Question

Approximate to the nearest tenth the real zeros of g(x)=2x^5+3x-2

Answer 1

@terenzreignz

Answer 2

Newton Rhapson method?

Answer 3

what's that?

Answer 4

i never met that method?
never heard of it

Answer 5

It's the only method I know to approximate zeroes :(

Answer 6

is there a formula?

Answer 7

If you've never heard of it, chances are it's not the correct way to solve this...

Answer 8

ahaha

Answer 9

do you know other methods in getting the real zeros

Answer 10

Nope :(

Answer 11

>.<

Answer 12

i'' ll pick another one

Answer 13

solve for b in the equation\[\sqrt{3^{b}}=2^{b +1}\]

Answer 14

First things first... you have to express them with the same base, right? Which one do you want, 2 or 3?

Answer 15

3

Answer 16

Okay, so... let's do this...
\[\huge 3^{\frac12b}=2^{b+1}\]

Answer 17

welp... actually, it might be easier to just get the natural logarithm of both sides...
\[\huge \ln3^{\frac12b}=\ln2^{b+1}\]

Answer 18

See where it goes from here?

Answer 19

what does ln stands for?

Answer 20

is that the same with log

Answer 21

@terenzreignz

Answer 22

It's the logarithm with base e, or approximately 2.718

Answer 23

ahhhhhhhh

Answer 24

So, due to the properties of logarithms, we get...
\[\huge \frac{1}{2}b\ln \ 3 = (b+1)\ln \ 2\]
and solve.

Answer 25

i don't know ? ? ?

Answer 26

ln 3 and ln 2 are just constants.
You can double both sides, giving you
\[\huge b\ln \ 3 = 2(b+1)\ln \ 2\]

Answer 27

you multiply both sides with 2 ?

Answer 28

@terenzreignz

Answer 29

I just did. To make it simpler, at least. Now let's rearrange a bit\[\huge b\ln \ 3 = (b+1)2\ln \ 2\]

Answer 30

Properties of logarithms again, to make it more pleasing to the eye...
\[\huge b\ln \ 3 = (b+1)\ln \ 2^2\]

Answer 31

So finally
b ln 3 = (b+1) ln 4
Now solve.

Answer 32

will i divide both sides by\[\ln 3\]

Answer 33

It's a start.
\[\huge b =(b+1)\frac{\ln \ 4}{\ln \ 3}\]

Answer 34

For simplicity, let's let
\[\huge k = \frac{\ln \ 4}{\ln \ 3}\]

Answer 35

k stand for CONSTANT, isn't it?

Answer 36

but where does the b+1

Answer 37

Just so that it's easier. Now solve for b
b = (b+1)k

Answer 38

\[\frac{ b }{b +1 }=k\]

Answer 39

We're not solving for k, we're solving for b.

Answer 40

\[\frac{ b }{ b +1}=\ln 4\div \ln 3\]

Answer 41

is it right?

Answer 42

@terenzreignz

Answer 43

It's right, but it doesn't help you, it's b you want to solve for.

Answer 44

what's the right thing to do?
i'm already comfused?

Answer 45

b = (b+1)k
Distribute k
b = bk + k
then bring bk to the left
b - bk = k
factor out b
b(1-k) = k
divide both sides by k
\[\huge b = \frac{k}{1-k}\]

Answer 46

will i substitute now the value f k ?

Answer 47

will i substitute now the value f k ?\[b =\frac{\frac{ \ln 4 }{\ln 3 }}{1+\frac{ \ln 4 }{\ln 3 }}\]

Answer 48

Yes.

Answer 49

0.557885891

Answer 50

@terenzreignz

Answer 51

don't have a calculator. as long as you keyed in the correct values, you should be fine.

Answer 52

TJ
i had an answer but this answer doesn't fit with it

Answer 53

our teacher had checked it. . .
my answer was right but i don't have the solutions. .
now, our teacher needed the solutions as our project

Answer 54

the answer there is -4.8188

Answer 55

@terenzreignz

Answer 56

Problem... I honestly don't know :/

Answer 57

?????????

Answer 58

I don't know how to get to your answer :(

Answer 59

@terenzreignz

Answer 60

Wait, I know now...
You put a plus sign instead of a minus sign.... Review the posts earlier :P

Answer 61

hahahaa
i see the difference

Answer 62

Good.

Answer 63

-4.818841679

Answer 64

i reallly got the answer
yepey . .
thanks or your help . . .
TJ

Answer 65

No problem :)

Answer 66

do you know somebody who's good in statistics

Answer 67

@terenzreignz

Answer 68

No... Just post in a new question, and start with the words "STATISTICS QUESTION"
That ought to attract anybody good in Stat :)

Answer 69

can you convince her or him to help me ??????please

Answer 70

ok
thanks 4 the advice
see you tomorrow
bye
good night
MY KNIGHT AND SHINING ARMOR

Answer 71

:)