How many real roots does the following equation have:\n\nx^5 + 7*x^3 + 3*x + 2 = 0.\n\na)0 \nb)1 \nc)3 \nd)5\n\nHow to solve this kind of question ?\n\nis there any rule /formula ?
198810
The idea is Descartes Rule of Signs.\n\nThe mathematican Descartes ( who invented coordinate geometry by the way ), introduced a "technique" by which one can possibly tell how many real roots to expect.  \n\nIts best illustrated by an example.\n\nSay Use Descartes' Rule of Signs to determine the number of real zeroes of:\n\nf (x) = x^5 – x^4 + 3x^3 + 9x^2 – x + 5. \n\nFor positive zeroes ( roots ) of f(x).\n\nThere are four sign changes. This number "four" is the maximum possible number of positive zeroes (x-intercepts) for the polynomial  f (x) = x5 – x4 + 3x3 + 9x2 – x + 5. However, remember that some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex. Then it may be that certain pairs of roots are not real, and therefore are not graphable as x-intercepts. Because of this possibility, you have to count down by two's. That is, while there may be as many as four real zeroes, there might also be only two, and there might also be zero (none at all).\n\nFor negative zeroes, you need to replace f(x) by f(-x).\n\nf (–x) = (–x)5 – (–x)4 + 3(–x)3 + 9(–x)2 – (–x) + 5\n\nf (–x) =  –x5 – x4 – 3x3 + 9x2 + x + 5\n\nThere is only one sign change, so there is exactly one negative root. In this case, you don't count down by two's, because you would end up with a negative number.\n\nSo the possibilities are :\nThere are 4, 2, or 0 positive roots, and exactly 1 negative root.\n\nOK. Thats basic background work. \n\nNow, for your question, there is no positive real roots for there is no sign changes.\n\nf( - x ) = (-x)^5 + 7*(-x)^3 + 3*(-x) + 2 \n         = -x^5 - 7*x^3 - 3*x + 2\n \nSo, there is one "sign" change. From [- 3*x ] to [ + 2 ].\n\nHence, we can conclude that the given equation has exactly one negative root.\n\nTo enumerate this root, is an altogether different matter.\n\nALTERNATE SOLUTION.( following your book )\n\nThe solution above is if you dont know calculus.\n\nWith calculus,\n\nf(x) is continuous\nf(x) is strictly increasing since the book showed \nf'(x) > 0 for all real x.\n\nf(0) = + 2  is positive\nf(-1)= - 9  is negative\n\nHence,\n=> we have one root in (-1, 0)\nWhy? Because think about it. This kind of deduction is only true if f(x) is continous. Continuity means that : \nwhen x = -1, f = -9 . So starting from a negative value, as x increases (from -1 to 0) , x = 0, f = 2, means that f has "increased from" -9 to 2. Now, by continuity, as f increased from -9 to 2, at some point say x = y, f MUST have "assumed" the value 0!. That is f(y)=0 because 0 lies between -9 and 2. And this y is described by "x increases (from -1 to 0)" and thus -1 =< y =< 0, or y falls in the interval (-1,0). \n\nWe have established the "existence" of y such that\nf(y) = 0 and -1 =< y =< 0.\n\nHow do we know they arent any other roots?\nBy the fact that f(x) is Increasing!\nWhy? Increasing means: \nas x increases f(x) increases \nor equivalently,\nas x decreases f(x) decreases\n\nSince there is only one root in -1 =< y =< 0, we need to consider and show NO roots for f(x) in x < -1 and x > 0.\n\nSo, for x < -1,\nIncreasing means (proved using calculus),\n"as x decreases f(x) decreases"\nSo, f(x) < f(-1) = - 9 < 0. \nThus for x < -1, always f(x) < 0. \nSo f(x) cannot equal zero at all, implying no root here.\n\nSo, for x > 0,\nIncreasing means (proved using calculus),\n"as x increases f(x) increases"\nSo, f(x) > f(0) = 2 > 0. \nThus for x > 0, always f(x) > 0. \nSo f(x) cannot equal zero at all, implying no root here too.\n \nConclusion is the same as before.\ny is the real root. Only one real root.\n\nAnd thats what the book assumed obvious.\n\nIf you wanted to know the nature of the roots (not just real roots),\nThe answer is 1 real root and 4 complex roots.\n\nI showed you both ways... with & without ca
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..
430373
You asked about the numerator before, didnt you get an answer?\nWith synthetic substitution, you should find that the roots of the numerator are -1, -1, -2, and 3.\nThis gives (x+1)(x+1)(x+2)(x-3)  see *.\n\nAre you sure you typed the denominator correctly?\nYou typed: \nx^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1\n(by the way, you need to insert spaces into long polynomials like this one or Yahoo does not display them after 30 spaces.  You can place the mouse cursor over them and see the complete polynomial, but many people dont know that, including the previous answerer.)\nThis is not factorable (over the rational numbers)\n\nHowever, if one digit is changed, namely if -30x^3 was written \n -20x^3, then the denominator becomes a very common one (and one that you may notice just by looking at it if you stay at this factoring stuff for a long time):\nnamely (x-1)^6 !!\n\n\n*\n1st root (-1):\n\n-1 || ..1... 1 ... -7 .. -13 ... -6\n...............-1 .... 0 ..... 7 ..... 6\n————————————\n..........1....0.....-7.....-6... || 0\n\nso (x+1) is a root\n\n2nd root (-1):\n\n-1 || ...1....0.....-7.....-6\n................-1 .... 1 .... 6\n——————————\n..........1...-1 ... -6 .. || 0\n\nso (x+1) is a root again.\n\nx^2 - x - 6 remains\n= (x-3)(x+2)\n\nInteresting correlative problem:\nhttp://answers.yahoo.com/question/index;_ylt=Aqz3W3s3NPcz.j__CQYxK0bsy6IX?qid=20061127003230AA2oHuA
432261
(x²+4x+4) / (2x-3) / (x²-4) / (2x²-7x+6)\n\nYou have 4 terms and three divisions, so it's a little unclear if you mean (a/b) / (c/d) or what.  I will try to help to the extent I can.\n\nLet's assume you mean:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ]\n\nFactor your x² equations to see if you can cancel terms:\n\nx²+4x+4 = (x+2)(x+2)\nx²-4 = (x+2)(x-2)\n2x²-7x+6 = (2x-3)(x-2)\n\nWoohoo - things are looking good!  Now we can rewrite our original problem using these factors:\n\n[ (x²+4x+4) / (2x-3) ] / [ (x²-4) / (2x²-7x+6) ] = \n\n[ (x+2)(x+2) / (2x-3) ] / [ (x+2)(x-2) / (2x-3)(x-2) ] =\n \n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\nnotice we can cancel out three terms: (2x-3), (x-2) and (x+2)\n\n[ (x+2)(x+2) / (2x-3) ] * [ (2x-3)(x-2) / (x+2)(x-2) ] =\n\n(x+2)\n\nHope that helped!\n\n~ ♥ ~
248630
quadratic equations of the form \n\nax^2+bx+c=0\n\ncan be used with the quadratic formula, see\n\nhttp://mathworld.wolfram.com/QuadraticFormula.html\n\nso just transform all your equations into the form\n\nax^2+bx+c = 0 (x can be called differently, of course).\n\nEdit: for example, let's take \n\n5t^2 + 20t - 60 = 0\n\naccording to the quadratic formula, the solutions are\n\nt_1 = (-20+sqrt(20*20-4*5*(-60))) / (2*5) = (-20+sqrt(1600))/10 = 2\n\nand\n\nt_2 = (-20-sqrt(20*20-4*5*(-60))) / (2*5) = (-20-sqrt(1600))/10 = -6\n\nNote: if the term below the root (sqrt) is negative, it is not a valid solution - sometimes the quadratic equation has only one solution, or none.
508145
You have to assume that all the coefficients are real numbers. Then, if z satisfies\na_n z^n+...+a_1 z+a_0=0,\ntake the complex conjugate of the equation to get\na_n (z*)^n+..a_1 z* +a_0=0.\nIn other words, if z is a root, so is z*, so roots occur in conjugate pairs.
370965
x^2-5x-14=0\n=>x^2-7x+2x-14=0\n=>x(x-7)+2(x-7)=0\n=>(x-7)(x+2)=0\nTherefore either x-7=0 or x+2=0\nIf x-7=0,then x=7,and if x+2=0,then x=_2\nThereforex=7 or -2 ans
484220
first into the second:\n\nx^2+(x-1)^2=25\n\nexpand:\n\nx^2+x^2-2x+1=25\n\nget 0 on the right side:\n\n2*x^2 - 2*x -24 = 0\n\na=2, b=-2, c=-24\n\nuse the root finding formula for quadratoic equation:\n\nx=(-b+/-sqrt(b*b-4*a*c))/(2*a)=(2+/-sqrt(4+4*2*24))/4=0.5 +/- 3.5\n\nyour solutions are: x = -3 and 4\n\ncheck:\n(-3)^2+(-4)^2 = 9 + 15 = 24, correct\n(4)^2+(3)^2 = 15 + 9 = 24, correct\n\nsubstitute to the first eqn: y = -4 for the 1st. x and 3 for the 2nd.
535276
if by this you mean\n\nx^2 - 2x - 2 = 0\n\nx^2 - 2x = 2\nx^2 - 2x + 1 = 3\n(x - 1)^2 = 3\nx - 1 = sqrt(3)\nx = 1 ± sqrt(3)\n\n-------------------------\n\nbut if you mean\n\n7x^2 - 2x - 2 = 0\n7x^2 - 2x = 2\nx^2 - (2/7)x = (2/7)\nx^2 - (2/7)x + (1/49) = (63/49)\n(x - (1/7))^2 = (9/7)\nx - (1/7) = (3/7)sqrt(7)\nx = (1/7) ± (3/7)sqrt(7)\nx = (1/7)(1 ± 3sqrt(7))
201108
You've done 90% of work.\n\n(x - 3m)(x - 2m) = 0\nIt can take place only if either x - 3m = 0 or x - 2m = 0.\nFirst equation means that x = 3m.\nSecond means that x = 2m.\n\nSo, there are two roots of equation: 3m and 2m.\nFor m = 3 we have roots: x = 9 and x = 6.
378565
x^2 + 4x + 4=7  (first put it in the form ax^2 + bx + c = 0, by subtracting 7 from both sides of the equation)\n\nx^2 + 4x - 3 = 0   (this is called a quadratic equation)\n\nNow substitute a =1, b = 4, c = -3 in the following formula\n\nx equals -b plus or minus the square root of (b^2 - 4ac) all over 2a\n\nBecause you are solving for x^2 you expect to get 2 answers.  The plus or minus in the middle of the formula gets you 2 answers as you solve the formula for both plus and minus the square root of (b^2 minus 4ac).\n\n[If you're going to have to do this in an exam, it's probably worth learning the formula off by heart.  It's boring to have to do so but you'll probably have more confidence in the exam if you see an quadratic equation and already know the formula.]\n\nGood luck for the working out - you can do this bit!
410651
√(t + 3) + √(2t + 7) = 1\nSo √(2t + 7) = 1 - √(t + 3) \nSquare both sides\n2t + 7 = 1 - 2√(t + 3)  + t + 3\nSo t + 3 = -2√(t + 3)\n\nSquare both sides again\nt² + 6t + 9 = 4(t + 3)\nt² + 2t - 3 = 0\n(t +3)(t - 1) = 0\nt = -3 or 1\n\nTest\n\nWhen t = -3\n√(t + 3) + √(2t + 7)\n= √(-3 + 3) + √(-6 + 7)\n= 0 + 1\n= 1 YESSSSSSSSSSSSSSSS!!\n\nWhen t = 1\n√(t + 3) + √(2t + 7)\n= √(1 + 3) + √(2 + 7)\n= 2 + 3\n= 5\n≠1 NOOOOOOOOOOOOOO!!!!!\n\nSo the only solution is t = -3\n\n(The extra solutions come from the fact that other numbers can come into the solution through squaring and so the solutions must be throroughly tested)\n\njoruaishi... was lucky because\n(√(t + 3) + √(2t + 7))² = t + 3 + 2√[(t + 3)(2t + 7)] + 2t + 7\n≠ (t + 3)  +  (2t + 7)\n\nThat is like saying (√4 + √9)² = 4 + 9 = 13\nwhen in fact (√4 + √9)² = (2 + 3)² = 25\n\nThe differewnce is 12 = 2 √4 * √9\nHis worked because t = -3 rendered one of the surds zero
478942
College alg help?\nsolve the equation by factoring\n\n3x^2+4x-15=0\n\n3x^2+9x-5x-15=0\n(3x-5)(x+3)=0\n3x=5 or x=5/3\nx=3=0 or x=-3\n\n2.4\n__\n2+3i\nmultiplying by 2-3i\n8-12i/(4+9)\n=(8/13)+(-4/3)i\n\n3.square root of -25 in a+bi form\n+/-5i\n0+5i or 0-5i\n\n4.x^2-4x-3=0\nx=[4+/-rt(16+12)]/2\n=[4+/-2rt7)]/2\n=2+/-rt7\n\n5.22x+11=(x+6)^2\n22x+11=x^2+12x+36\n=>x^2-10x+25=0\n=(x-5)^2=0\nx=coincidental roots5,5
494039
Let y = x^2\n\nSo you have y^2 + 3y - 2\nYour discriminant (b² - 4ac) = 3² - 4*1*(-2) = 17\n\nSo that means you have two real roots for y^2\ny = x^2 = (-3 +/- sqrt(17))/2\ny = x^2 = 0.561552813 or -3.56155281\n\nWhen you solve for x, you have:\nx = sqrt(0.561552813) or sqrt( -3.56155281)\n\nSo you'll have two real roots:\nx = +/- 0.749368276\n\nAnd two imaginary roots:\nx = +/- 1.88720768 i
547962
1. zero means roots >> x-intercepts\n     so u let f(x) = 0\n     x - 6 = 0\n     ∴ (c) x = 6\n\n2. using the formula ( y - a )² + ( x - b )² = R²\n     given center is origin, y² + x² =  R²\n     since it passes thru ( 2 , 1 ) then we use pytha. rule to find R, which is 5^(1/2) \n      ∴ (c) the equation is .x²+y²=5
385259
Do a google search on "Descartes Rule of Signs" and you will learn all you need to know.\n\nAlso, the complex roots come in matched pairs known as "complex conjugates."  To make up an example, suppose your roots are x = 3 + 2i and x = 3 - 2i.\n\nThen x - (3 + 2i) = 0 and x - (3 - 2i) = 0.\n\n[x - (3 + 2i)] [x - (3 - 2i)] = 0\n\n[(x-3) + 2i] [(x-3) - 2i] = 0\n\n(x-3)^2 - (2i)^2 = 0\n\n(x^2 - 6x + 9) -  4i^2 = 0\n\n(x^2 - 6x + 9) - (-4) = 0\n\nx^2 - 6x + 13 = 0\n\nThis quadratic has no real roots.  Its pair of roots are called "complex conjugates."  If you graph this function y = f(x), you'll see that it never crosses the x-axis.  To get the minimum, take the derivative:\n\ndy/dx = 2x - 6 = 0  ==>  x = 3\n\nf(3) = 3^2 - 6 x 3 + 13 = 9 - 18 + 13 = 4 > 0\n\nYou can solve by the quadratic formula or by completing the square:\n\nx^2 - 6x + 9 = -13 + 9 = -4\n(x - 3)^2 = +/- sqrt(-4) = +/- 2i\nx = 3 +/- 2i\n\nThis solution may be represented as two points in the complex plane, (3, 2i) and (3, -2i), or as vectors 3i +/- i2j.  The magnitude of these vectors is sqrt(3^2 + 2^2) = sqrt(13).  Therefore, they can be expressed as complex numbers\n\nsqrt(13)(sin theta +/- i cos theta)\n\nwhere the angle theta is determined by tan theta = 2/3.\n\nFinally, to directly answer your question, "are complex roots like fake roots, but what does that mean?", I'd say that the complex roots are roots that satisfy the original equation.  If you plug in the complex roots, the original equation is satisfied.\n\nA complex number, such as 3 + 2i, typically has a real portion and an imaginary portion.  The imaginary portion is not a real number, and it involves the square root of negative one.  Nevertheless, complex numbers can be graphed in the complex plane, and complex numbers have properties that make them useful in mathematics and, for example, in the theory of electrical circuits.  If you study transient effects in electromagnetics, you learn the value of complex numbers in a big hurry.\n\nAnd one final thought about graphing.  You can graph a complex function in three dimensions, with x, y, and z coordinates.  Let x be the real portion, y the imaginary portion, and z the magnitude, i.e., sqrt(x^2 + y^2).  When you do that, you'll see that these complex solutions <do> cut some of the axes.
229974
Are you sure it is x^3 and not x^2? If you're sure, then there is one relatively easy way to solve the equation:\n\n(x^3)-4x=24\n\nNow, this equation is of the form: x^3+px=q (where p=-4, q=24). The solution to this equation was given by the Italian mathematician "Cardan" to be:\n\nx=(sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3) - (sqrt((p^3)/27 + (q^2)/4) + q/2)^(1/3)\n\n\nFor your case, if you plug in the values for p,q you will get:\n\nx = (√(-64/27+144)+12)^(1/3) - (√(-64/27+144)-12)^(1/3)\n\nx = 2.888 - (-0.4629) = 3.3434\n\n\nThus, x=3.3434 is the real solution to the equation.