Thread: Some more problems..(solving for x)

1. Some more problems..(solving for x)

I don't know why I just can't get these stupid problems..it seems that every problem has something different about it and I just don't get what I'm doing wrong...

Okay, first one..I worked it out all the way but it just looks really wrong and none of my answers work when I plug them in, so I don't know what I did wrong:

8(m-4)^4 - 10(m-4)^2 + 3 = 0

I used substitution..and made y=(m-4)^2 and y^2 = (m-4)^4

8y^2 - 10y + 3 = 0
8y^2 - 4y - 6y + 3 = 0
4y(2y - 1) - 3 (2y-1) = 0
4 y - 3 = 0
y = 3/4 and 2y - 1 = 0 ---> y = 1/2

Then I plugged the y's in

y = (m-4)^2
1/2 = (m-4)^2
1/sqroot(2) = m - 4
m = 4 + 1/sqroot(2)
simplified to: 4 + or - 1/sqrt(2)

and

3/4 = (m-4)^2
sqrt(3)/4 = m-4
m = sqrt(3)/2 + 4
m = 4sqrt(3) + 3

Yeah, it looks confusing =/ but the point is that neither of my answers worked out..Can anyone point out what I did wrong? If it's not too hard to read?

2. 4(x+1)^1/2 - 5(x+1)^2 + (x+1)^5/2 = 0

I used substitution again

x = (x+1)^1/2
x^4 = (x+1)^2
x^5 = (x+1)^5

4x - 5x^4 + x^5 = 0
x(4-5x^3 + x^4) = 0

and once I get here I'm pretty lost as to what to do next =/

3. (x^2 - 9) / (x^2 - 2x - 3) = 3/2

So I assume I should factor everything first:

(x-3)(x+3) / (x-3)(x+1) = 3/2
(x+3) / (x+1) = 3/2
then..I guess I multiply by...the denominators..to ..get rid of them?

and got

2(x+3) = 3(x+1)
2x+6 = 3x+3
x=3
but when I plug it in it doesn't work...and everything it doesn't work I just feel like I did something wrong

Sorry for the long post..appreciate any help I can get

2. 8(m-4)^4 - 10(m-4)^2 + 3 = 0
=> 8(m-4)^4 - 6(m-4)^2 - 4(m-4)^2 + 3 = 0
=>(4(m-4)^2 -3)(2(m-4)^2-1) = 0
=>((m-4)^2 -3/4)((m-4)^2-1/2) = 0
=>(m-4-sqrt(3)/2)(m-4+sqrt(3)/2)(m-4-1/sqrt(2))(m-4+1/sqrt(2))=0
=> m = 4+sqrt(3)/2, 4-sqrt(3)/2, 4+1/sqrt(2), 4-1/sqrt(2)

3. (x^2 - 9) / (x^2 - 2x - 3) = 3/2
=> 2(x-3)(x+3) = 3(x-3)(x+1)
=> 2(x-3)(x+3) - 3(x-3)(x+1) = 0
=> (x-3)(2x + 6 - 3x -3) = 0
=> (x-3)(-x + 3) = 0
=> (x-3)^2 = 0 ;
=> x = 3
But this can not be the solution to the equation as the function (x^2 - 9) / (x^2 - 2x - 3) is not defined at x = 3.

Therefor no solution.

4. Thank you! I guess I had done it correctly

Can anyone help me out with the second one?

2. 4(x+1)^1/2 - 5(x+1)^2 + (x+1)^5/2 = 0

I used substitution again

x = (x+1)^1/2
x^4 = (x+1)^2
x^5 = (x+1)^5

4x - 5x^4 + x^5 = 0
x(4-5x^3 + x^4) = 0

and once I get here I'm pretty lost as to what to do next =