how to solve cubic root with microsoft excel?
820052
I do not understand how the "problem" has to do with the question. But to answer the first part...\n\nThe cell should contain \n=POWER(number, power)\n\neg. =POWER(5, 2) would display 25\n\nA cubic root would look like this...\n=POWER(number, 1/3)
412542
x = -b +(or)- sqrt(b^2 - 4ac)\n    -------------------------\n               2a
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
411332
While Albert has found the correct solution using calculus, the question asked how to find the solution with graphing.  So, if you don't care how you find the answer, use Albert's.  But if you need to graph it to find the solution, you use the function that Albert got V(x)=x(6-2x)(8-2x) and graph the resulting cubic function (explained well by Just Wondering).  You will see a graph that increases from -infinity up to about 24 (on the V), then decreases to about -3.5, and then increases to infinity.  You want to find the x value and V value at that first "hump," or maximum.  If you are using a graphing calculator you can use the "Maximum" function to see that Albert's answer is approximately correct.
795657
1. Move the constant to the right side of the equation:\n3x^2 - 8x = 6\n\n2. Divide out the x^2 coefficient:\nx^2 - 8/3x = 2\n\n3. Divide the coefficient of x (in this case, -8/3) by 2 and square it, then add it to both sides:\nx^2 - 8/3x + 16/9 = 34/9\n\n4. Factor the left side of the equation to (x + a)^2, where a is the number you just squared:\n(x - 4/3)^2 = 34/9\n\n5. Take the square root of both sides, remembering that this will produce both a positive and a negative root:\nx - 4/3 = +/-sqrt(34/9)\n\n6. Move the constant on the left to the right and simplify:\nx = 4/3 +/-sqrt(34/9) = 4/3 +/-(sqrt 34)/3 = (4 +/- sqrt 34)/3
967957
There really isn't any way to do this other than guess and check. If you find that you substitute some value in and find f(a) = 0, then (x-a) will be a factor. You can repeat that a few times until you have found some roots, and then do a long division to something which you *can* solve by hand (a quadratic or linear).\nNote also that you only need to check things which are factors of the constant term, and there aren't too many.\nIn the first case, you will find that substituting in -2, -1 and 3 will all give 0, but you may have trouble finding the last root. So you will have to do some long divisions (divide by (x+2)(x+1)(x-3)), and you will end up finding that the -1 was a repeated root - so the answer is (x+2)(x+1)^2(x-3).\n\nSame thing for the second: you will find -1 as one root. You can then do a long division and find -1 is repeated, and again, and again, and again. The actual factorisation is (x+1)^5 - you may have noticed this if you knew much about pascals triangle and recognised the coefficients are an entire row - 1,5,10,10,5,1.
929261
y/9 = 250x^3 + 150 x^2 + 55x + 70\n\nTo get the x-intercepts, we set y=0:\n\n250x^3 + 150 x^2 + 55x + 70 = 0\n\n50x^3 + 30x^2 + 11x + 14 = 0\n\nLet f(x) = 50x^3 + 30x^2 + 11x + 14\n\nWe want to see where f(x)=0.  We note that f(0) = 14, and that for all positive x, f(x) > 0, so there are no positive roots.  Also, f(-1) = -50 + 30 - 11 + 14 = -17, so there's a root between -1 and 0.  Also, for all x < -1, f(x) < 0, so there are no roots less than -1.  All the roots are between -1 and 0.\n\nWe have two points, (0,14) and (-1,-17).  Let's try x = -0.5:\n\nf(-0.5) = 9.75  ==>  (-0.5, 9.75)  (I'm sort of interpolating)\n\nf(-0.7) = 3.85  ==>  (-0.7, 3.85)\n\nf(-0.8) = -1.2  ==>  (-0.8,-1.2)\n\nf(-0.78) = -0.06  ==>  (-0.78, -0.06)\n\nThat's pretty good, although we could go to a third decimal if we wanted to.  Just as a check, f(-0.785) = -0.34.  Let's go with one root at x = -0.78.\n\nNow let's divide that root out, reducing the cubic to a quadratic:\n\nf(x) = (x + 0.78)(50x^2 - 9x + 18) = 0\n\nThe roots of the quadratic g(x) = 50x^2 - 9x + 18 = 0 are imaginary.  Using the quadratic formula,\n\nx = 0.01 [9 +/- sqrt(81 - 3600)] = .01 [9 +/- 3 sqrt(9 - 400)]\n\nx = .03 (3 +/- i sqrt 391)\n\nso your only real root is x = -0.78.  (Answer)
358817
- Open MS-Excel;\n- Start adding all the equations you want;\n- Save the file once done...\n\nEquation Example:\nA simple adding equation:\n- type the number 1 in cell A1\n- type the number 2 in cell A2\n- type =A1+A2\n\nand you are done\n\nTo type, click on the cell with the mouse, then type, or double-click on the cell and type\n\nTo confirm text and exit editing it, hit the Enter button\n\n_________________________\n\nTips:\n=====\n- you can use any of the following: + - * /\n- to start an equation, it must start with the = sign\n- ^ is the power sign\n- pi is used in geometry [EX: type pi() to get its value]\n\n\nHope this helps a little...
789077
in excel all you have to do is to put = before any equation you want to solve and then press ok \nyou can use squares instead of the no.s
1187142
maybe something like 7^(1/3) over 7^(1/2)\n\nthat's the same as the cube root of 7 over the square root of 7.\nTo solve it, you would just subtract the exponents and you'd get 1 over 7^(1/6)\n\nI hope that is what you are looking for.
386376
My computer gives me this 4 real results, there are two more Imaginary results.\n\n1) -1\n2) 1.8698\n3) -0.260\n4) -2.688\n\nImaginary roots are\n\n5) 1.039 + j1.40\n6) 1.039 - j1.40
712942
x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0\nx^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 - 10x^3 = 0\nSo (x - 1)^6 - 10x^3 = 0\nSo ((x - 1)²)³ - (³√10x)³ = 0\n\nSo ((x - 1)² - ³√10x)((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0\n\nSo ((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0 N0 real roots (Min value near x = 0.2 just more than 0.87)\n\nOR (x - 1)² - ³√10x = 0 \nie x² - (2 + ³√10)x + 1 = 0\nie x = ½[(2 + ³√10) ± √(4 + 4³√10 + ³√100 - 4)\n= ½[(2 + ³√10) ± √(4³√10 + ³√100)]\n\nSOLUTION\nx = ½[(2 + ³√10) ± √(4³√10 + ³√100)]
737743
The positive square root is 11.  The cube root is about 4.946 (it's not an integer).
231877
ax^2+bx+c=0\nx^2+3x-9=0\n\nformula:\nfirst root:\n(-b+sqrt(b^2-4ac))/2a==> (-3+sqrt(3^2+4*9))/2=1,85\nsecond root:\n(-b-sqrt(b^2-4ac))/2a==> (-3-sqrt(3^2+4*9))/2=4,85
264348
The best way to start is to create a function based on the equation. Do this by setting the equation equal to 0. For instance, the first step to solving the equation x^3=5 would be to subtract 5 from both sides to get x^3-5=0. This creates the function f(x)=x^3-5.\n\nNow, look at the graph of the function. Are there any obvious places it appears to cross the x-axis? If the function appears to cross at x = 2 then perform synthetic division, dividing the cubic equation by x-2 (in other words, put a +2 in the synthetic division "box"). If you get a remainder, then it wasn't a root. If there's no remainder, you've factored out the root, and you'll get (x-2)(ax^2+bx+c), where a, b, and c are the numbers that come out of the synthetic division you just completed.\n\nNow you can factor ax^2+bx+c (if it's factorable) or use the quadratic formula (if it's not) to get the other two solutions. There will be 3, even if some of them are imaginary (have negatives inside radicals).\n\nIf none of the x-intercepts are rational (none of the synthetic division worked), pick an x-value close to each x-intercept on the graph and apply Newton's method to each one to estimate each root as accurate as you'd like.\n\nClick on the link below to see how I apply Newton's Method to solve f(x) = x^3-5. It's a big graphic, and your browser may smoosh it to make it fit in the screen, which makes it unreadable. Usually, clicking on the image or holding your mouse over the image until a resize box appears in the lower right corner will let you expand it to the real size. Either that or save it to your computer and double-click it there.\n\nBest of luck.
297409
This doesn't make any sense. You need to put down the numbers you're working with. If you're talking about finding cube roots, that's one thing; taking any number to the third power (x, for instance), is just: x*x*x.
443349
We are given:\nx^1/2 + y = 11 ---- eqn. 1\ny^1/2 + x = 7 ----- eqn. 2\n\nLet a = x^1/2. Therefore x = a^2\nLet b = y^1/2. Therefore, y = b^2\n\na + b^2 = 11 ==== eqn. 3\na^2 + b = 7 ==== eqn. 4\n\nSub. (3) into eqn. (4):\n\n=> (11 - b^2)^2 + b = 7\n=> 121 - 22b^2 + b^4 + b = 7\n=> b^4 - 22b^2 + b + 114 = 0\n\nUsing trial and error, we can arrive that (b-3) is a factor.\nTherefore, by factorisation:\n=> b^4 - 22b^2 + b + 114 = 0\n=> (b-3) (b^3 +3b^2 - 13b - 38)=0\nTherefore, b = 3  (there are no other possible root values for b because the cubic equation lies completely above the x-axis.)\n\nSince b = 3, therefore, y = 9. And back into eqn. 1, x= 4.