HARDCORE Math lol!

[aisha]

solve (4x^2-9)^2-11(4x^2-9)+24=0
I tried simplifying this and got 16x^4-44x^2+204=0 im not sure if this is right, and I dont know what to do to solve. Please help! :confused:

[Gokul43201]

Is there not a simple substitution that will make things easier ?

[aisha]

Is there not a simple substitution that will make things easier ?

There is nothing but the question, what do I do?

[Gokul43201]

Actually, I was making a suggestion there. Try a substitution...look for a repeating quantity and call it something.

[elote]

let W = (4x^2-9), solve for W using some method, and then replace W with (4x^2-9), and now solve for x

:)

[aisha]

:smile: Thanks I get how to do it but want to make sure I did it right can you please check? Here is what I did:
(4x^2-9)-11(4x^2-9)+24=0
Let w=(4x^2-9)
w^2-11w+24=0
factored to become (w-8)(w-3)=0
sub in w value (4x^2-9-8)(4x^2-9-3)=0
(4x^2-17) (4x^2-12)=0

so solving for x
4x^2-17=0
4x^2=17
x=the square root of +or- 17/4

or 4x^2-12=0
4x^2=12
x=the square root of +or-3

[Tom McCurdy]

off the subject aisha, did you get your name from the ebaumsworld video aisha, I am sure it is on otherwebsites as well, but it is the video fo the skinny dude singing about some girl in want to be backstreet boys way?

[Gokul43201]

:smile: Thanks I get how to do it but want to make sure I did it right can you please check? Here is what I did:
(4x^2-9)-11(4x^2-9)+24=0
Let w=(4x^2-9)
w^2-11w+24=0
factored to become (w-8)(w-3)=0
sub in w value (4x^2-9-8)(4x^2-9-3)=0
(4x^2-17) (4x^2-12)=0

so solving for x
4x^2-17=0
4x^2=17
x=the square root of +or- 17/4

or 4x^2-12=0
4x^2=12
x=the square root of +or-3

Correct...except that you mean x=+or- the square root of 17/4 NOT x=the square root of +or- 17/4

And Tom : 'Aisha' is a reasonably common name; originally of Persian origin, it has spread into other cultures as well.

[aisha]

Thanks EVERYONE!!! :smile: