algerbra help please?

Question

I only have 6 problems left to do and i am stuck...can someone please help..I need the work shown because I don't know where I am messing up at. Ok here goes...
(1) solve by factoring; x^2 - 9x = -8
(2) solve using sq root property; 2x^2 - 5 = 45
(3) use sq root property; (x+3)^2 = 36
(4) solve by completeing the square; x^2 + 4x + 2 = 0
(5) solve using quadratic formula; x^2 - 3x = 2x - 1
and (6) use quadratic formula; x^2 - 5x - 3 = -7
Please this is the last of 100 problems that I had to do for a review and I really need them all finished. Please help.

Answer 1

1) x^2 - 9x +8=0 -----> (x-8) (x-1)

2) 2x^2 =45+5 = 50
2x^2=50
x^2=25
Therefore x=+ or - 5

3) (x+3)^2 =36
Taking square root on both sides
x+3=6 and x+3= -6

so x=3 and x=-9

4) x^2 + 4x + 2 = 0
is in the form a^2+2ab+b^2=0
This can be rewritten as (a+b)^2
so ans is (x+2)^2=0 that is (x+2) (x+2)=0

so the two solutions for x are -2,-2

5) x^2 - 3x = 2x - 1
can be rewritten as x^2-5x+1=0
here a=1 b= -5 and c=1

the quadratic formula is {-b+(-)root of b^2-4ac}/2a
substituting the two solutions are {5+root(21)}/2 and
{5-root(21)}/2

6) x^2 - 5x - 3 = -7
can be rewritten as x^2-5x+4=0
here a=1 b=-5 and c=4

the quadratic formula is {-b+(-)root of b^2-4ac}/2a
substituting the two solutions are (5+3)/8 and (5-3)/8

that is 1 and 1/4..!

Hope this was helpful.

Answer 2

Here is #2

2x^2 -5 = 45
2x^2 = 50
x^2 = 25
x = 5

Answer 3

yes