# Completing The Square

#### debbie29

##### New member
Hey everyone,

I've been trying to grasp the concept of completing the square, but it still confuses me. I am completely stumped on these two questions. It is probably a stupid mistake I'm making or something-- I just keep getting the wrong answers. Any help/tips would be appreciated.

(1)3x^2-5x-2=0

(2){} = sq rooted
{2x-1} - {x+3}=1

#### Denis

##### Senior Member
debbie29 said:
Hey everyone,
I've been trying to grasp the concept of completing the square, but it still confuses me. I am completely stumped on these two questions. It is probably a stupid mistake I'm making or something-- I just keep getting the wrong answers. Any help/tips would be appreciated.
(1)3x^2-5x-2=0

(2){} = sq rooted
{2x-1} - {x+3}=1 Thanks for your time.
HOW to complete the square: a classroom session?
Or what the solutions to those 2 problems are?

(1) is fairly simple; factors to:
(3x + 1)(x - 2) = 0
x = -1/3 or x = 2
Do you understand that?

(2) is sqrt(2x - 1) - sqrt(x + 3) = 1
square both sides, then square once more; you'll get:
7x^2 + 30x - 37 = 0; factors to:
(7x + 37)(x - 1) = 0
x = -37/7 or x = 1

So asking again: are you able to solve them not using
complete the square?

#### debbie29

##### New member
They have to be done with completing the square.
Just wanting a walk through for the questions because I can't seem to get them right.

#### ChaoticLlama

##### Junior Member
(1)
3x² - 5x - 2 = 0

First factor out the 3 from the 'x' terms.

3(x² - [5/3]x) - 2 = 0

Now take the coefficient of the x term (in this case, 5/3) and divide it by 2, then square the result.
You will end up with 25/36
To make sure that you do not change the equation (it must remain equal at all times, right?) You both add AND subtract this number from the equation like so.

3(x² - [5/3]x + 25/36 - 25/36) - 2 = 0

Now multiply the term -25/36 by 3 to remove it from the brackets.

3(x² - [5/3]x + 25/36) - 2 - 25/12 = 0