radicals: solve 64x = sq root of 1 - 96x by factoring

[onthesnap55]

I have racked my brain.
The directions say: Solve by factoring.
64x=the sq root of 1-96x (with the 1-96x under the radical)
I tried squaring both sides and putting it in standard form, but then I end up with:
4096xsq + 96x - 1=0
I'm not sure how to factor this from here. Because of the -1.

Also can you help me with factoring
3x^2/3 - 17x^1/3 -28=0
These must be factored, so I am sure its possible, just not sure how. Please help.

 

[Loren]

Re: Factoring radicals

4096 = 2[sup:fzon1fqc]12[/sup:fzon1fqc].

2 x 2048...2048 - 2 = 2046
4 x 1024...1024 - 4 = 1020
8 x 512...512 - 8 = 504
16 x 256...256 - 16 = 240
32 x 128...128 - 32 = 96 <<< BINGO

(128x - 1)(32x + 1) = ???

 

[onthesnap55]

Re: Factoring radicals

Thank you, but I thought the middle number had to equal the numbers on the end. I will solve from here.

 

[Loren]

Re: Factoring radicals

The bottom line is that when the factors are multiplied together, the result equals the original expression.

(128x - 1)(32x + 1) = 4096x[sup:kkr8l92z]2[/sup:kkr8l92z] + 128x - 32x -1 = 4096x[sup:kkr8l92z]2[/sup:kkr8l92z] + 96x -1.

 

[Deleted member 4993]

I have racked my brain.
The directions say: Solve by factoring.
64x=the sq root of 1-96x (with the 1-96x under the radical)
I tried squaring both sides and putting it in standard form, but then I end up with:
4096xsq + 96x - 1=0
I'm not sure how to factor this from here. Because of the -1.

Also can you help me with factoring
3x^2/3 - 17x^1/3 -28=0

substitute:

x^(1/3) = u

then your equation becomes

3u^2 - 17 u - 28 = 0

Now follow the rule shown by Loren for the other problem.....

These must be factored, so I am sure its possible, just not sure how. Please help.