algebra help!?

Question

I have to complete a summer packet, and I'm having trouble with a few of the problems on it.
One of the questions says: solve by completing the square: x^2 + 6X - 10
I don't get it!!

Another question says: solve and state restrictions: 3 - 2/c^2 = 5/c
I've solved it, and I have a hazy idea of what restrictions are, but I still can't answer it.

Answer 1

Question 1
x² + 6x - 10 = 0
(x² + 6x + 9) - 9 - 10 = 0
(x + 3)² = 19
(x + 3) = ± √19
x = - 3 ± √19

Question 2
3c² - 2 = 5c
3c² - 5c - 2 = 0
c = [ 5 ± √(25 + 24) ] / 6
c = [ 5 ± 7 ] / 6
c = 2 , c = - 1 / 3 [ c ≠ 0 ]

Answer 2

x^2 + 6x - 10 = 0
x^2 + 6x = 10
x^2 + 6x + 9 = 19
(x + 3)^2 = 19
x + 3 = +/- sqrt(19)
x = -3 +/- sqrt(19)

You use the x^2 and x terms to construct a quadratic polynomial that is the square of a binomial, set equal to a real number. Then the binomial is equal to the positive or negative square root of the real number. Once I got x^2 + 6x on one side of the equation, I added 9 to both sides because the constant term in a quadratic must be the square of half the coefficient of the x term in order for it to be the square of a binomial. That is, the coefficient of x was 6, so I took the square of half of that. (6/2)^2 = 3^2 = 9.

For the second question, I multiplied every term by c^2 and solved as I would any quadratic. I got c = (-1/3) or 2, and they both checked out. I'm not sure what restrictions there would be; it sounds like something specific to the packet, and perhaps the term is explained in there.

Answer 3

For the second question, numbers divided by zero are undefined, so your restrictions are that c^2 and c cannot equal zero. Therefore c cannot equal zero.