explain what the discriminant means to the quadratic function, y = ax^2 + bx + c?
925102
Discriminant is part of the solution.  Just considering it tells you the types of solutions that you have.  \n\nThere are three possibilities:\n1. 2 real roots,\n2. 2 complex roots,\n3. 1 repeated real root.
698387
y= -(2/3)x^2 + (2/3)x + 4\n\n\n(P): y = ax^2 + bx + c\n\nPoints (0;4)  (-2;0)  (3;0) are element of (P)\n-->  4 = 0a + 0b + c --> c = 4\n0 = 4a - 2b + c\n0 = 9a + 3b + c\n\n--> 5a + 5b = 0 --> b = -a\n--> 4a + 2a + 4 = 0 --> a = -2/3,  b=2/3
945878
I'm assuming you are looking for values of b that will give rational roots. That is, the discriminant will be a perffect square or 0. If this is not what you wanted, I'm sorry.\n\n\n3x^2 + bx -3 =0\nb^2 + 36= perfect square if b=0 or b=8\n\n5x^2 +bx +1=0\nb^2-10 =perfect square for no value of b\n\n-3x^2 +bx -3=0\nb^2 -36 = perfect  square fo r b=6 and b=10
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)).
1171466
If you have an equation\n\nax2+bx+c = 0\n\nthe roots of the equation are found by the formula:\n\n \n\nThe expression b2 - 4ac is called the discriminant. Its value tells you whether you have no roots, 1 root or 2 roots.\nCase 1: b2 - 4ac <0\nIf it is less than 0, that means you are trying to find the square root of a negative number. Ergo, no roots. An example is x2-2x+5.\nCase 2: b2 - 4ac = 0\nThe square root of 0 is 0. One root. Period. An example is x2-2x+1. The only root is x = 1.\nCase 2: b2 - 4ac > 0\nHere we have 2 roots. An example is 2x2-x-15. We have a = 2, b = -1 and c = -15. Plugging these numbers into the formula, one root is 3 and the other is -5/2.\n\nyou can still use that shock absorber for the positive in contraction as well .
593998
When you have a polynomial, the coefficients are integers.  For example:\n\nax^2 + bx + c = 0\n\nThis is the general formula for a second degree polynomial (x is to the 2, hence second degree).  In this case, a, b, and c are the integers.\n\nAlso, if you add other terms, such as x^3, they would also be multiplied by integers.
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.
338193
The equation represents a straight line graph whose gradient is 3 , and intercepts the the ordinate axis(i.e y-axis) at the -2 position.\n\nthe easiest way to arrive at this is to draw a table of values. Such that you find the values of y, using values of x from -6 to +6.\n\nYour graph will confirm my story, \nBut from mere looking at the equation I can make a good prognosis, when i compare it with Y=mx + c .where Y represents the value of the equation, X represents the value of x which made y what it is.\nM stands for the gradient of the graph, while c stands for the intercept on the Y-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, ...
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
740067
Let's start from the beginning with a quadratic.\n\nLet y=f(x)=ax^2+bx+c, where a is not equal to zero.  The y-intercept is the point where the parabola intersects the y-axis.  So, the point is (0,f(0))=(0,c).\n\nIf the parabola opens upward (a positive) and its vertex is above the x-axis or opens downward (a negative) and its vertex is below the x-axis, there aren't any (real) x-intercepts.  \n\nOtherwise, if the quadratic has x-intercepts, solve ax^2+bx+c=0.  \n\nTry factoring first and if that doesn't work easily, then try using the quadratic formula,\n\nx=(-b &#92;pm &#92;sqrt{b^2-4ac})/(2a).\n\nIf a parabolar has x-intercepts it may have one or two so be sure to check your work carefully.\n\nExamples:\n\ny=x^2+1 has y-intecept (0,1) by no x-intercepts\n\ny=x^2-2x+1=(x-1)^2 has y-intercept (0,1) and x-intercept at (1,0)\n\ny=x^2-x-2=(x-2)(x+1) has y-intercept (0,-2) and x-intercepts (2,0) and (-1,0).\n\nA good idea:  Given a quadratic either try to factor it so it looks like y=(a x+b)(c x+d) or try to write in the form y=a(x-b)^2+c.  In the second case, (b,c) is the vertex and the sign of a tells you whether it goes up or down.  The first case shows that the x-intercepts are x=-b/a and x=-d/c.\n\nExcellent question! :)
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
707752
1) plug y=4-x in for y in the equation of your circle.  foil it out and bring everything to the left hand side.  now you have a quadratic equation that you can solve by using your quadratic formula.  once you have values for x, the corresponding value of y is just 4-x.  you should get (3,1) and (5,-1) as your answers.\n\n2) for this one, try to do the same thing you did in (1).  this time plug in x = 3-2y for x, just because y is the easier variable to try and solve for.  when you use your quadratic formula, you'll see your discriminant is negative, and thus you get no real solutions.  therefore the 2 equations don't intersect\n\n3) find the points of intersection of the circle and the line, exactly as you did in problem (1) and tried to do in problem (2).  i.e. plug in x = 2y-4 in for x, in the equation of your circle, and solve for y.  then, go back and solve for x and you should have 2 ordered pairs (the 2 points of intersection).  the length of the chord is just the distance between these 2 points.  remember your distance formula:\n\nd = sqrt( (x1 - x2)^2 + (y1-y2)^2 )
818628
4x^2 + kx + 9 = 0.\nTwo methods: a) there are two equal roots if the discriminant, which is b^2 - 4ac, equals 0.\nSo k^2 - 4*4*9 = 0, so k = 12. (since k>0)\nb) there are two equal roots if the quadratic is actually a perfect square. You need a 4 = 2^2 on the front, and a 3 = 3^2 on the end, so it must equal (2x+3)^2 (since the middle bit must be > 0).That is 4x^2 + 12x + 9, so k = 12.
798499
Given ax² + bx + c = 0, the quadratic formula tells us that:\n\nx = [ -b ± √(b²-4ac) ] ÷ 2a\n\nIn your problem, a=3, b=12, c=1  --  simply plug these values into the quadratic formula to solve for x.
516234
The discriminant = b^2-4ac = 1 - 4.2.(-4) = 33 > 0 \n⇒ We can get two unequal, real but irrational roots.\n\nThis is so, because, by actually calculating the roots of the equation applying formula. we find that:\nx= [1±√33]/4
732775
y = -16x^2 +80x +3\n\nFirst of all, it is a negative quadratic function, so, you expect this quadratic function to open downwards. Now that you know the general shape, you need a few points on the graph in order to help you graph it. So, let's find the x and y intercept. \n\nTo find the y intercept, you sub x = 0 into the above equation, which leads to 3. Now, you know one point (0,3).\n\nTo find the x intercept, the easiest way is to use the quadratic formula:\n\nFor an equation y = Ax^2 + Bx + C\nx = [- B ± sqrt(B^2 - 4AC)] / (2A)\n\nIn your equation, A = -16, B = 80, C = 3. Sub these information into the quadratic formula, you get\n\nx = [- 80 ± sqrt(6592)]/(-32)\n\nFurther simplify this expression, you get\n\nx = [80 ± 8*sqrt(103)]/32\nx = [10 ± sqrt(103)]/4\nThis leads to two answers, x ~ 5.04 and x ~ -0.037. So, the two x intercepts are (5.04, 0) and (-0.037, 0).\n\nNow you have three points: (0,3), (5.04, 0), and (-0.037, 0). Put these on a coordinate, with the general shape in mind (quadratic facing downwards), it should be very obvious how the function should look like. Hope that helps.