I have some algebra q's...such as solving t^4 - 5t^2 + 4 =0 and more?

Question

1)t^4 - 5t^2 + 4 =0

2) p^-2 - 2p^-1 - 1 =0

3) .5m > -1 and 5 - m> 1

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

Answer 1

1.) t^4-5t^5+4

combine the powers and their numbers

4t+4=0

-4

4t=4

divide by four on both sides

t=1

g2g sry i cant finish but that is how you do the rest of them.

Answer 2

1) This is just like any other factoring problem, don't let the t^4 scare you.  Think of it as x^2 and x, where x is t^2.  So then we have x^2-5x+4, we know that the end form will be (x-?)*(x-??) because the middle term is negative and the end term is positive. The two question marks have to multiply and get 4 and add and get 5.  So that gives you 4 and 1.  (4*1=4, 4+1=5). So the answer is (x-1)*(x-4) where x=t^2 --> so you get (t^2-1)*(t^2-4)=0.  If you set each part equal to 0, you get t^2 = 4 and t^2 = 1, so the answers are (-2, 2, -1, 1).

2) This one is tricky, I would start by multiplying both sides by p^2 (I picked that to cancel out the first term).  That should give you 1-2p-p^2=0 multiplying by (-1) and rearranging gives p^2+2p-1.  Unfortunately this won't factor.  (The only option is -1 and +1 which would make the middle term disappear.) So you would need to use the quadratic equation.

3) These work almost like = signs. So multiplying by 2 in the first one gives m>-2 and adding m to both sides gives 5>m+1 or 4>m so we know that -2<m<4.  Any number between these two will solve the equaiton.

4) I'm not sure how to do this one off the top of my head.  Sorry