Rational equation help?
829201
x+1 / 2x = 2/3\nmultiply both sides by 2x; giving:\nx+1 = 4x/3\nmultiply both sides by 3; giving:\n3x +3 = 4x\nsubtract 3x from both sides; giving:\n3 = x\n____________\n9/2x-1 = 3/x\nmultiply both sides by 2x-1; giving:\n9 = (6x-3)/x\nmultiply both sides by x; giving\n9x = 6x-3\nsubtract 6x from both sides; giving:\n3x = -3\ndivide both sides by 3; giving:\nx = -1
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
858168
1. 14x^2 +21x\n\na good way to start would be to factor out any common terms:\n\n7(2x^2 + 3x)\n\n7x(2x + 3)
941825
First, check for rational roots. If this doesn't have\nany we are in trouble, but fortunately it does have some.\nIf r/s is a rational root of this equation, then r must\ndivide -35 and s must divide 2.\nBy trial we find that -1 is  a root of this equation,\nand so x + 1 is a factor. But x+1 is also a root\nof the quotient, so x+1 is a double root.\nNow work on the resulting cubic.\nWe  find that x = 5/2 is one of its roots.\nThe final quadratic is\nx² - 4x + 7 = 0\nwhich you can now solve by the quadratic formula.\nHope that helps a bit!
729449
It will not factorise in the rational number field so there are no rational solutions\n\nIrrational roots are -1.72107 and 0.269209 (approximately) using Quickmath
813596
Okay, the goal here is to get values of x that fit the equation.  These are not like easy linear ones, they have x^2 in them.  We can still solve them.  \n\nx^2=16\nWhat is the reverse sign of the square?  It's the square root.  Square root both sides and get\nx=sq(16)\nSo x=+/-4\nBoth work because (-4)^2 becomes 16 as well.  Those are your answers.\n\n\n2x(x+3)=0\nThe goal here is to make this equation equal zero.  Do you see how there are two "factors?"  Things that are multiplied together?  Well, to equal zero, one of them must equal zero.  Either one can equal zero, so you have different solutions that work.\n\nTake 2x:\n2x=0\nx=0\n\nNow x+3:\nx+3=0\nx=-3\nSo\nx=0,-3\n\n\nHere we have to "factor" this equation:  Find two things that, when multiplied together, equal this.  y^2 and 8y both have y, so factor out the y and get:\ny(y+8) \nWe can do as we did before, and get\ny=0\nand \ny+8=0, so\ny=-8\nOur answers are\ny=0,-8\n\nThe next two are harder:\n\nFirst bring the 28 over to the other side:\n\nx^2+3x-28=0  \n\nThe factored form of this actually has two factors, but none with just x.  To factor this, we need two factors of -28.  When one of them is multiplied by x and added to the other one multiplied by x, they equal 3x.  Now, a negative times a positive is a negative so we are looking for one positive and one negative factor of 28.  Let's try these two:\n\n7 and -4.\n7(-4)=-28\n7x+(-4x) = 7x-4x=3x!\n\n-4 and 7 are our factors, so:\n(x+7)(x-4) is our factored form.  Solve the components:\nx+7=0\nx=-7\n\nx-4=0\nx=4\n\n x=-7,4 are our two answers.\n\n\n\n2x^2+9x-5\n\n\nThe next one is similar, but more complicated.  We have a 2x^2.  No matter.  We need to find two factors of -5, and one of them will be multiplied by 2x instead of x.  The only factors of -5 are +/-1, and +/-5.  It looks like the 5 will be positive since 9x is fairly big and positive.  Let's try -1 and 5:\n2x(5)=10x\nx(-1)=-x\n10x+(-x)= 10x-x=9x!\n\nSo 2x will be multiplied by 5, and x by -1.  In factored form, this is:\n\n (2x-1)(x+5)\n\nBecause you use FOIL (first outside, inside last) so the 2x is not multiplied by -1, but by 5.  Now solve these components.\n\n2x-1=0\n2x=1\nx=1/2\n\nx+5=0\nx=-5\nSo\n x=1/2, -5 are our two answers. \n\nAll done.  Hope this helps.
1130519
Seperate the constants - you get:\n\n2x = -3\n\nDivide both sides by 2\n\n\nx = -3/2\n\ntake the cube root of -3/2\n\nIts an irrational about -1.2599
60599
This is not an equation.  Its garbage.  Both side of the equation have to equal each other.\n-35=.000075 this does not compute
508029
The equation is:\n(x+a+b+c)^2 = 2x^2+(x+a)^2-(x+b)^2-(x+c)^2\nA simple rearrangement yields\n(x+a+b+c)^2 - x^2 = (x+a)^2-(x+b)^2 + x^2-(x+c)^2\nFactorize using the identity a^2-b^2=(a+b)(a-b)\n(x+a+b+c+x)(x+a+b+c-x) = (x+a+x+b)(x+a-x-b) + (x+x+c)(x-x-c)\nwhich can be written as\n(2x+a+b+c)(a+b+c) = (2x+a+b)(a-b)-c(2x+c)\n==> 2x(a+b+c)+(a+b+c)^2 = 2x(a-b)+(a-b)(a+b)-2xc-c^2\n==> 2x(a+b+c-a+b+c) = (a+b+c)^2 + a^2-b^2-c^2\n==> 4x(b+c) = a^2+b^2+c^2+2ab+2bc+2ca+a^2-b^2-c^2\n            = 2(a^2+ab+bc+ca) = 2(a+b)(c+a)\n==> x = (a+b)(c+a)/[2*(b+c)]
202779
This is quadratic. A quadratic equation looks like:\nax^2+bx+c = 0\n\nSo a=4, b=-1, and c=3\n\nSo by using the quadratic equation: \n\nx= (-b+sqrt(b^2-4ac))/2a\nand\nx= (-b-sqrt(b^2-4ac))/2a\n\nTry to get the answer from that.... note that imaginary numbers may be involved!
859744
(x + 4y)(x - 4y)\nI answered 2 of your questions. Do the rest of the questions yourself.
791694
The rational roots of any polynomial:\n\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) \n\nare in the form ±p/q, where p divides a(0) and q divides a(n).\nIn your case, a(0) = -1 and  a(n) = 1, so the only rational roots possible are ±1.\nSubstitute ±1 into the equation to see if either is a root. (neither is)\n\nExample:\nx^3 + 4x^2 + 2x - 4 = 0\nThe only possible rational roots are ±1, ±2, ±4 (These numbers divide a(0), and only 1 divides a(3)=1 ).  Substituting each of these into the equation (synthetic substitution works well), we find -2 the only rational root.\n\n*****************************************\nProof by contradiction.\nLet the rational number p/q be a reduced root (i.e, q  and p are relatively prime integers) to the rational polynomial equation:\na(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) = 0\nand such that either p does not divide a(0) or q does not divide a(n).  [Note: without loss of generality we can assume all coefficients are integers by multiplying both sides by the least common multiple of the denominators of all the a(n)'s]\n\nThen we know that \na(n)p^n/q^n + a(n-1)p^(n-1)/q^(n-1) + ... + a(1)p/q + a(0) = 0\nmultiply both sides by q^n to get:\n\na(n)p^n + a(n-1) qp(n-1) + a(n-2) q^2p^(n-2) + ... + a(1) q^(n-1)p \n+ a(0)q^n = 0\n\nor a(n)p^n + q * f(a,p,q) = 0 for some function f of a,p,q.\nand a(0)q^n + p * g(a,p,q) = 0 for some function g of a,p,q.\n\nThis means \na(n)p^n = -q * f(a,p,q)\na(0)q^n = -p * g(a,p,q)\n\nBy the first equation, q must divide a(n) * p^n.\nBut q has no factor in common with p, so it cannot have a factor in common with p^n, so all the factors of q (ie, q itself) must divide a(n), and we assumed that q does not divide a(n), so p must divide a(0) * q^n (the only possibility left)\nBut likewise with the second equation, p must divide a(0) * q^n. But p does not have any common factors with q, so it cannot have common factors with q^n, so p must divide a(0) and we assumed that p does not divide a(0), so a contradiction arises.\nTherefore no such p/q can exist, therefore the roots are irrational.\n\nQ:\nwhy shouldn't q divide a(n) or p divide a(0)?\n\nBecause otherwise if both did then the roots COULD be rational (p/q could be a root).  If thats the case you prove/disprove it by example.  In other words, its easy to show that in your example the roots are all irrational because the only POSSIBLE rational ones are 1 and -1.  So you just substitute them in to see that they're not roots. My proof just shows that IF there are rational roots, they must be p/q where p|a(0) and q|a(n).\n\nThere is no general way to determine  by inspection that a random polynomial has all irrational roots. The only way is to check the POSSIBLE rational ones by substitution.
974937
please complete the equation
530142
use the quadratic equation:\nx=-b +/- the square root of b^2 - 4ac all over 2a\na=1,b=7,c=10 so plug it in\nx=-7 +/- the square root of 49-40 all over 2\n-7 +3 = -4 divided by 2 x=-2\n-7-3=-10 divided by 2 = -5\nso x = -2 or x=-5
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
754366
You need to use the quadratic equation as it is not factorable into rational expressions.\nThe form of the equation is:\nax^2 + bx + c = 0 \nSo in your case,\na = -2, b = 6, c = -3\nand the solution is:\nx = [ -b ± √(b^2 - 4ac) ] / (2a)\n\nso substitute and reduce.\n\nx = [ -6 ± √(6^2 - 4(-2)(-3)) ] / (2(-2))\n\nx = [ -6 ± √(36 - 24) ] / (-4)\n\n= [ -6 ± √12 ] / (-4)\n= [ -6 ± 2√3 ] / (-4) \n= (3 ± √3)/2
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too