what are examples of quadratic equations or applications with imaginary numbers?
83855
y = a*x^2 + b*x + c\n\nThat is, y = constant times x squared plus constant times x plus a third constant. The solution is Eqn 5 on the link below.\n\nIf the stuff under the radical in the solution is negative then you will have an imaginary number on your hands.
953700
The discriminant\n\n√b² - 4ac portion of the quadratic Formula\n\nx = - b ± √b² - 4ac/ 2a is negative , no real number solutions \n\n- - - - - -s-
371301
1. Solving AC circuits that have resistors, inductors, and capacitors in them.\n2. Signal processing\n3. Solving differential equations.\n\nA smart guy by the name of Euler showed that e^(i*x) = cos(x) + i*sin(x), thus allowing the conversion of a complex number in rectangular form (a + b*i) into polar form (r*e^(i*x)). It also allows us to write sinusoids in terms of complex exponentials, as in:\ncos(x) = (1/2)(e^(i*x)-e^(-i*x))\nIt turns out, perhaps unexpectedly, that complex exponentials are oftentimes easier to deal with then sinusoids. It also turns out, according to a guy by the name of Fourier, that most "well-behaved" functions can be written in terms of sinusoids (and thus can be written in terms of complex exponentials as well). This paved the way for a lot of neat applications. Google "complex exponential" (or any of the terms or names I dropped here) if you want more details.
410829
The commonly taught method for computing square roots with pencil and paper can be found in the Wikipedia article, "Methods of computing square roots".\n\nScroll down to the heading "An exact long-division-like algorithm".
654434
Yes, there is. The square root of negative 1 is the number i, which is an "Imaginary Number." \n\nDespite the name, imaginary numbers are real quantities that are used in equations to get real answers. However, they are a somewhat confusing concept, mathematically. \n\nOne way to visualize them is to consider that multiplication by an imaginary number rotates a function orthogonally on a number axis. The answers given by performing this procedure once may not seem to be applicable to real situatons, but if you multiply by an imaginary number again, in rotates the function back onto the initial axis, yielding real number results. \n\nI am including links to a number of good sites on imaginary numbers. Although it can be difficult to understand the implications of imaginary numbers beyond the fact that their squares are negative, if you spend some time on these sites, I'm sure you'll come to understand them.
713002
Quite an old example, and one of the best-known, is the so-called Pell equation X^2 - D.Y^2 = -1 or +1, where D is a square-free integer and X, Y are both to be integers.  For example, when D = 3, there is no solution when -1 is on the right-hand side, but the solutions when it is +1 are (X, Y) = (2, 1), (7, 4), (26, 15), (97, 56), . . . an infinite series of integer pairs constructed by a simple rule starting from the first one (X' = 2X + 3Y, Y' = X + 2Y).
820932
Logarithms, when applied to imaginary numbers can have many results. This is because exponentiation, when applied to imaginaries, is defined as a trig function:\n\ne^(ix) = cos x + i sin x\n\nSo, the logarithm is an inverse trig function. And we know inverse trig functions can have many results. ln e^(2 pi i) can be 2 pi i, but it could also be 0, or -2 pi i, or basically any ix where (cos x + i sin x = 1). In this case 0 fits with the real logarithm on the right side, since ln(1) = 0.
1087511
This is a cubic equation, not a quadratic. Most of the time, it will be very difficult (although never impossible) to find the solution. In this case, it so happens to be straitforward because you know that x=9 is a solution. That means that the polynomial \n\nP(x)=x^3-25x-504 can be rewritten as:\n\nP(x)=(x-9)Q(x), where Q is a polynomial of degree 2. So now that you know that x=9 is a solution, your next task is to find Q(x) and solve Q(x)=0. Indeed, in order for a product of factors to be =0, it is sufficient for any product to be =0. \n\nTo find Q(x), you divide P(x) by x-9. Doing this gives:\n\nQ(x)=x^2+9x+56. So now you are solving the quadratic equation\n\nx^2+9x+56=0.\n\nDiscriminant: d=9^2-4*56= some negative number.\nSince d<0, Q(x)=0 has no real solutions. (It does however have imaginary solutions-- if you are interested in those, simply go ahead with the computation).\n\nSince Q(x)=0 has no solutions, P(x)=0 if and only if x=9. As a result, your equation has one and ONLY ONE real solution, namely x=9.\n\n----\nNow this method does not always work, because most of the time it is not possible to eyeball one possible solution to the equation (here we "eyeballed" that x=9). In those situations, there is a much more obscure and messy method of computing the solutions, and this method is known as Tartaglia's method. \n\nIf you are interested, look at the following pdf file on the web:\nhttp://www.m-a.org.uk/docs/library/2059.pdf
839760
The roots are 0 and -16/5\n\nTo solve any quadratic equation eg. ax^2+bx+c, you can apply formula x= (-b+-sqrt(b^2-4ac))/2a
458634
The equation becomes x^(-4)=32+22; or x^(-4)=54; x^4=1/54\n.: x^2=±1/√54; so that x=±1/∜54 are two real roots - that is regular numbers. There are two more roots, x=(±1/∜54).√(-1), which are special numbers called imaginary numbers.\n\nQuadratic eqn is an equ of the form ax^2+bx+c=0. where a, b, c are known, x is unknown. if you put x^(-2)=u, the eqn becomes u^2=54 which is a quadratic eqn. solve it and put back the value of u. What you get can be written as another pair of quadratic eqn. Solve this to get the value of x. The end.
750403
I have to go teach my class in about 4 minutes, but when I return, if no one has done it, I can show you how.  It requires factoring out a specific value that makes the coefficients "a" and "b" turn into two values that satisfy x^2 + y^2 = 1.  In other words, I know it starts out by factoring out sqrt(a^2 + b^2) from \n\na sin(Bx)+b cos(Bx)\n\n\nto get \n\n\nsqrt(a^2 + b^2) [ a/sqrt(a^2 + b^2) * sin (Bx) + b/sqrt(a^2 + b^2) * cos(Bx)]\n\n\nEDIT: Okay here we go.  And the "orthogonal rotating vectors" proof below is usually over most people's heads, or at least, most people who would ask this question on Yahoo answers.\n\n\nFirst, you factor out SQRT(a^2+b^2) as I mentioned above.  For the sake of simplicity, I will let A = SQRT(a^2+b^2).  So factor A out of  \n\na sin(Bx)+b cos(Bx)\n\nto get\n\nA [a/A sin(Bx) + b/A cos(Bx)]\n\n\n\n\nSince (a/A)^2 + (b/A)^2 = 1 (you might need to actually verify this for your purposes), there exists some angle C such that\n\ncos (C) = a/A and sin (C) = b/A \n\n(also, you should point out that both a/A and b/A have absolute values less than one, so that C is real and not complex with a nonzero imaginary part).\n\nSo make the substitutions to get\n\nA [cos(C) sin(Bx) +  sin(C) cos(Bx)].\n\n\n\n\n\nAccording to the formula \n\nsin(y+z) = sin y cos z + cos y sin z, \n\nthe expression in the brackets must equal sin(Bx + C).  So now the whole expression becomes\n\nA sin (Bx + C).\n\n\n\n\n\nSo in conclusion, \n\na sin(Bx)+b cos(Bx) = Asin(Bx+C),\n\nwhere A = SQRT(a^2+b^2), and C = arccos(a/A).  \n\n\n\n\nThere is a better representation for C, however.  \n\nSince cos C = a/A and sin C = b/A, you have\n\ntan C = (sin C)/(cos C) = (b/A)/(a/A) = b/a.\n\nThus, tan C = b/a, and C = arctan (b/a).  This definition of C does not rely on the creation of a third variable (A).\n\n\n\n\n\nHope this helps!
309372
If you have a quadratic equation of the form ax squared + bx + c = 0, the discriminant is\nb squared minus 4 times a times c, or b^2 - 4ac\n\nIt's the part under the square root in the Quadratic Formula.  It tells if the roots  are real, rational, or imaginary.\n\nSo in your first example, it's 9 - 4X1X-4, or 9 - -16, or 25.
545628
---THIS IS THE CORRECT ANSWER!!---\n\nIn algebra, the letter "i" relates to the term imaginary.  "i" actually is a shorthand for the square root of -1. Real numbers includes any number positive, negative, or zero, rational or irrational. Basically, any possible number you can think of on a number line. A non-real number would then be imaginary, a complex number whose square is a negative real number.
992331
The trick here is to note that it looks like a quadratic formula.  Let's do a little manuevering to see this:\n\nx^4 = (x^2)^2\n\nwhich means that we can re-write the problem as (x^2)^2 + x^2 + 1.  Now, substitute another letter, say t for x^2.  We get:\n\nt^2 + t + 1\n\nAll I've done is replace the two x^2 values with 't'.  Now, it's just a quadratic equation of the form:\n\nax^2 + bx + c\n\nTo factor this, you'll need to find two values that multiply to the last term (1, in this case) and add to the coefficient of the second term (which is also 1, in this case).  \n\nWith a little investigation, you'll find that two integers do not exist with such properties, which for most math books, means that, though the problem has a solution, it can't be factored.\n\n***THE ANSWER STOPS HERE if you haven't learned about imaginary numbers.***\n\n*However*, in case you really want to factor this, you simply use the quadratic formula to find the values of x that solve this equation:\n\nt = (-b +/- sqrt(b^2 - 4ac))/2a\n\nSo, we get that t = (-1 + sqrt(-3))/2 and t = (-1 - sqrt(-3))/2, which are imaginary numbers.  These two solutions I will call solutionone and solutiontwo.  \n\nSo the factorization for f(t) is:\n\n(t - solutionone)(t - solutiontwo) where solutionone and solutiontwo are the values that come out of the quadratic formula:\n\nThen, you plug x^2 back in for t and get:\n\n(x^2 - solutionone)(x^2 - solutiontwo)\n\nThese need to be factored as well (ugh) because a linear factorization implies that all the factors have x instead of x^2.  \n\nThank goodness, though, that these are really easy.\n\nIt is always the case that x^2 - a can be factored like this:\n\n(x - sqrt(a))(x + sqrt(a)).\n\nAlways.\n\nSo, this means that the complete factorization for the problem is:\n\n(x - sqrt(solutionone))(x + sqrt(solutionone))(x - sqrt(solutiontwo))(x + sqrt(solutiontwo))\n\nYeow!  Good luck!
394036
Try to solve the eqs with quad rule:\na*x^2 + b*x + c=0\nx=(-b +- Sqrt(b^2-4*a*c))/(2*a)\n\nx^2+2x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+5=0 => Get two Complex answers, \nso no it can't be factorized.\n\nx^2+4x+3=(x+3)*(x+1)=0; \nSo, yes, this can be factorized. \nSolutions are x=-3 and x= -1\n\nx^2+2x-3=(x+3)*(x-1)=0;\nSo, yes, this can be factorized. \nSols are x= -3 and x=1
26326
A procedure for finding the quadratic factors for the complex conjugate roots  of a polynomial P(x) with real coefficients.\n\n[x-(a+ib)][x-(a-ib)]=\nx^2+2ax+(a^2+b^2)=\nx^2+Bx+C.
435410
2i or -2i, where i = sqrt(-1)\n\nIf you haven't learned imaginary numbers, then your answer would be that the sqrt(-4) is not a real number.