Question

Solve 4x^2+4x-2=0 by completing the square.

Answer 1

I have already solved for the discriminate and received a solution of 48 which told me my reply will be irrational and that there are two potential solutions.

Answer 2

Well first to complete the square, you must make sure the leadig coefficient is 1

Answer 3

So you can't have 4x^2, you hve to divide the whole equaion by 4 first so that you have x^2

Answer 4

leading coefficient*

Answer 5

Okay. That will give me \[(x^2+x-1)/2=0 \] correct?

Answer 6

no

Answer 7

Yes, very god, but make sure you keep the (-1/2) in the parentheses

Answer 8

4x^2+4x+1-3=0,
(2x+1)^2-3=0,
2x+1=\[\sqrt{3}\],
x=(√3-1)/2=0.366025

Answer 9

I don't understand why you would change the -2 to being +1-3. Why would you do that?

Answer 10

is it for the sake of factoring?

Answer 11

because when I factor the equation, I get (2x+2)(2x-1)=0 which when worked gives me \[4x^2-3+4x+1=0\]

Answer 12

or the original equation.

Answer 13

(2x+2)(2x-1) <- not completing square.

Answer 14

I am pretty confused right now. I have no idea how guhuan ended up with a \[\sqrt{3}\] in his answer. I want to understand HOW to get the answer, and why it works the way it does, and the instructor didn't respond to my question via email in time for me to complete this part of my assignment.

Answer 15

Sorry im back but before where we were at \[x^2+x-\frac{1}{2}=0\] ll wegotta do i add 1/2 to both sides to get \[\Huge x^2+x=\frac{1}{2}\]

Answer 16

Now we are going to add something to both sides so that the left side becomes a perect square trinomial, i.e., a trinomial that facotrs to (x+a)^2, you've seen those before right? A trinomial such as \[x^2+4x+4\] that factors to
\[(x+2)(x+2)=(x+2)^2\]?

Answer 17

Yeah, we covered that as part of this chapter I believe.

Answer 18

Ok, where thats what we're going to do becuase then if we take the square root of both sides we're left with x plus or minus a number and then we have a radical number on the right hand side of the equation, and then we'll add or subtract whateer number is next to x and we'll have evaluated x, this may seem confusing but lets go through with your question

Answer 19

And to think I originally wanted to major in accounting....

Answer 20

We have \[x^2+x=\frac{1}{2}\] Well the trick here is we take the b of ax^2+x+c (a standard trinomial) in this case it ax^2 the a is 1 and the bx is also 1 and the c is (-1/2) we just moved it to the right the standerd trinomial is x^2+x-(1/2) ut anyways we're gonna take one half of b and we're gong to square it so in this case we're gong to take 1 (our b in this case) and divide it by 2, which gives us 1/2, then we're going to square it, to get (1/2)^2=1/4 and now we're going to add it to both sides \[x^2+x+(\frac{1}{4})=\frac{1}{2}+(\frac{1}{4})\]

Answer 21

b of ax^2+bx+c*************

Answer 22

which is then \[x^2+x+1/4=3/4\] right?

Answer 23

Correct

Answer 24

Now we can factor the left side into \[\Large (x+\frac{1}{2})^2=\frac{ 3 }{ 4 }\]

Answer 25

Okay. Now this is where I would then square both sides of the equation? or am I jumping ahead?

Answer 26

or rather find the square root?

Answer 27

Nope that exactly correct, you have
"completed the square" (made it possible to make a perfect square trinomial, I find this name to be REALLY lame but yes this is where you take the square root of both sides, and remember the right side has two square roots, a positive and a negative, this will relate to your two solutions

Answer 28

How would I isolate the x? I see how the answer would be \[x=\sqrt{3}-1/2, \sqrt{3}+1/2\] now but I am not sure how to show that I isolated the x.

Answer 29

Its not exactly that, you got rid of the denominator in sqrt(3)/2 for te sqrt of 3/4 Here: \[x+\frac{1}{2} = \sqrt{\frac{3}{4}}\]

Answer 30

So \[x=\frac{ -1 }{ 2 }\pm \frac{ \sqrt{3} }{ 2}\]

Answer 31

OKAYYYYYYYY I see now. the square root canceled the exponent then I was able to subtract 1/2 from each side

Answer 32

Yep

Answer 33

Isolating the x. And because 2 negatives = a positive when multiplied, that is why it would be \[\frac{ -1 }{ 2 }\pm \frac{ \sqrt{3} }{ 2 }\] or \[\frac{ -1\pm \sqrt{3} }{ 2 }\]

Answer 34

Thank you so much for all of the help. It really helped it click for me. Understanding HOW to get the answer is what I needed more than just what the answer was and you really helped me understand that.

Answer 35

You're welcome, that's what were here for

Answer 36

mind writing a testimonial?

Answer 37

For sure.

Answer 38

Thanks, you're awesome