n\n{(b^2) - (4*k*m)} = 0\n\nb^2 = 4mk\n\nb = + sqrt(4mk)  <or>  b = - sqrt(4mk)\nIn our equation 'b' is the damping coefficient. Therefore, the above two values are the critical damping values. The former is the upper critical value & the latter is the lower critical value.\n\n\nHope this helps.
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
814311
The issue lies in the fact that the original equation is 0=0.\n\nThen, if you put in any value for x, you are dividing by 0 which as we all know, is a "no no."
341524
Second order polynomial equation in a single variable is called quadratic equation.\nSo the answer is no.
738210
The first answer is not exactly right. The roots of the quadratic are\n(-b(+/-)sqrt(b^2-4*a*c)) / (2*a) (not -b^2...) \nHowever the solution is correct; the discriminant is b^2-4*a*c which for the first quadratic is 1^2 - 4*5*2 = -39. Per the ref., a negative discriminant means both roots are complex. They are: (-1(+/-)sqrt(39)*i ) / 10.\nFor the second quadratic, the discriminant is 4*4 - 4*4*1 = 0. As a result there is one real root: -4 / 8 or -0.5.\nRe your followup question, the summary is:\n1st equation, disc. = -39, there are no real roots and 2 complex roots, \n(-1+sqrt(39)*i ) / 10 and (-1-sqrt(39)*i ) / 10.\n2nd equation, disc. = 0, there is one real root (0.5) and no complex roots.
577215
This again is a direct theorem from the field axioms: http://mathworld.wolfram.com/FieldAxioms.html . \nSuppose a*b = 0. If a is non-zero, multiply both sides by a^(-1) to get a^(-1)ab = a^(-1)0, and thus b = 0, since the product of 0 and any number is 0.\nIf b is non-zero, multiply both sides by b^(-1) to get the similar result a = 0. Thus, either b or a must be 0. \nThe final case is that both a and b are 0, which is trivial.\nFurther discussion of this theorem can be found in the source.\nThis theorem can be used to solve equations by getting 0 on one side of the equation and factoring the other side completely, then using the theorem to imply that at least one of the factors must be zero. For example, x^2 = -3x - 2 can be rearranged to be x^2 + 3x + 2 = 0. We factor the left side to read (x + 1)(x + 2) = 0. From the theorem we know that either x + 1 = 0 or x + 2 = 0 or both. Thus we have two solutions, x = -1 or x = -2. It is obvious that in this case, only one factor could be 0.
500175
Quadratic polynomials are usually in the form ax^2 + bx + c. They are usually factorable (which means you can find it's simplified roots). These roots should look like (x-F) and (x-G) where F and G are numbers so that when multiplied together using FOIL (First Outside Inside Last), you get your original polynomial. The roots are also where the polynomial = 0, assuming you've set it to zero. If the polynomials do not factor nicely, then you have to use this formula:\n\n-b +/- root(b^2-4ac)\n--------------------\n      2a\n\n\nWhich will give you the roots. Sometimes you get imaginary roots where in the equation above you will be taking the root of a negative number.\n\nHope that helps.
134065
Your statement translates to the following algebraic equation.\na+b=a-b=a*b\nSolving a+b=a-b, we get b = 0; \nSubstituting b=0 and solving a+b=a*b, we get a = 0;\nSo the two numbers are zeroes.
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)).
700548
So x4-5*y2*x2+4*y4=0; or (x^2-y^2)*(x^2-4y^2)=0; or (x-y)*(x+y)*(x-2y)*(x+2y)=0\nIt splits into independent linear equations: (a) y=x; (b) y=-x; (c) y=0.5*x; (d) y=-0.5*x;\nNow exclude (a) & (c); thus (x+y)*(x+2y)=0; did you mean it Don?