fine the value of the discriminant and discribe the nature of the roots of each quadriatic equation?
824781
9u^2-24u+16=0\ndiscriminant\n=(-24)^2-4*9*16\n=0\nsince it is = 0\nso its roots are real and\nequal
808695
If r and s are the roots of ax² + bx + c = 0, then we can rewrite it as\nx = r or x = s.\n\nTransposing, we get\nx - r = 0 or x - s = 0\n\nThen\n(x - r)(x - s) = 0\n\nMultiply\nx² - (r + s)x + rs = 0\n\nWe divide a from the quadratic\nx² + b/a x + c/a = 0\n\nTherefore,\nr + s = -b/a and rs = c/a\n\nNow, we need to find\nr^-1 + s^-1\n\nSince an exponent of -1 means the reciprocal, then\n= 1/r + 1/s\n\nGetting the LCD,\n= (r + s)/rs\n\nNow we substitute:\n= (-b/a)/(c/a)\n\nWe cancel the a's:\n= -b/c\n\nTherefore,\nr^-1 + s^-1 = -b/c\n\n^_^
405166
Try this:\nhttp://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut9_bar.htm
317499
The discriminant is b^2 - 4ac where a is the number in front of the x^2, b is the number in front of the x and c is the constant.  In this case, the constant is negative.\n\nSo, b^2 - 4(a)(c) will always be positive because b^2 is positive and -4(a)(-n) will also be positive.  Since both are positive, the sum will be positive.
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, ...
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.
837237
Hi\nI think you're referring to one of my answers.\nhttp://answers.yahoo.com/question/index;_ylt=ApZ1XohHNUdL5xkBg8bXQEfsy6IX?qid=20061124162535AAURdJ5\n\nHy is correct, this was done in order to eliminate the x^3 term in the quartic equation.  It made the subsequent solution much easier to handle, since the actual values of a,b,c,d or p,q,and r didn't matter:  the solution wasn't looking for an answer to a specific quartic but for a generalization about quartics in general.\n\nConverting a quartic into reduced form is also done when finding the roots of any quartic equation by an old method (hardly used anymore since computers).  After placed into reduced form, the 'cubic resolvant' is formed:\n\nz^3 + 2pz^2 + (p^2 - 4r)z - q^2 = 0\n\nFinding the three roots (z1,z2,z3) of this equation (using Cardano's or other methods), will yield the four roots of the reduced form quartic:\n\nx1 = (√z1 + √z2 + √z3) / 2\nx2 = (√z1 - √z2 - √z3) / 2\nx3 = (-√z1 + √z2 - √z3) / 2\nx4 = (-√z1 - √z2 + √z3) / 2
539132
both roots are complex numbers
58489
In an equation of the form p(x)=ax^2+bx+c, where a,b,c are constants and x is the variable, solving for p(x)=0 yields the two solutions:\n\nx=(-b(+ or -)sqrt(b^2-4ac))/(2a)\n\nsqrt represents taking the square root.\n\nThe discriminant (b^2-4ac) tells you how many distinct real solutions you will have.\n\nDiscriminant >0 implies 2 distinct real solutions.\nDiscriminant =0 implies 1 distinct real solution.\nDiscriminant <0 implies 0 real solutions.
742818
simple just find out DISCRIMINANT= b^2 - 4ac\n\nif Dis is +ve then there are two real distinct roots\n\nif Dis is 0 then there is exactly one root.\n\nIf the discriminant is negative, there are no real roots. Rather, there are two distinct (non-real) complex roots, which are complex conjugates of each other\n\nin your question\n\ncheck for solving quintic equations and  bring it in the form of \nquadratic equation.
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
858220
Depends on what variable you're differentiating with respect to:\ndV/dr : 2/3*pi*r*h\ndV/dh:  1/3*pi*r^2
201685
If y is a root of this and b is in F_p, then (y+b)^p -(y+b)=y^p +b^p -y-b=a, so y+b is another root. Hence the complete list of roots is y, y+1, ....y+p-1.\n\nThus we have two cases, the first where the polynomial splits in F_p (a=0) and the other where we get a degree p extension, so F_(p^2).\n\nNotice that we are using the algebra student's dream (x+y)^p=x^p +y^p which is justified because we are working over a field of characteristic p.
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
870960
Do you mean that the roots of x^2+pX+r are α  and β and you need to find quadratics in terms of p and q?\nIf so you can use the following:\nIn any quadratic ax^2 + bx + c = 0, the sum of the roots = -b/a and the product of the roots = c/a.\nUsing this,  -(α + β) = p, and αβ= r.\n\nAnswers are for terms b and c in the quadratic x^2 + bx + c\ni)\nb = -(1/α + 1/β) = -(α + β)/αβ = p/r\nc = 1/αβ = 1/r\n\nii) \nb = -(1/α^2 + 1/β^2) =  -(α^2 + β^2)/(αβ)^2 = -[(α + β)^2 - 2αβ])/(αβ)^2 = -(p^2 - 2r)/r^2\nc = 1/(αβ)^2 = 1/r^2\n\niii) \nb = (α^2β + αβ^2) = αβ(α + β) = -pr\nc = -α^2β * -αβ^2 = (αβ)^3 = r^3
868180
the roots are (2a-1)x(2a-1)
894899
Horners' method of synthetic division is used to find the roots of polynomials. \nwe have to know one of the roots to proceed with this. we can check if 1or -1 are roots of the polynomial using some easy tricks. \n1) if the sum of all the coefficients is equal to 0, then 1 is a root\n2) if the sum of coefficients of even terms is equal to the sum of the coefficients of the odd terms, then -1 is a root.\nnow, for the first equation, we already know that -2 is a root, because the question is to find the quotient when the polynomial is divided by x+2. \nin synthetic division, u first put down a 0 beneath the first coefficient, ie., 3. now add these two. the result is 3. \nnow multiply 3 with the root, ie., -2. the result is -6. write this beneath the second coefficient, ie., 2. the result is -4 after addition. \nrepeat the same steps until u reach the last coefficient.\nnow next we write 8 below -1. result is 7.\nnext we write 7*-2 ie., -14 below 5. the result is -9.\nlastly, we write -9*-2=18 below . the result is 12, but the result should be zero.\nso, the given equation is not divisible by x+2.\n\ncoming to the second equation, follow the same procedure as above and u will get the resultant coefficients at each stage to be 2,6,22,61 respectively. so it is not divisible by 3