Completing the square (How to)

[danielle36]

Here's the problem I'm currently working on:
x^{2} + 6x + 1 = 0
Solve.

Now I'm supposed to be learning completing the square here and the textbook I have to work with doesn't quite explain it in much detail - so I don't know that I've even got the whole concept on how to complete the square.

This is what I've done, which will end up giving me the wrong answer.
x^{2} + 6x + 1 = 0
x^{2} + 6x + 1 + 8 = 8
(x + 3)^{2}= 8
\sqrt{(x+ 3)^{2}} = + or -\sqrt{8}
x = \sqrt{8} - or + 3

I'm really just following an example given from the textbook, to tell the truth I have no idea if the concept actually applies to this situation or not.

The answer is supposed to be
x = -3 + or - 2 \sqrt{2}

 

[yenchin]

when you have

x+3 = +/- sqrt(8),

the next step is x = -3 +/- sqrt(8)

so I don't understand how you end up with sqrt(8) +/- 3

By the way sqrt(8) = 2sqrt(2).

 

[Kurdt]

Completing the square comes from:

(x+a)^2 = x^2+2ax+a^2

(x+a)^2 - a^2 = x^2+2ax

I'm sure the text probably introduced it in that manner. In that case what are you stuck on specifically?

 

[VietDao29]

...
(x + 3)^{2}= 8
\sqrt{(x+ 3)^{2}} = + or -\sqrt{8}

This line is wrong. You don't square root it. Square root something will only return you a non-negative value.

Say, if a² = 16, then \sqrt{a ^ 2} = |a| = 4

What you should do is just leave out the square sign, like this:

(x + 3) ^ 2 = 8
\Rightarrow x + 3 = \pm \sqrt{8}

i.e, (x + 3) can have 2 different values (i.e, \pm \sqrt{8}), both of which if you square, you'll get 8.

Can you get it? :)

 

[rocomath]

standard form is

ax^2 + bx + c

which term do you use in "completing the square," ax^2 or bx ? what would your next step me?

 

[VietDao29]

standard form is

ax^2 + bx + c

which term do you use in "completing the square," ax^2 or bx ? what would your next step me?

Well, he did show his work in the first post. =.=" He just asked where he had gone wrong. :)

 

[rocomath]

i don't see why you kept working with him, it's wrong from the 2nd line?

 

[VietDao29]

i don't see why you kept working with him, it's wrong from the 2nd line?

Err, sorry, but where is the mistake in the second line? I cannot seem to find one.

 

[rocomath]

x^2 + 6x + 1

(6/2)^2

 

[VietDao29]

x^2 + 6x + 1

(6/2)^2

1 + 8 = (6/2)² = 9, isn't that correct? He adds 8 to both sides, and comes up with the second line, to get 1 + 8 = 9, as required. Well, I see nothing wrong there. :)

 

[rocomath]

i PM'ed u so that OP doesn't see my solution.

 

[VietDao29]

x^2 + 6x + 1

(6/2)^2

x^2 + 6x + 1 + 9 = 9

x^2 + 6x + 9 = 8

(x+3)^2 - 8 = 0

...
f(-3) = (-3)^2 + 6(-3) + 1 = -8

Vertex at (-3,-8)

Just read what the problem asks, does it ask you to find the vertex of the parabola? >"<

Btw, doesn't your third line, and his second line, look exactly the same? They are twins, can't you see? :)

 

[rocomath]

i know they are, i was typing out all my steps (obviously)

ok by reading his again, i thought he had "added" 8 to both sides, my bad

btw, i was "confirming" my answer, get over it.

anyways OP, post 4 is good. i got confused with your 2nd line, oops.

 

[cristo]

i know they are, i was typing out all my steps (obviously)

ok by reading his again, i thought he had "added" 8 to both sides, my bad

But he has added 8 to both sides. There are many ways to do this, all equivalent, but some people prefer other methods. The OP has added 8 to both sides, whereas others would perhaps add 0(=8-8) to the LHS.

btw, i was "confirming" my answer, get over it.

There's no need to take that tone.

 

[rocomath]

There's no need to take that tone.

Btw, doesn't your third line, and his second line, look exactly the same? They are twins, can't you see? :)

oh yes, and since I'm human i took this a little offensive as in implying I'm an idiot not even realizing my lines were the "twins." it's not like i cussed him out. he threw something small at me, so i threw something equivalent, no harm :-]

like i said, get over it. come on now, be realistic. don't make it something bigger than it is.

 

[danielle36]

This line is wrong. You don't square root it. Square root something will only return you a non-negative value.

Say, if a² = 16, then \sqrt{a ^ 2} = |a| = 4

What you should do is just leave out the square sign, like this:

(x + 3) ^ 2 = 8
\Rightarrow x + 3 = \pm \sqrt{8}

i.e, (x + 3) can have 2 different values (i.e, \pm \sqrt{8}), both of which if you square, you'll get 8.

Can you get it? :)

Well that certainly helped, I had no idea why it was being done in the original example in my text - now it makes sense!

Ok so if \sqrt{8} is the same as 2\sqrt{2} then my
answer was basically right except, well except where it was wrong.

So it would look more like:

x^{2} + 6x + 1 = 0
x^{2} + 6x + 1 + 8 = 8
(x + 3)^{2} = 8
(x + 3) = +/- \sqrt {8}
x = +/- \sqrt {8} - 3

Would that be correct?

 

[Kurdt]

Thats correct. Of course you can simplify \sqrt{8} = 2 \sqrt{2} to get the answer from your text. Have you seen the method I posted before? Sometimes that's easier to use I find if you're not comfortable with the method presented in your book.

 

[danielle36]

Thanks! Your efforts are definitely appreciated! :)

And I did take a look at the other method too, I've got it written down here I'm going to try it out on some more practice problems... once I get beck to my work... I'm in the midst of procrastinating