what is the quadratic formula?
538847
For any quadratic of the form ax^2 + bx + c = 0, the roots are\n\nx = (-b +/- sqrt(b^2-4ac)) / 2a\n\nSqrt here is shorthand for "the square root of".
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
693558
On a graph with x and y-axes , if you plot out the values of y calculated from an equation involving x squared you draw out a parabola.  A quadratic equation is one in the form y = a.x-squared + b.x + c \n a can be positive or negative (not zero or you don't involve x-squared). If negative, the parabola is upside down (open end downward).\nb and c can be positive negative or zero.\nWhen you are asked to "solve" a quadratic, you are finding where the parabola cuts the x-axis, that is to say the values of x where the equation calculates out to y = 0
85829
I used the quadratic formula x={-b+-sqrootof b^2-4ac}/2a in order to find the factors on both the numerator and denominator\n\nthat leaves {(2x-1)(x-3)}/{4(x-3)(x-3)}\n\nClear (x-3) from both sides, that leaves\n\n(2x-1)/(4(x-3)) your answer
955903
3x^4-2x^2-8=0\nQuadratic formula with x^2 instead of x\n\na=3, b=-2, c=-8\n[2+-sqrt(4-4(3)(-8)]/6\n(2+-10)/6\nx^2=2 or x^2=-4/3\n\nx=+2 or -2\nx can also equal +sqrt(-4/3) or -sqrt(-4/3) but these are complex numbers
810321
you will need to use the quadratic formula since you cannot easily find factors\nfrom the given equation above a=1  b= -7  c=7 (where a,b,c are the co-effiecients of the given eqn respectively)\n\nusing the quadratic formula x= -b(plus or minus) under the square root sign(bsquared- 4ac) all divided by 2a\n\nthen x= -(-7) (plus or minus) under square root sign (-7)(-7) - 4 (1*7) divided by 2(1)\n\nx= 7+ or - square root 49-(4*7) /2\n\nx=7+ root of 49-28 all divided by 2\n\nx=7 +or - root of 21 all divided by 2\n\nx=7+ 4.58 divided by 2\nthen x=5.79=5.8\nor\nx=7-4.58 divided by 2\nx=1.21=1.2\ntherefore\nx=1.2, x=1.8\ni ope u understand wat i ve done\nthe needed symbols and signs arent availabe that is y i wrote sum of the words out\nall the best
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
781910
not possible to factor easily ... try the quadratic formula.
626330
1) 2x^2 + 3 = x\n2x^2 - x + 3\n\nQuadratic formula:\n[1 +/- sqrt(1-4*2*3)]/(2*2)\n[1 +/- sqrt(-23)]/4\n\nx= 1/4 + sqrt(23)/4 * i\nor\nx= 1/4 - sqrt(23)/4 * i\n\n2) (2m - 7)^2 - 2(2m - 7) - 24 = 0\nSay X = 2m-7\nSo: x^2 - 2x - 24 = 0\n(x-6)(x+4)=0\nx=6,-4\n\n2m-7=6\nm=13/2\n\nor\n\n2m-7=-4\nm=3/2
827891
analyzemath.com/quadraticequations
1053962
I am assuming this is an "I hate math so I'm gonna complain about its lack of relevance" question and not a real question, but I'll answer it anyway.  I can't tell you every use of it, I suppose I've seen several, but here's one I see a lot.  If you have an object thrown in a trajectory, and you need to know when it will cross a certain height, or when it will be going a certain speed (I dunno, in case you're lobbing hand grenades or missiles someday when you're bored), the quadratic equation will come up in the solution, and it will give you BOTH of the answers, one for when the object is on the way up and one for when it's on the way down.
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
1015721
Pythagoras:\nAC squared = AB squared + BC squared\nTherefore\nAC = square root of quantity (AB^2 + BC^2)\nYou can do the math.
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
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
443949
Use quadratic formula and do your own homework.
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)).