Discriminants of equations?
368115
discriminant formula square root of b^2-4ac\n\nD=square root of 6^2-4(1)(-16)\nD=square root of 36+64\nD=square root of 100 \nD=10\n\nD=square root of 5^2-4(2)(-7)\nD=square root of 25+56\nD=square root of 81\nD=9\n\nif the discriminant is less than zero it has no solution which means the line doesn't touch the x-axis\nif the discriminant is equal to zero,it has one solution meaning the line touches the x-axis once\nif the discriminant is more than zero, it has two solutions which means the line touches the x-axis twice
887510
Differential equations.\nLinear algebra.\nNumerical analysis.
533429
well..write them down ...try to derive them (if their derivation is simple enough or else skip it )......\n\ntry to connect each formula with other one with a hint ....a small hint will help u to recollect other one ....\n\ntry to do at least one problem based on each formula
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.
978709
The first set of differences between your y-values is 1, 2, 4, 7, 11\nThe second set of differences is 1, 2, 3, 4\nThe third set is 1, 1, 1, so you know that you're dealing with a cubic equation.\n\ny = ax^3 + bx^2 + cx + d\n\nSubstitute in four points to get a 4 x 4 system of equations:\n\n1 = d\n2 = a + b + c + d\n4 = 8a + 4b + 2c + d\n8 = 27a + 9b + 3c + d\n\nYou can solve this either by hand or using matrices, but it should give you a = 1/6, b = 0, c = 5/6, d = 1 if it matches the result that you say is the answer.
82007
Just type in quadratic equation and maybe roots in a serch engine. You will have your pick of dozens of sites.
803954
check the discriminant:\n81-4(5) = 81 -20 =61 >0,\nso there are 2 real solutions to this problem\n`
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
293371
The equation has 0 real solutions, but 2 complex solutions.\n\nThere is a very fundamental theorem of algebra which states that any polynomial equation of degree n has n roots. In your case, the degree is 2, so the equation has to have 2 roots.\n\nHowever, roots can be either real, complex, or repeated roots. For example, x^2=0 appears to only have one solutions (x=0) but x=0 is a repeated root, occurring twice. For polynomials of degree 2, you can find out about the solutions by using the discriminant. An equation of the form ax^2+bx+c=0 has discriminant b^2-4ac. The discriminant can be one of three things:\n\n>0 This means that both roots are REAL roots.\n=0 This means that there is ONE REAL root, and it is a repeated root.\n<0 This means that both roots are COMPLEX roots and there is no real solution.\n\nThe methods for general degree n polynomials are much more difficult, if not impossible, to classify.
463511
Use the double angle identities.
803558
1.\nc=(xy^2-1)/y\ndifrentiate w.r.t x\n(y(y^2 + 2xyp)-p(xy^2-1))/y^2=0\ny^3 + 2xy^2p - pxy^2 + p=0\ny^3 +xy^2p +p=0\nwhere p=(dy/dx)\n2.\nsimilarly diffrentiate \n(x+y^2)/x^2=-c\nto get\n2(x^2)yp - x^2 - 2xy^2=0\nthanks
750837
1--d\n2--a\n3--b\n4--a
190850
also try  http://mathworld.wolfram.com/ImplicitDifferentiation.html
143223
a) (6x-7y)(6x+7y)\nb) y(4xy-7)\nc) (x-8)(x+3)
933066
Look for a pattern, then write your equation to fit the pattern.  One typical way is to find the differences between each consecutive pair of terms (called the first differences) and then find the difference of consecutive first differences (called second differences).  If the second differences are all the same, you know you have a quadratic pattern, with x^2 in it.
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
846447
http://www.bbc.co.uk/schools/gcsebitesize/chemistry/classifyingmaterials/formulae_and_equationsrev4.shtml