Question

What is the first step in solving x2 + 2x + 3 = 0 by completing the square?
A Factor the left side.
B Add 3 to each side of the equation.
C Divide both sides of the equation by 2
D Subtract 3 from each side of the equation.

Answer 1

\[x^2 +2x +3=0\] You want to group your like terms together. \[(x^2+2x)+3=0\] Now you will basically be completing the square inside the parenthesis. Tell me what you get when you have" (2/2)^2

Answer 2

Is that how you were told to solve i?

Answer 3

it*

Answer 4

those are the answer choices, I think that it may either be A or B but im not sure how to solve this

Answer 5

Well, essentially when we "group" our like terms together, in a way we are completing the square by factoring, or creating a quadratic equation within the parenthesis.

Answer 6

B does notmake any sense.

Answer 7

The only time you add numbers to both sides of the equation is when you have fully factored what is inside the parenthesis.

Answer 8

Are you understanding?

Answer 9

Yes now I see that adding 3 cannot be the solution and I think we have to factor it out before solving

Answer 10

B and C are random, unnecessary choices, I would go with either A or D.

Answer 11

I agree I think that the answer is A

Answer 12

But I guess I am letting the way I complete the square interfere with the methods of completing the square are. so before factoring, you have to isolate your variables onto one side of the equation. how would you so that?

Answer 13

We want our x's on one side of the equation and our #s on the other.

Answer 14

X^2+2X=-3

Answer 15

yes, and which answer choice do you think that is?

Answer 16

Now I think that D is it because that should be done before factoring the equation

Answer 17

Yes!
Good job :)

Answer 18

Did you want to also learn how to complete the square?

Answer 19

Yes thank you so much! :)

Answer 20

Alright, so you have \[x^2 -2x = -3\] yes?

Answer 21

yes

Answer 22

in completing the square, we are trying to find a value of c that will put our equation in the form : \(ax^2 +bx+c=0\)

We already have our \(ax^2+bx\).

In order to find our \(c\) value, we shall take our coefficient of \(b\), divide it by 2, and square the entire quantity. We can see that in this format: \[c=\left(\frac{b}{2}\right)^2\]

Answer 23

So what is our value of b?

Answer 24

2

Answer 25

try again :)

Answer 26

Remember, the equation you have can be written: \(x^2 +(-2x)=-3\)

Answer 27

So it would be written as a negative?

Answer 28

yes :) -2 is our b value.

Answer 29

So what will we get when we plug it into that equation for finding c? \[c=\left(\frac{-2}{2}\right)^2\]

Answer 30

More specifically, what does c equal?

Answer 31

1

Answer 32

Good :) c=1. Now we have completed the square on the left hand side, but one more thing you have to remember is, whatever you do to the left side of the equation, you must also do to the right side. we will have: \[x^2 -2x+1 = -3-1\]

Answer 33

Do you know what the left side of the equation can be reduced to?

Answer 34

x^2-2x=-4

Answer 35

Not quite. I'll tell you a little trick.

Answer 36

When we were solving for c, we had reduced what was inside the parenthesis before squaring it, yes?

Answer 37

yes

Answer 38

So we had \[c=\left(\frac{-2}{2}\right)^2 = (-1)^2\] Taking that -1 value inside the square root, we can rewrite our equation on the left hand side as so: \[(x^2+c) =-4 \iff (x+(-1))^2 =-4 \iff (x-1)^2=-4\]

Answer 39

Let me know if you have any questions with this. It can be a little tricky but it's really useful

Answer 40

I think I get this better now...That helped so much! :)

Answer 41

another way is to factor \((x^2-2x+1)\)

Answer 42

this might be a little easier to understand. Like to see?

Answer 43

yes

Answer 44

Okay. What are 2 numbers that multiply to give +1 but add to give -2?

Answer 45

-1

Answer 46

good :)

Answer 47

\[(x-1)(x-1)\]Since these are both the same terms, we can write them as \((x-1)^2\)

Answer 48

Does that make things more clear as to how i did it the first way?

Answer 49

yes

Answer 50

Awesome :) Don't forget to medal the user who you think helped you understand :)

Answer 51

Thank you so much! :)

Answer 52

no problem :) glad you understand!