Solve the following equation.... (2 Viewers)

[Smile12345]

Hello All....

If I had to solve the following equation:

x^6 - 7x^3 - 8 = 0

Would we let a (or whatever pronumeral) = x^3 for a start?

Thanks in advance...

 

[Menomaths]

Yeah

 

[Kurosaki]

Yeah, that breaks it down to a quadratic which you can easily solve for.

 

[Smile12345]

Ok Good... Thanks.

 

[Smile12345]

And this one....

4^x - 3 . 2^x + 2 = 0 (the . is a 'times')

I would say a = 2^x ??

 

[HSC2014]

Ultimately you want to get it into the quadratic form ax^2 + bx + c = 0
Would letting a = 2^x achieve that?

 

[HSC2014]

Why not? What were you thinking of when you first thought of a = 2^x? Justify your thought processes. You really have to be more independent Smile12345!!

 

[Smile12345]

Why not? What were you thinking of when you first thought of a = 2^x? Justify your thought processes. You really have to be more independent Smile12345!!

Sorry HSC2014....

 

[Menomaths]

I think Smile is correct by making a=2^x correct me if I'm wrong though

4^x-6^x+2= 0
2^2x-6^x+2= 0
a²-3a+2= 0
a= 2 or a= 1

 

[Parvee]

Cos for the 2nd part I'd get -3a ... Am I correct here?

So it would be 2(a)^2 -3 (a) + 2 = 0

Are you sure its 2(a)^2?

 

[HSC2014]

Are you saying a,b,c cannot be negative in ax^2 + bx + c? Because they definitely can be; go back to the definition of the expression: a,b,c are constants where a=/= 0

So is -3 a constant? Does it fit the criteria? Yes! there is nothing wrong with letting a = 2^x
Another way to see it is (2^x)^2 -3 (2^x) + 2 = 0
See how it looks like the general form?

 

[HSC2014]

I think Smile is correct

Correct, but not confident; and I want to help bring out the self-thought process required to grow. I don't believe a teacher should teach with answers, but guidance.

 

[Parvee]

I think Smile is correct by making a=2^x correct me if I'm wrong though

4^x-6^x+2= 0
2^2x-6^x+2= 0
a²-3a+2= 0
a= 2 or a= 1

Don't forget to solve for x :L

 

[Menomaths]

Don't forget to solve for x :L

She can do it

 

[Smile12345]

I think Smile is correct by making a=2^x correct me if I'm wrong though

4^x-6^x+2= 0
2^2x-6^x+2= 0
a²-3a+2= 0
a= 2 or a= 1

Thanks.

 

[Smile12345]

She can do it

Yup I can!!

2^x = 2 .... So therefore x = 1

 

[Menomaths]

Yup I can!!

2^x = 2 .... So therefore x = 1

There's one more.

 

[Smile12345]

Correct, but not confident; and I want to help bring out the self-thought process required to grow. I don't believe a teacher should teach with answers, but guidance.

Understand your point HSC2014.... Thanks for your continued guidance...

It's good...

 

[Smile12345]

There's one more.

Yup so 2^x = 1.... Hmmm.

 

[Menomaths]

Do you know how to solve that?