+0

# help

0
138
3

If x = sqrt(3) - 1, then find x^4 + 4x^3 + 6x^2 + 4x - 7.  Show all your steps please!

Nov 15, 2019

### 3+0 Answers

#1
+109752
+1

( √3 - 1)^4  + 4 ( √3 -  1)^3  + 6 (√3 - 1)^2  + 4 (√3 - 1)   -   7    =

[(√3)^4  - 4 (√3)^3  + 6(√3)^2 - 4(√3) + 1]  +4 [ (√3)^3  - 3(√3)^2 + 3(√3)  - 1  ]  +

6 [ (√3)^2 - 2(√3) + 1 ] + 4 ( √3 - 1)  - 7  =

[ 9 - 12(√3)  + 18 - 4√3  + 1 ]  +  [12√3 - 36 + 12√3 - 4 ] + [ 18 - 12√3 + 6 ] + 4√3 - 4 - 7   =

[ 28 - 16√3] + [ 24√3 - 40 ] + [ 24 - 12√3] + 4√3 - 11  =

Nov 15, 2019
#2
+1078
0

The same method as CPhill but you make x equal 0.732050808.

0.732050808^4+4*0.732050808^3+6*0.732050808^2+4*0.732050808-7 = 1.00000001 which is basically 1.

You are very welcome!

:P

Nov 16, 2019
#3
0

Very welcolme for what ?

A largely useless and irrelevant answer ?

You have not answered the question.

You begin with an approximation, the best that you can hope for is an approximation to the exact result.

All that you have succeeded in showing, with the aid of a calculator,  is that the value of the expression is somewhere in the neighbourhood of 1.

You don't seem to have much of an idea of what mathematics is about.

When looking at a problem like this, the first thing that you should do is to ask whether there is something significant about the numbers chosen

or whether any old set of coefficients/numbers would have done.

If you had stopped to do that you might have seen the significance of the coefficients 1 4 6 4.

$$\displaystyle x^{4}+4x^{3}+6x^{2}+4x-7\\=x^{4}+4x^{3}+6x^{2}+4x+1-8\\ =(x+1)^{4}-8\\= (\sqrt{3}-1+1)^{4}-8\\=9-8=1.$$

Guest Nov 16, 2019
edited by Guest  Nov 16, 2019
edited by Guest  Nov 16, 2019