Question

geq%200%20%5C%5C%5C%5C%20-y%5E3-27y%2By%5E2%2B27%2B81%20%5Cgeq%200%20%5C%5C%5C%5C%20-y%5E3%2By%5E2-27y%2B108%20%5Cgeq%200%20%5C%20%5C%20%7C%28-1%29%20%5C%5C%5C%5C%20y%5E3-y%5E2%2B27y-108%20%5Cleq%200" id="TexFormula1" title="81-(y-1)(y^2+27) \geq 0 \\\\ 81-(y^3+27y-y^2-27) \geq 0 \\\\ -y^3-27y+y^2+27+81 \geq 0 \\\\ -y^3+y^2-27y+108 \geq 0 \ \ |(-1) \\\\ y^3-y^2+27y-108 \leq 0" alt="81-(y-1)(y^2+27) \geq 0 \\\\ 81-(y^3+27y-y^2-27) \geq 0 \\\\ -y^3-27y+y^2+27+81 \geq 0 \\\\ -y^3+y^2-27y+108 \geq 0 \ \ |(-1) \\\\ y^3-y^2+27y-108 \leq 0" align="absmiddle" class="latex-formula">
And what should I do from now on? I have to find the solutions for y... Please, a little help here :3

Answer 1

First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.

You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...

The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so $y_{n+1}=y_n-\frac{f(y)}{f'(y)}$ where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.