???????????

EDIT: Woops Dragonlance is correct my answer should go like this. 

 

$4a^2-4a+1-6a^2-28\textcolor{red}{+}2a^7=0$

 

$2a^7-2a^2-4a=27$

 

Or is the general form this?

 

$2a^7-2a^2-4a-27=0$

 

 

 

On a different note can you still edit your previous posts? 

(1-2a)^2/6-a^2-(14-a^7)/3,

 

After a number of algebraic steps you get:

 

a =1.521282721...........etc., which is the right answer.

Assuming it is equal to 0...

 

$(1-2a)^2/6-a^2-(14-a^7)/3=0$

 

$(1-2a)(1-2a)/6-a^2-(14-a^7)/3=0$

 

$(4a^2-4a+1)/6-a^2-(14-a^7)/3=0$

 

$4a^2-4a+1-6a^2-2(14-a^7)=0$

 

$4a^2-4a+1-6a^2-28-2a^7=0$

 

$2a^7+2a^2+4a=-27$

 

... but from here I'm not sure what to do

 

Expand

$\frac{(1-2a)^2}{6}-a^2- \frac{14-a^7}{3}$

 

Find common denominator. it is 6

 

$-\frac{2(-a^7+14)}{6}-\frac{6a^2}{6}+\frac{(-2a+1)^2}{6}$

 

 

$\frac{-2(-a^7+14)-6a^2+(-2a+1)^2}{6}$

 

Expand this part

 

$(-2a+1)^2$

to this

$4a^2-4a+1$

then expand this

$-2(-a^7+14)$

to this

$2a^7-28$

then combine the parts

$2a^7-6a^2+4a^2-4a+1-28$

simplefy it and put it over the 6 and make equal to 0

$\frac{2a^7-2a^2-4a-27}{6} =0$

change to this

$\frac{2a^7-2a^2-4a}{6} =\frac{27}{6}$

change to this

$2a^7-2a^2-4a =27$

 

I out of time. Next post I solve it (or try to).

This is strange. This question do not show on the question page but it do show on the answer page. Do any one know why?

 

Any way, I play with this for a long time and I still not know how to solve it.

I plug both the first equation and then mine into Wolframalpha and it give a=1.52128... 

so I know I do the rearranging correctly but I still not know the steps to get to the answer.

 

If one of the mods would do it that it would be great because I would like to know how.

ha we both got to the same point and couldn't do it :)

Polynomials comes to mind but I haven't learnt that yet.

It is a degree 7 polynomial so there are 7 solutions for it. It is a odd degree so at least 1 solution is a real number. This is the number wolfram give.

 

I think the way this is solved is with the bisection method. It look easy to do, but graphing it look even easier. I already do that but it only give the answer to 1 decimal place.

 

 

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 After a short time or if someone make another post then there do not seem to be a way to edit an older post.