What number would you add to both sides to complete the square in this trinomial?

Question

there are a couple of these kind of questions on this test I am doing:

I was wondering if you can herlp me with to of them so that I will really get how to do it so ill do the rest myself!!
thanks 4 everything!!

1: What number would you add to both sides to complete the square in this trinomial?
x2 + 6x = -5
6
9
12


2: What number would you add to both sides to complete the square in this trinomial?
2x2 - 12x = -2
9
12
36

thanks again... these tests are soo stressful!!

Answer 1

This is the format that we can use for this problem:
a(squared)x + bx = c
To solve these problems, you divide the 'b' value by 2 and then square the halved value.

For example, on the first question::
x2 + 6x = -5
The 'a' value is 1, the b value is 6 and the c value is -5.
So you would take the 'b' value, 6, and divide it by two to get 3. Now you take this number, 3, and square it, to get 9. This is the number you add to complete the square, so in the end the problem would look like this.
x2 + 6x + 9 = 5

For the next problem:
2x2 -12x = -2
The a value is 2, the b value is -12 and the c value is -2. Take the b value, -12, and divide it by two to get -6. Now square -6 and you get 36. This is the number you add to complete the square, and this is what the problem would look like in the end:
2x2 - 12x + 36 = -2

Answer 2

I hope this doesn't confuse you but I will show you how to do it in general then we can plug in the numbers for your problems.

In general we have

ax^2 + bx = c

and we want to complete the square.

Step 1: divide both sides by a

x^2 + (b/a)x = c/a

Step 2: complete the square. The idea here is to find a number, d (say), so that when we expand (x + d)^2 the parts that have an "x" in them match what we found in Step 1. So (x + d)^2 = x^2 + 2dx + d^2 and what we found in Step 1 was x^2 + (b/a)x. For the parts containing an "x" to match up we need to have b/a = 2d. So solving for d we get d = b/(2a). Finally, the number we have to add to both sides is d^2.

1. x^2 + 6x = -5. Here a = 1 and b = 6. So d = 6/2 = 3 and the number we have to add to both sides is 3^2 = 9.

2. 2x^2 - 12x = -2. Now a = 2 and b = -12 so d = -12/(2*2) = -3 and again the number we have to add to both sides is 9 (-3^2).

Math Rules!

Answer 3

1: That's half of six which is three squared which is nine

2: This one you would have to divide through by 2 first:
x^2 - 6x = -1
Then proceed as in number one you would get the same answer.

Answer 4

to end the sq. you pick (x+6)^2 to get the x^2+ 12x. this suggests you pick 36 on the LHS which potential you upload 36. so (x+6)^2=27 So x+6=rt27 or x+6==rt27. So x=rt(27)- 6 or x=-rt(27)-6