Yet again more maths help

[SoSolid]

Ok maths orientated people. I do not want anyone to solve this question for me, I just want an indication as to what method I need to use in order to solve the following equation: (12+4x)^-3/2 - (12-4x)^-3/2 = 0

I basically need to solve this for X, what method do i use?

 

[DB_SamX]

Expand, then send all x on one side, non x to the other, finally simplify and solve.

No homework qu. though.

 

[Hamzter]

Well there's one really obvious solution (I'm not gonna spoil it for u ), give me a second and I'll think whether there's another....

Edit: nope there's only 1 solution, and it's about the simplest answer possible, you don't need to do anything tricky like expanding or anything (which would be really tedious as I think you'd get an infinite series). It's not hard, think about it for a minute...

 

[yer_averagejoe]

yup since the side are almost symetrical, it can be solved in a jiffy quite a dissapointing answer! You could get the anser just by looking at it actually..

 

[Brizzles]

Even i got that one, and i'm stoopid.

The answer is 6.43 or something like that.

 

[Al Vallario]

The answer is 6.43 or something like that.

I'm afraid you get a bit fat zero for that answer

*av

 

[Alan!]

[]

 

[Brizzles]

I'm afraid you get a bit fat zero for that answer

*av

Fat zero would have been my second guess

 

[SoSolid]

I am slightly confused as to how to solve an equation for x such as the following:


e^-2x [-10x + 11] = 0

The method i have been using is to do ln e to bring down the -2x and of course ln e is 1, so I have just then multiplied out of the brackets and solved it. However I am confused as to whether or not this is the correct method, because when I substitute my value of x back into the original equation for one of my values (x=0) i do not get the the solution to equal 0 as e^0 is 1. So this is really confusing me, can anyone clarify this matter for me please?

 

[carvegio]

What you do to one side you have to do to the other ... that is your problem ....

Misread it first, is [-10x + 11] not part of the power of the exp.

If it isn't why don't you divide both sides by something...

 

[SoSolid]

It's not part of the power.

 

[carvegio]

see how to do this now?

 

[Cold_And_Miserable]

I am slightly confused as to how to solve an equation for x such as the following:
e^-2x [-10x + 11] = 0

I'm not the best mathematician ever but I'm quite sure you can't remove the exponential function, certainly not to make it simpler.
In(e^-2x[-10x + 11] = In[0] simplifies to
-2x[-10x + 11] = -O0 (minus infinity) which is not a real number and therefore you get stuck at this point.
But obviously:
e^x = 0 has no solution, so the answer is no solution.

 

[Ricochet J]

Ha ha I know how to do this!
I did it last week

Hint: Just remember the binomial expansion rules.

[Cough]n![/cough] (and no, it's not a hint that penski knows the answer )

 

[carvegio]

there is a solution, it is the solution of [-10x + 11] = 0.

As it isn't part of the exp.

His eqn is exp(-2x)*(-10x + 11) i think?

 

[SoSolid]

Yes that is correct, I got x = 1.1 before, but my question was more so that I am confused as to why we can disregard the e^-2x.

 

[SoSolid]

It's ok guys I understand it now thanks a lot people for the input.

 

[Arcade Fire]

Just remember the maxim - "what you do to one side, you do to the other".

The first thing to notice is that if you have two things multiplied by each other and they equal zero, then one of them must be zero, i.e. if AB=0 then either A=0 or B=0.

In your case, you either have exp(-2x)=0 or you have (-10x+11)=0. But exp of anything is never zero (think about the graph of it), and so you must have -10x+11=0, which gives x=1.1.