Thread: please help me with these problems.

please help me with the following problems:
1.For x2- 25 = 0, x=

2.Add and simplify: 5/6+1/4

3.Simplify the square root of 27 + the square root of 75.

4.solve:3xsquared-2x-2=0

5.Find N for log2N=3

6.Find N for logb32=5

thanks so much for your help!!!

Originally Posted by yellowjello

please help me with the following problems:
1.For x2- 25 = 0, x=

2.Add and simplify: 5/6+1/4

3.Simplify the square root of 27 + the square root of 75.

4.solve:3xsquared-2x-2=0

5.Find N for log2N=3

6.Find N for logb32=5

thanks so much for your help!!!

1) Factor. Difference of squares.
2) Common denominator.
3) Common factor. sqrt(27) = sqrt(3*9) = sqrt(3)*sqrt(9) ... Same thing with sqrt(75)
4) Factor or quadratic formula
5) Definition of logarithms.
6) Where's the N?

okay so i'm trying to factor #3
and i'm getting as far as (3x-2) (x-1) I know the signs are probably messed up and the 1 &2 are messed up, I know that the #'s that go in the spots of the 1 & 2 are supposed to multiply to equal -2 and add to equal the other -2 I just don't knwo how to get there. help?

yes i know the quad form where do i plug everything in? that i do not remember

Originally Posted by yellowjello

yes i know the quad form where do i plug everything in? that i do not remember

if you knew the formula, you would know where to plug everything in

the formula says the solution(s) for is/are given by



now, what is a, b and c in your problem? then what would x be?

yeah nevermind i just tried it, is it no solution?
that's what i'm getting...
as long as this is right that's the answer i'm getting
a=3
b=2
c=-2

Originally Posted by yellowjello

yeah nevermind i just tried it, is it no solution?
that's what i'm getting...
as long as this is right that's the answer i'm getting
a=3
b=2
c=-2

according to what you typed originally, b = -2. solutions exist in that case

I had b=-2...



so the top is set up as -2 plus or minus the sqrt of 2squared -4 (3)(-2)
and the bottom as 2(3)
is this right?

Originally Posted by yellowjello

I had b=-2...



so the top is set up as -2 plus or minus the sqrt of 2squared -4 (3)(-2)
and the bottom as 2(3)
is this right?

yes. and that has 2 solutions.