Can someone refresh my memory?

[melissa.ross23]

For some reason I'm having a heck of a time solving this.....

3(2x+1)^4-4(2x+1)^2-4=0


Can someone gimme a hand???? It would be appreciated...

 

[tkhunny]

Would it look any different if you first substituted:

Y = (2x+1)^2

??

Recognizing the Quadratic FORM is a valuable skill.

 

[melissa.ross23]

OK....but why that? I'm sorry ...I'm kind slow when it comes to math

 

[tkhunny]

3(2x+1)^4-4(2x+1)^2-4=0

Y = (2x+1)^2

Making this substitution yields:

3Y^2 - 4Y - 4=0

This will factor

(3Y+2)(Y-2) = 0

Giving

Y = 2 or Y = -3/2

One might be tempted to stop, here, but we still haven't solved for 'x'.

Substituting back

Y = 2 or Y = -3/2
(2x+1)^2 = 2 or (2x+1)^2 = -3/2

If we have to use Real Numbers, we can discard the second one. Do you see why? This leaves only:

(2x+1)^2 = 2

Expand

4x^2 + 4x + 1 = 2
4x^2 + 4x - 1 = 0

I'll let you suggest how to solve that for 'x'.

 

[melissa.ross23]

Use the quadratic Formula right?

 

[tkhunny]

That will do. Good call.

 

[melissa.ross23]

Thanks for the help tkhunny