Melody

1 - 1/(20e^5)

Yes you are right.

It is my experience that even very sophisticated calcs can't handle roots of negative numbers. 

I think Alan went into more detail on this once before. Maybe he will comment later. :/

You can get superscript and subscript formats using the ribbon.

But Heurka and I both use LaTex to write our math speak in.

The Latex gateway is in the ribbon too.

If you decide you want to have a look at LaTex there is a rather messy thread on it in the sticky notes.  Some of our youngest members have learned it in the past but it does take a little effort. :)

Sorry for confusing you guest.  That was a small mistype.  It was =0 not =1

I have fixed it in my answer.  My answer is correct.  :))

I like Heureka's answer better though.

Thanks Heureka :))      

Why can't I take (-1)^1/2

lets think about this.

This means

\(\sqrt{-1}\)

Now

\(\sqrt{25}=5\\ because\;\; 5*5 = 25\\ so\\ think\;\;about\;\; \sqrt{any\;negative\;number} \)

a positive * a positive = a positive

a negative * a negative = a positive

Mmm

you cannot multiply any real number by itself and get a negative answer.  That is impossible !!

So

\(\sqrt{-1}\)

has no real solution!!!

That is why you cannot do it!!

\(\sqrt{-1}\)

is the basis of the imaginary number system.  It is called i (or sometimes j)

\(i=\sqrt{-1}\)

\((x^2-5x+2)^2-(x^2-5x)=8\\ (x^2-5x+2)(x^2-5x+2)-x^2+5x-8=0\\ x^4-5x^3+2x^2-5x^3+25x^2-10x+2x^2-10x+4-x^2+5x-8=0\\ ...\\ x^4-10x^3+28x^2-15x-4=0\\ \)

Now I am going to look for factors.  I am using the remainder theorem.  

If I sub in x=1,  which was just the first number that I tried.  I get

1 - 10 +28 - 15 - 4 and that DOES =0  so that means that one of the factors is   (x-1)

so now I am going to do an algebraic division to see what (x-1) would be multiplied by

            x^3 - 9x^2 + 19x + 4

        ---------------------------------------------

x-1  |   x^4 - 10x^3 + 28x^2 - 15x - 4

           x^4  - x^3

           --------------------------

                   - 9x^3+28x^2

                    - 9x^3 +9x^2

                      ------------------------

                               19x^2  - 15x

                               19x^2  - 19x

                               ---------------------

                                             4x - 4

                                             4x - 4

                                            --------

                                                   0

                                             -------

So we have

\((x-1)(x^3-9x^2+19x+4)=0\)

I tested  +1,-1,+2,-2,+4,-4     I chose these because they are all factors of 4

x=4   made that second bracket = 0  therefore another factor is (x-4)

I did another algebraic division the same as before and found that 

\((x^3-9x^2+19x+4)\div (x-4)=x^2-5x-1\\ so\\ (x-1)(x-4)(x^2-5x-1)=0\\ so\\ x-1=0\qquad or \qquad x-4=0 \qquad or\qquad x^2-5x-1=0\\ x=1\qquad or \qquad x=4 \qquad or\qquad x=\frac{5\pm\sqrt{25-\;-4}}{2*1}\\ x=1\qquad or \qquad x=4 \qquad or\qquad x=\frac{5\pm\sqrt{29}}{2}\\ x=1\quad or \quad x=4 \quad or\quad x=\frac{5+\sqrt{29}}{2}\quad or \quad x=\frac{5-\sqrt{29}}{2} \)

\(x=1 \quad or \quad x=4 \quad or \quad x\approx5.193 \quad or \quad x\approx -0.193\)

NOW I have not checked thes BUT if you substitute them back into the initial equation they should all make the equation true.

Let's have  a look at the  graph    y=(x^2-5x+2)^2-(x^2-5x)-8    and see where y = 0

Thanks, that would be great :))

Good I am glad.  

I already figured that you liked maths.

I hope you stick about :))

\(e^5x-e^{-5}x\\ =e^5x-\frac{x}{e^5}\\ =e^5x-\frac{x}{e^5}\\ =\frac{e^{10}x-x}{e^5}\\ =\frac{x(e^{10}-1)}{e^5}\\\)

I would consider the second last answer as the most simple but the last one looks better :))

That is right,  you answered your own question LOL