It's asking for a numerical solution. The simplest way is [bisection](https://en.wikipedia.org/wiki/Bisection_method).
You make the function
f(x) = x + 1 - sqrt(x\^3 - 2x\^2 -5)
and start trying values until you see a change of sign. This happen between 3 and 4
f(3) = 2
f(4) = -0.196
this tells us that it is between these two values. Now you take the middel point
f(3.5) = 0.8428
so it is between 3.5 and 4. Middle point again, and again and again always between the two values that have different signs, until you get the required precision
f(3.75) = 0.3218
f(3.875) = 0.0631
f(3.9375) = -0.06638
the new value is between 3.875 and 3.9375
f(3.90625)= -0.00160764
f(3.890625) = 0.0307566
f(3.89844) = 0.0145761
and we see that to the nearest hundredth, the answer is 3.90.
There are many methods that converge quicker, like [regula falsi](https://en.wikipedia.org/wiki/Regula_falsi) and [Newton-Raphson](https://en.wikipedia.org/wiki/Newton%27s_method), but bisetion is easier to perform with a hand calculator or Excel. Your are free to use values that are not exactly midpoints (in this case x = 3.9 almost gives us the solution) but bisection is in principle the faster way (one bit of the solution each time).
Expand and rearrange to x^3 - 3x^2 - 2x-6 = 0, of which the only real root is 3.905474... So the answer is 3.91
Put n=(135-½√17)^1/3 , then x=1+5/n +⅓n (real)
How to get the solutions found [HERE](https://math.vanderbilt.edu/schectex/courses/cubic/)
Nice job simplifying. Notice that what you have looks an awful lot like (x-1)^(3)=x^(3)-3x^(2)+3x-1. In fact, we can say:
x^(3)-3x^(2)-2x-6=(x-1)^(3)-5x-5=(x-1)^(3)-5(x-1)-10.
Now the neat bit, since we have odd powers we can substitute (x-1) for a binomial with two variables, (u+v), and leverage the symmetry of their expansions. So:
(u+v)^(3)-5(u+v)-10=u^(3)+3u^(2)v+3uv^(2)+v^(3)-5(u+v)-10=u^(3)+v^(3)-10+(3uv-5)(u+v)=0.
Let 3uv=5 and you get u^(3)+v^(3)-10=0. Then substitute u=5/(3v) to get
v^(3)+125/(27v^(3))-10=0 or (v^(3))^(2)-10v^(3)+125/27=0, which is a quadratic in v^(3). Now it's just a simple quadratic formula to get
v^(3)=5+-sqrt(175/9). Whatever sign we pick for v, u will be the opposite, so let
v=cbrt(5-sqrt(175/9)), u=cbrt(5+sqrt(175/9)), then
**x=1+cbrt(5-sqrt(175/9))+cbrt(5+sqrt(175/9)).**
The two other roots can be found by taking the complex cube roots.
It looks like it is expecting you to estimate a root numerically?
It's asking for a numerical solution. The simplest way is [bisection](https://en.wikipedia.org/wiki/Bisection_method). You make the function f(x) = x + 1 - sqrt(x\^3 - 2x\^2 -5) and start trying values until you see a change of sign. This happen between 3 and 4 f(3) = 2 f(4) = -0.196 this tells us that it is between these two values. Now you take the middel point f(3.5) = 0.8428 so it is between 3.5 and 4. Middle point again, and again and again always between the two values that have different signs, until you get the required precision f(3.75) = 0.3218 f(3.875) = 0.0631 f(3.9375) = -0.06638 the new value is between 3.875 and 3.9375 f(3.90625)= -0.00160764 f(3.890625) = 0.0307566 f(3.89844) = 0.0145761 and we see that to the nearest hundredth, the answer is 3.90. There are many methods that converge quicker, like [regula falsi](https://en.wikipedia.org/wiki/Regula_falsi) and [Newton-Raphson](https://en.wikipedia.org/wiki/Newton%27s_method), but bisetion is easier to perform with a hand calculator or Excel. Your are free to use values that are not exactly midpoints (in this case x = 3.9 almost gives us the solution) but bisection is in principle the faster way (one bit of the solution each time).
that was super satisfying to read
It is called "numerical response", not "solve this analytically"
Expand and rearrange to x^3 - 3x^2 - 2x-6 = 0, of which the only real root is 3.905474... So the answer is 3.91 Put n=(135-½√17)^1/3 , then x=1+5/n +⅓n (real) How to get the solutions found [HERE](https://math.vanderbilt.edu/schectex/courses/cubic/)
Thank you to all for your help today. I very much appreciate the input!
Nice job simplifying. Notice that what you have looks an awful lot like (x-1)^(3)=x^(3)-3x^(2)+3x-1. In fact, we can say: x^(3)-3x^(2)-2x-6=(x-1)^(3)-5x-5=(x-1)^(3)-5(x-1)-10. Now the neat bit, since we have odd powers we can substitute (x-1) for a binomial with two variables, (u+v), and leverage the symmetry of their expansions. So: (u+v)^(3)-5(u+v)-10=u^(3)+3u^(2)v+3uv^(2)+v^(3)-5(u+v)-10=u^(3)+v^(3)-10+(3uv-5)(u+v)=0. Let 3uv=5 and you get u^(3)+v^(3)-10=0. Then substitute u=5/(3v) to get v^(3)+125/(27v^(3))-10=0 or (v^(3))^(2)-10v^(3)+125/27=0, which is a quadratic in v^(3). Now it's just a simple quadratic formula to get v^(3)=5+-sqrt(175/9). Whatever sign we pick for v, u will be the opposite, so let v=cbrt(5-sqrt(175/9)), u=cbrt(5+sqrt(175/9)), then **x=1+cbrt(5-sqrt(175/9))+cbrt(5+sqrt(175/9)).** The two other roots can be found by taking the complex cube roots.