Algebra Help Please?

Question

I have a couple questions that I'm stumped on. I'm supposed to solve these by using the completing the square method. However, some of these don't seem to work, but there has to be an answer and I don't know what I'm doing wrong. If any advandced algebra people could help me out and give me a detailed explanation, I'd bee really grateful. Ok so here are the problems.
1) p2 + 3p = 1
2) r2 + 5r = 2
3) t2 - 3t = -2
4) w2 - 8w = 3
5) x2 + 6x + 4 = 0
Yeah its just those five and the 2 next to the variables should actually be superscripted to be powers. Thanks!

Answer 1

Tell ya what, I'll explain how to do one in detail, and then I'll do one more, and you do the other three, cool?

First step: get all the terms with variables on one side and the numbers on the other side. All but #5 are already in this form.

p² + 3p = 1

Next step: Take the number in front of the "second" term (3 in this case), divide it by 2 (3/2), square it (3/2*3/2 = 9/4) and add that to both sides:

p² + 3p + 9/4 = 1 + 9/4 = 4/4 + 9/4 = 13/4

Now the left is a perfect square, where the second part of the binomial is the middle coefficient (3) divided by 2:

(p + 3/2)² = 13/4

Now, take the square root of both sides of the equation, remembering that the square root can be either positive or negative:

p + 3/2 = ±√(13/4) = ±√13/2

And then get p by itself on the left:

p = -3/2 ±√13/2 = (-3 ±√13)/2

Since you can't simplify farther, that's your answer.

The rest work exactly the same way. I'll do one more, number 5.

x² + 6x + 4 = 0
x² + 6x = -4
x² + 6x + 9 = -4 + 9 = 5
(x + 3)² = 5
x + 3 = ± √5
x = -3 ± √5