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How do you solve x cubed -x equals 20? - Answers

You can solve this using Newton's Method. First, solve the equation for zero, and use it to define the function f(x): Let f(x) = x3 - x - 20 It's derivative then would be: f'(x) = 3x2 - 1 Now let's take a rough guess at what x would be - we'll go with three - and plug it into the system xn+1 = xn - f(xn) / f'(xn): x0 = 3 x1 = x0 - (x03 - x0 - 20) / (3 * x02 - 1) = 2.8461538461538461539 x2 = x1 - (x13 - x1 - 20) / (3 * x12 - 1) = 2.8371684181740047662 x3 = x2 - (x23 - x2 - 20) / (3 * x22 - 1) = 2.8371386689493412305 x4 = x3 - (x33 - x3 - 20) / (3 * x32 - 1) = 2.8371386686239233464 x5 = x4 - (x43 - x4 - 20) / (3 * x42 - 1) = 2.8371386686239233464 So x is approximately equal to 2.8371386686239233464. If we plug that into the original equation, we can see that it is correct: 2.83713866862392334643 = 22.8371386686239233464 22.8371386686239233464 - 2.8371386686239233464 = 20



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How do you solve x cubed -x equals 20? - Answers

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You can solve this using Newton's Method. First, solve the equation for zero, and use it to define the function f(x): Let f(x) = x3 - x - 20 It's derivative then would be: f'(x) = 3x2 - 1 Now let's take a rough guess at what x would be - we'll go with three - and plug it into the system xn+1 = xn - f(xn) / f'(xn): x0 = 3 x1 = x0 - (x03 - x0 - 20) / (3 * x02 - 1) = 2.8461538461538461539 x2 = x1 - (x13 - x1 - 20) / (3 * x12 - 1) = 2.8371684181740047662 x3 = x2 - (x23 - x2 - 20) / (3 * x22 - 1) = 2.8371386689493412305 x4 = x3 - (x33 - x3 - 20) / (3 * x32 - 1) = 2.8371386686239233464 x5 = x4 - (x43 - x4 - 20) / (3 * x42 - 1) = 2.8371386686239233464 So x is approximately equal to 2.8371386686239233464. If we plug that into the original equation, we can see that it is correct: 2.83713866862392334643 = 22.8371386686239233464 22.8371386686239233464 - 2.8371386686239233464 = 20



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_x_cubed_-x_equals_20

How do you solve x cubed -x equals 20? - Answers

You can solve this using Newton's Method. First, solve the equation for zero, and use it to define the function f(x): Let f(x) = x3 - x - 20 It's derivative then would be: f'(x) = 3x2 - 1 Now let's take a rough guess at what x would be - we'll go with three - and plug it into the system xn+1 = xn - f(xn) / f'(xn): x0 = 3 x1 = x0 - (x03 - x0 - 20) / (3 * x02 - 1) = 2.8461538461538461539 x2 = x1 - (x13 - x1 - 20) / (3 * x12 - 1) = 2.8371684181740047662 x3 = x2 - (x23 - x2 - 20) / (3 * x22 - 1) = 2.8371386689493412305 x4 = x3 - (x33 - x3 - 20) / (3 * x32 - 1) = 2.8371386686239233464 x5 = x4 - (x43 - x4 - 20) / (3 * x42 - 1) = 2.8371386686239233464 So x is approximately equal to 2.8371386686239233464. If we plug that into the original equation, we can see that it is correct: 2.83713866862392334643 = 22.8371386686239233464 22.8371386686239233464 - 2.8371386686239233464 = 20

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      You can solve this using Newton's Method. First, solve the equation for zero, and use it to define the function f(x): Let f(x) = x3 - x - 20 It's derivative then would be: f'(x) = 3x2 - 1 Now let's take a rough guess at what x would be - we'll go with three - and plug it into the system xn+1 = xn - f(xn) / f'(xn): x0 = 3 x1 = x0 - (x03 - x0 - 20) / (3 * x02 - 1) = 2.8461538461538461539 x2 = x1 - (x13 - x1 - 20) / (3 * x12 - 1) = 2.8371684181740047662 x3 = x2 - (x23 - x2 - 20) / (3 * x22 - 1) = 2.8371386689493412305 x4 = x3 - (x33 - x3 - 20) / (3 * x32 - 1) = 2.8371386686239233464 x5 = x4 - (x43 - x4 - 20) / (3 * x42 - 1) = 2.8371386686239233464 So x is approximately equal to 2.8371386686239233464. If we plug that into the original equation, we can see that it is correct: 2.83713866862392334643 = 22.8371386686239233464 22.8371386686239233464 - 2.8371386686239233464 = 20

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