Simplify the following:

EbenezerSon · 2014-08-27 07:03:42

Polynomials are in the book, but I am yet to learn them properly.

EbenezerSon · 2014-08-27 07:10:36

By the way, if you can supply me with a link that teaches or deals with polynomial happy will I be
Besides, those problems are treated under indices

Thanks

Last edited by EbenezerSon (2014-08-27 07:13:11)

bobbym · 2014-08-27 07:15:07

Hi;

I am unable to solve either of those equation separately without using factoring or the quadratic formula. I will leave the question here and maybe someone else can do it. I am drawing a blank, sorry.

Can anyone solve these equations?

1)

2)

EbenezerSon · 2014-08-27 07:21:34

At the back of the book it gave these as it answers; {x : x = -1, -2}

I will try to solve, I doubt if will get those answers.

Thank you Bobbym

Last edited by EbenezerSon (2014-08-27 07:24:52)

bobbym · 2014-08-27 08:00:35

Yes, those are the correct answers. Sorry, I could not do them.

EbenezerSon · 2014-08-27 09:30:36

Hi;
I have endeavored to solve them, I literally used hours, which I had {-1, -2} for the first one and {-1} for the second problem. At the back of the book it has {-1} as the answer for the second problem, so I think I am correct.
Did you say you used factorization and quadratic to solve them, and what are your answers?
Thanks

bobbym · 2014-08-27 23:45:11

Those are the correct answers by factorization. How did you do the first one without it?

EbenezerSon · 2014-08-28 03:59:03

Actually with respect to the first one I did used factorization in the course of manipulation, see;

4 3^(2x+1) + 17 3^X -7 = 0
= 4 (3^x)^2 3^1 + 17 (3^x) - 7 = 0
Let 3^x = M
= 4 M^2 3 + 17M - 7 = 0
= 12M^2 - 4M + 21M - 7= 0
= 4M(3M - 1) + 7(3M - 1) = 0
=(4M + 7) (3M - 1) = 0
Let M = 3^x.
4M + 7 = 0
= 4M = -7
=M = -7/4
Again,
3M - 1 = 0
3M = 1
M = 1/3 Let 3^x = M
3^x = 1/3
3^x= 3^-1.
x = -1

I used the same method to do the first one

Last edited by EbenezerSon (2014-08-28 04:01:08)

bobbym · 2014-08-28 04:05:07

That is what I did! I am sorry, I thought you would not understand it so I did not post it.

EbenezerSon · 2014-08-28 04:25:32

The first one;
proceeding from the question:

3^2x 3^3 - 4 (3^x 3) + 1 = 0
= 3^2x 27 - 4(3^x 3) + 1 = 0
= (3^x)^2 27 - 4(M 3) + 1 = 0
Let 3^x = M
(3^x)^2 27 - 4 (M 3) + 1 = 0
= 27M^2 - 12M + 1 = 0
= 27M^2 - 9M - 3M + 1 = 0
= 9M(3M - 1) - 1(3M - 1)= 0
= (9M - 1) (3M - 1) = 0
Let M = 3^X
9M - 1 = 0
M = 1/9
3^X = 1/9 = 3^ -2 = -2
X = -2

3M - 1 = 0
M = 1/3
3^X = 1/3 = 3^-1 = -1

{ X : X = -1, -2}

bobbym · 2014-08-28 04:31:11

That is very good.

EbenezerSon · 2014-08-28 04:34:10

Hi Bobbym please help with this;

Find the value of x in the following equation

2^x - 2^-x = 2
This is similar to the one we solved previously, I tried and had [0] as an answer yet don't understand

bobbym · 2014-08-28 04:41:04

Multiply both sides by 2^x and do what you did before.

EbenezerSon · 2014-08-28 05:03:07

Following the question;

= 2^2x - 1 = 2^(1+x)
= (2^x)^2 - 2^(1+x) = 0
Let 2^x = N
= N^2 - 2N - 1 = 0
Factorizing impossible!
What do you say?
If factorization is impossible should one use log? Or the method of completing the square?

bobbym · 2014-08-28 05:15:33

That can not be factored, you are correct. We need another way.

EbenezerSon · 2014-08-28 05:28:48

The book has at it back zero [0] as its answer, please do it for me to see by using log, I am certain you can get by log application
Thanks!

bobbym · 2014-08-28 05:31:35

Have you copied the problem correctly?

2^x - 2^-x = 2 does not have zero for a root.

EbenezerSon · 2014-08-28 05:36:52

Yes that is exactly what's in the book, I have double check it.

bobbym · 2014-08-28 05:37:55

Please plug in x = 0 and tell me what you get?

EbenezerSon · 2014-08-28 05:42:10

2^0 - 2^ 0 =
1 - 1 = 0 I got zero

bobbym · 2014-08-28 05:44:24

Does 0 equal 2?

EbenezerSon · 2014-08-28 05:47:49

The book is wrong, I will be inclined to say that, what do you think should be the method that will eventually give you the correct answer?

bobbym · 2014-08-28 05:49:45

I think the same method to factor but use the quadratic formula to get the roots will get the answer.

I also think the book meant to solve this problem instead.

2^x + 2^-x = 2

EbenezerSon · 2014-08-28 05:55:19

bobbym wrote:

I think the same method to factor but use the quadratic formula to get the roots will get the answer.
(I don't get you here)
I also think the book meant to solve this problem instead.
I am very sure!
2^x + 2^-x = 2

bobbym · 2014-08-28 06:00:56

You will get the polynomial when you use the substitution of y = 2^x of y^2 - 2y - 1 = 0.

Math Is Fun Forum

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