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Solve the following equation.... (1 Viewer)

Smile12345

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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... :)
 

Smile12345

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And this one....

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

I would say a = 2^x ??
 

HSC2014

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

HSC2014

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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!! :(
 

Menomaths

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

4x-6x+2= 0
22x-6x+2= 0
a2-3a+2= 0
a= 2 or a= 1
 

HSC2014

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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?
 

Smile12345

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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...:)
 

Menomaths

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Do you know how to solve that?
 

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