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Hi, I was wondering with an equation such as this : (x^2 - 1 )^3 = 27 do I times the 3 into the brackets and cube the last term (27)?
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Hi Mariner;
That is what you have? To solve it you would take the cube root of both sides first:
Now you just have a quadratic to solve. Using inspection you can see that one root is x = 2. Can you get the other?
Or are you trying to expand that out?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Hi Mariner;
That is what you have? To solve it you would take the cube root of both sides first:
Now you just have a quadratic to solve. Using inspection you can see that one root is x = 2. Can you get the other?
Or are you trying to expand that out?
From the last step you have there I did:
x^2 - 1 = 3
X^2 = 3+1
X^2 = 4
x= root 4
x= 2
I know it should be 2,-2 but how do I get the negative 2?
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First you should know that (-2)^2 = 4 just the same as 2^2 = 4. Also I encourage people to think about that problem before running to the methods of algebra.
But to do it in a more formal way. When you got here:
Use small x for the variable x instead of the capital x.
x = 2
x = - 2
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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excellent! thanks mate!
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Glad to help!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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