There are three real answers (which can be shown through graphing). Two of them are 2 and 4. The other one is a negative real number. I know how to find it numerically (through successive approximation), but would like to know a way of solve this and other similar equations analytically. \n\nRelated question: How to call this type of equations, exponential-polynomial?\n\nExtension: How to solve for complex solutions of the equation as well?
210358
I believe that you will need to use Numeriacl Analysis to solve an equation such as this. It has been many, many years since I was a student in Math but I recall that your approach seems to be the way to go. Your method of graphing was a good analytic approach. Next using the successive approx was good. \n\nAs an alternative, You could try to expand 2^x- x^2=0 into a series and see if there is an answer in closed form. You will have to show limits etc.  I think that you will discover that you can't solve this in closed form. The equation that you have to solve is:\nx^2-exp(x*ln(2))=0\n\nAn example of functions that can't get a closed form answer is trying to integrate this function: \nf(x)=exp(x^2). You will discover that you can Numerically integrate the exp(x^2) but can't get an answer in Closed Form.\n\nMy conclusion is that you can't solve the third solution in closed form (analytically). The third answer seems to be bwtween -1 and -1/2. I did this quickly in my head. Your graph probably shows this if my calculations in my head are correct. So, if you try to solve this using one of the Numerical Analysis Methods, start the first approx with maybe -.75. You can do this with a Calculator. If you have Fortran or basic on your PC, you can write a simple program. When I went to school, we used Fortran & Basic. Is that used today? Or do you use Pascal? Well, you can write a simple program to solve this using your Computer.
486304
all the questions relate to the same basic problem. When a function p(x) changes sign in the interval [a,b] it has to cross the real line (y=0). Therefore it has a real solution there. So, first question is solved since p(1) is negative and p(2) is positive.\n\nthe second question has the same answer, only more complicated but just a change of sign when you consider the polynomial function f(x) - g(x).\n\nthird. all non-real solutions come in pairs therefore an odd degree polynomial has at least one real solution since the number of solutions equals the number of roots. (Fundamental theorem of algebra.)
493851
16x^2 = 8x - 1\n16x^2 - 8x = -1\n16x^2 - 8x + 1 = 0\nFactorise the polynomial to get all possible values of 'x'\n16x^2 - 8x + 1 = 0\n16x^2 - 4x - 4x + 1 = 0\n4x(4x - 1) -1(4x - 1) = 0\n(4x - 1)(4x - 1) = 0\nObviously 4x - 1 =0\n4x - 1=0\n4x = 1\nx = 1/4
326735
a) Let x+y = q -----eqn1\n       x² - 2x + 2y² = 3 -----eqn2\n\nsub x= q-y into eqn2:\n(q-y)^2 -2(q-y) + 2y^2 =3\n\nq^2 -2qy +y^2 -2q +2y +2y^2 =3\n\ncomparing the coefficients of the eqn with that of a standard quadratic eqn, ax^2 +bx +c, we have\na= 3\nb= -2q+2\nc= q^2 -2q -3\n\nAnd to check for the nature of the roots of this eqn, we need to find the determinant, b^2 -4ac.\n\n(-2q +2)^2 -4(3)(q^2 -2q -3)\n\nsolving this and you will eventually get \nq^2 -2q -5\n\nHowever, given that q^2 > 2q +5, we know that q^2 -2q -5 > 0.\n\nSince the determinant is greater than 0, it shows that there are 2 real and distinct roots of the eqn given.\n\nb) the approach is the same, but you must know that since the two curves dont intersect, we would need to find or show b^2 -4ac < 0, ie the determinant is negative.
449036
well assuming you just want to break it down...\n1: you put the -2 power on both the 4 and a^3.  then it'd be (4^-2)(a^3^-2).\n2: remember that when it's an exponent to an exponent you multiply the numbers together so it becomes (4^-2)(a^-6)\n3: a negative exponent means it is one over itself so it would be 1/(4^2)(a^6)\n4: just simply from there to make it 1/ (16a^6)\n\nhopefully that was what you were asking!!
183477
Use remainder theorem and polynomial division for the first one and then either the quadratic formula or, if you know your maths, logics. Just plot the second one in a graph and you'll see the answer. The next few lines I just don't understand what you are writing, perhaps if you wrote it more clearly. As for the simplify bits, you don't really need help there, it's a matter of seeing how you can create similar terms. The last bit, once again, I do not know what you mean. \n\nEdit: corrected some spelling mistakes (it's quite late over here).
344775
I think you are looking too much into it, but here we go. \n\nIf you have the equation x^2 and say we pick points x= -1, and 1. \n\nWhen you plug the x's into the equation you will get a positive answer. This is why whenever the exponent is even, it will always bounce off the x- axis, because there is no way for it to go negative.\n\nfor the second question, say we take x^2 and plug in the same points above.\n\n-1 ^ 3 gives you -1. This would mean that ever x = -# will have a negative y value which means it will cross the x- axis\n\nTry it with more equations and you will understand
492617
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
501967
For the first one, the zero exponent should not be disregarded.  2x^0 is just one so it just eliminates that second 2x.  For the cubed part, the three needs to be distributed to the 2 and the x.  Is the negative one only on the x or is it on both the 2 and the x?  If it is only on the x, then your answer will be (2^3)(x^-3)=8x^-3 or 8/x^3.  If it is on both then you have (2^-3)(x^-3) which is 1/(8x).\n\nFor the second one, first flip the fraction, you get (4/3)^2 and then distribute the power to both the top and the bottom, youget (16/9).\n\nTHE NEGATIVE IN THE EXPONENT HAS NOTHING TO DO WITH A NEGATIVE IN THE BASE.  So just because there is a negative power, DO NOT put a minus sign in the base.\n\nFor the third one, just solve for y.\ny+2=3/8(x-16)\nget rid of the parenthesis and we have\ny+2=3/8x-6\nand then subtract 2 from both sides\ny=3/8x-2
463871
D ( 2 or 0)\n\nThis is because the coefficients of the given cubic polynomial are real numbers and so any real zero needs be accompanied by its complex conjugate and so as stated  earlier MUST come in pairs so either there are 2 or there cannot be any.
616454
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
499316
The discriminant is b^2 - 4ac. If it looks familiar, it's in the quadratic formula.\n\nThe quadratic is in the form ax^2 + bx + c. Plug in your values for a, b, and c.\n\nIf your result is:\n....positive, there are two real solutions\n...zero, there is one real solution (that is repeated)\n...negative, there are two imaginary solutions.\n\nHope that helped.
136288
Hello.  First you have to determine if you are going to find 'real' answers.  use b^2 - 4*a*c, where b=-3, a=1, c=5 (from ax^2 + bx + c).  \n\nNow, b^2 - 4*a*c = 3^2 - (4*1*5) = 9-20=-11, hence, you will have imaginary roots in your answer.\n\nNote that sqrt(-1) = i, then,\n\nx^2 - 3x + 5 = 0\n(x - [3/2])^2 + 5 - (3/2)^2 = 0 ;(Completing the roots)\n(x - [3/2])^2 = -11/4\n(x - [3/2])^2 = (i^2) * 11/4 ; (note that i^2 = -1)\n\nx - 3/2 = +/- i * 0.5 * sqrt(11)\nx = (3/2) +/- i * 0.5 * sqrt(11)\n\n------------\nAlways remember to check if b^2 - 4*a*c > 0 or < 0 then you will know if there is imaginary i in the answer..
90163
1) \nxy - x + 4y - 4\nx(y-1) + 4(y-1)\n(x+4)(y-1)\n\n2)note: x^3 means x cubed; x^2 means x squared\nx^3 -6x^2 + 2x - 12\nx^2(x-6) + 2(x-6)\n(x^2 + 2)(x-6)\n\n3)\n6tx + r - 3t - 2xr\n3t(2x - 1) - r(2x - 1)\n(3t - r)(2x - 1)\n\n4)Are you sure it wasn't -x^2 - 2x - xy - 2y?\n-x^2 - 2x - xy - 2y\n-x(x + 2) - y(x + 2)\n(-x - y)(x + 2)\n\nTo "factor by grouping," you must split the original polynomial that you are given into two different binomials by "grouping" them together.  This is done with perenthesis in the problems above.  Then, you factor out the "greatest common factor" from each of the binomials.  After factoring, the two quantities still within the perenthesis should be the same.  You could call these quantities the "common binomial."  Lastly, you factor out the common binomial to get the final answer.
535145
sin(X)+sin(2x)+sin(3x)+....+sin(nx)\n= img[exp(ix)+exp(i2x)+exp(i3x)+...+exp(inx)]\n=img[(exp(i(n+1)x) -1) / (exp(ix) - 1)]\n=img[(exp(i(n+1)(x/2)) -1) /exp(i(n+1)(x/2)) * (exp(i(n+1)(x/2)) -exp(-i(n+1)(x/2))) / (exp(i(x/2)) - exp(-i(x/2)))\n=[sin((1/2)(n+1)x)/sin((1/2)x)]* img(exp(inx/2))\n=[sin((1/2)nx)/sin((1/2)x)]*[sin((1/2)(n+1)x)/sin((1/2)x)]/[sin((1/2)x]\n\n[exp is exponential, img is imaginary part, 'i' is usual imaginary i in complex numbers] I also suggest you write down this so that you will be more clear with the equations. Sorry but without equation editorits bit difficult to type math.\n\nIf you want an infinite sum, please take the limit as x-->infinity of the last term obtained.\n\nI hope this answeres your question. All the sest.
96179
x^4 + 4\n\nWell, you can add 0 in the form of +a - a, so\nx^4 + 4 +4x^2 - 4x^2\n\nRegrouping them, we get\nx^4 + 4x^2 + 4 - 4x^2\n\nUsing factoring, we get\n(x^2 + 2)^2 - (2x)^2\n\nThen,\n[x^2 + 2 + 2x][x^2 + 2 - 2x]\n\nRearranging them for a more beautiful sight, we get\n{x^2 + 2x + 2)(x^2 - 2x + 2)\n\nThanks For the Problem ^_^
395910
where did you get THIS problem?\nThe (real-valued) solutions are:\n\n3π/2 + 2πn and ≈ 2.945023353 + 2πn where n = 1,2,3,...\n(radians, of course)\n\n2sin2x - 2cosx - sinx = 1\n4sin(x)cos(x) - 2cos(x) - sin(x) = 1\n2cos(x)*[2sin(x) - 1] - sin(x) = 1\nlet y = sin(x) then cos(x) = √(1-sin^2(x)) = √(1-y^2)\nso,\n2√(1-y^2) *[2y - 1] - y = 1\nor\n2√(1-y^2) *[2y - 1] = 1 + y\nsquaring both sides,\n4(1 - y^2)(2y - 1)^2 = (1 + y)^2\n4(1 - y)(1 + y)(4y^2 - 4y + 1) = (1 + y)^2\nwe can divide both side by (1+y) noting that this shows that\ny = -1 is a solution, or sin(x) = -1, x = 3π/2 + 2πn ; n = 1,2,3,...\n\nthen we have\n4(1 - y)(4y^2 - 4y + 1) = (1 + y)\nexpanding,\n16y^3 - 32y^2 + 21y - 3 = 0\n\nThis is why I asked where you got this problem.  This cubic equation is not easily solved.  There are two complex conjugate solutions as well as y ≈ 0.195305854...(I wont show a derivation.  Use Mathematica or something similar. Plug it in-it works. You could also use Cardano's formula to find an exact value but I'm sure not doing that here and now.  Maybe if you paid me.)\n\nSo, sin(x) ≈ 0.195305854\nThen x ≈ 0.1965696 or π - 0.1965696 ≈ 2.945023353 \nBut the first one gets rejected since cos(x) must be negative.(This spurious solution comes from squaring the original equation)  Also, we can add even multiples of 2π to the solution to get another.  So,\nx ≈ 2.945023353 + 2πn where n = 1,2,3,...\n\nThese are the real solutions.  As I indicated, there are two complex conjugate solutions which will yield additional complex solutions for x. \nsin(x) ≈ 0.90234707 ± 0.3818408i\n\nYou can figure out x in these cases!!!