What is the Solution? x4 + x3 + x2 + x = 0 ?
Woodsman is correct. 0 is a relatively obvious member of the solution set -1 can be seen fairly easily now, consider when x = i i^2 = -1 i^3 = -1 * i = -i i^4 = -1 * -1 = 1 Adding them all together: i + -1 + -i + 1 = 0 The same reasoning is applies for x= -i, except all the signs change: -i + 1 + i + -1 = 0
Which values of x are solutions of the equation x^3+x^2-2x=0?
Since x is a common factor of all terms on the LHS, factoring it out yields: x^3 + x^2 - 2x = 0 ==> x(x^2 + x - 2) = 0 ==> x(x + 2)(x - 1) = 0, by factoring the inner quadratic. Then, by the zero-product property, the solutions are: x = 0, x + 2 = 0, and x - 1 = 0 ==> x = 0, 1, -2. I hope this helps!
How can one solve the following by using the method of factorization: [math]\frac{4}{x-3}=\frac{5}{2x+3}[/math]where [math]x[/math] is not equal to [math]0[/math] or [math]-\frac{3}{2}?[/math]
4/x-3=5/2x+3Multiply both sides by (x-3)(2x+3)4(2x+3)=5(x-3)=8x+12=5x-15Subtract 12 from both sides8x+12 -12 =5x-15 -12= 8x=5x-27Subtract 5x from both sides.8x -5x =5x-27 -5x= 3x=-27x=-9
Find the solution of the equation 2x^3+e^x=0 using three iterations of Newton's method.?
f(x) = 2x^3+e^x f'(x) = 6x^2+e^x Start off with a guess value x0=1 x0 = 1.000000000 x1 = x0 - f(x0) / f'(x0) = 1 - (4.71828183) / (8.71828183) = 0.458805999 x2 = x1 - f(x1) / f'(x1) = 0.458806 - (1.77534375) / (2.84520139) = -0.165172240 x3 = x2 - f(x2) / f'(x2) = -0.16517224 - (0.83873526) / (1.01143889) = -0.994421807 x4 = x3 - f(x3) / f'(x3) = -0.99442181 - (-1.59677991) / (6.30318566) = -0.741092776 x5 = x4 - f(x4) / f'(x4) = -0.74109278 - (-0.33745091) / (3.77190384) = -0.651628431 x6 = x5 - f(x5) / f'(x5) = -0.65162843 - (-0.03219207) / (3.06891402) = -0.641138704 x7 = x6 - f(x6) / f'(x6) = -0.6411387 - (-0.00039913) / (2.99304536) = -0.641005353 x8 = x7 - f(x7) / f'(x7) = -0.64100535 - (-6e-8) / (2.99208975) = -0.641005332 x9 = x8 - f(x8) / f'(x8) = -0.64100533 - (0) / (2.9920896) = -0.641005332 x = -0.641005332 http://www.wolframalpha.com/input/?i=sol...
Use Newton's method to estimate the requested solution of the equation.?
Use Newton's method to estimate the requested solution of the equation. Start with given value of x0 and then give x2 as the estimated solution. x^4 - 4x + 2 = 0; x0 = 0; Find the left-hand solution.
Use Newton's method to find all roots of the equation.?
Use Newton's method to find all roots of the equation correct to six decimal places. (Enter your answers as a comma-separated list.) 5 cos x = x + 1 Thanks