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Another quadratic equation...

by autodidude
Tags: equation, quadratic
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autodidude
#1
Feb17-11, 04:00 AM
P: 333
1. The problem statement, all variables and given/known data
Solve 2kx^2 - (k+1)x + 4 by completing the square


2. Relevant equations

2kx^2 - (k+1)x + 4

3. The attempt at a solution

What I have so far:

-> x^2 - (k+1/2k)x + 4/2k = 0

-> x^2 - (k+1/2k)x + ((k+1/2k)/2)^2 - ((k+1/2k/2)^2 + 2k = 0

-> x^2 - (k+1/2k)x + (k+1/4k)^2 - (k+1/4k)^2 + 2k = 0

-> x^2 - (k+1/2k)x + (k^2+1/16k^2) - (k^2+1/16k^2) + 2k = 0

-> (x - (k+1/4k))^2 = (k^2+1/16k^2) + 2k

What next?

Thank you
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tiny-tim
#2
Feb17-11, 04:49 AM
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tiny-tim's Avatar
P: 26,148
hi autodidude!

(try using the X2 icon just above the Reply box )
Quote Quote by autodidude View Post
-> x^2 - (k+1/2k)x + 4/2k = 0

-> x^2 - (k+1/2k)x + ((k+1/2k)/2)^2 - ((k+1/2k/2)^2 + 2k = 0
what happened to the 4/2k ?
-> x^2 - (k+1/2k)x + (k+1/4k)^2 - (k+1/4k)^2 + 2k = 0

-> x^2 - (k+1/2k)x + (k^2+1/16k^2) - (k^2+1/16k^2) + 2k = 0
(k+1/4k)2 + 2k isn't (k2+1/16k2)
What next?
take the square-root () of both sides
eumyang
#3
Feb17-11, 04:52 AM
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P: 1,347
Quote Quote by autodidude View Post
2. Relevant equations

2kx^2 - (k+1)x + 4
This is not an equation. This is an expression. I'm assuming that there is a "= 0" at the end.

Quote Quote by autodidude View Post
What I have so far:

-> x^2 - (k+1/2k)x + 4/2k = 0
I am reading this as
[tex]x^2 - \left( k + \frac{1}{2k} \right) x + \frac{4}{2k} = 0[/tex]
I know what you mean but if you can't use LaTeX, add grouping symbols, at least, like this:
x^2 - ((k+1)/2k)x + 4/2k = 0
It's hard to be precise, but one has to be,

Quote Quote by autodidude View Post
-> x^2 - (k+1/2k)x + ((k+1/2k)/2)^2 - ((k+1/2k/2)^2 + 2k = 0
4/2k = 2k? That's wrong. You mean
[tex]\frac{4}{2k} = \frac{2}{k}[/tex]

Quote Quote by autodidude View Post
-> x^2 - (k+1/2k)x + (k+1/4k)^2 - (k+1/4k)^2 + 2k = 0

-> x^2 - (k+1/2k)x + (k^2+1/16k^2) - (k^2+1/16k^2) + 2k = 0
There are still issues with parentheses. Also, from the bolded part, you state:
(k + 1)2 ≠ k2 + 1
... which is wrong. Use the square of a binomial pattern, or FOIL:
(k + 1)2 = k2 + 2k + 1

Fix these errors and repost, please.

EDIT: Beaten to it.

autodidude
#4
Feb21-11, 09:14 PM
P: 333
Another quadratic equation...

Fixing the silly '2k' error made everything work out! But now I'm having trouble with solving it by using the quadratic formula. It works fine when I just plug in the values as they are but if I simply first to get

x^2 - ((k+1)/2k)x + (2/k) = 0

and then plug THOSE values in, after working it all out, I end up with:

[k+1√[((k^2+1)/4k^2) - (64/k^2)]] / 4k

There's probably another stupid error in there somewhere but I just don't see it

By the way, the answer is [k+1√(k^2 +1-32k)] / 4k right? The book says -30k instead of -32k

Thanks a lot and sorry about my imprecision lol
Mentallic
#5
Feb22-11, 01:51 AM
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P: 3,531
You've forgotten to expand (k+1)2 correctly again. If you did, you would get -30k instead of -32k.

By the way, the question asked for you to solve it by completing the square. You've used the quadratic formula, which you could've used right at the start anyway before dividing through by 2k.
autodidude
#6
Feb23-11, 07:55 PM
P: 333
Quote Quote by Mentallic View Post
You've forgotten to expand (k+1)2 correctly again. If you did, you would get -30k instead of -32k.
Ah yeah I got it for completing the square now, the stupid mistake above was squaring 4ac, and not expanding (k+1). After reworking everything, I still get:

[k+1√((k^2+1-30k)/4k^2)]/4k

The only way I can think of getting rid of the 4k^2 is by taking the square root of it first before multiplying the 2k under k+1 with 2. This gives 2k, so I can write it as

[(k+1)√(k+1-30k)]/2k and that whole expression over another 2 which would give the answer though I'm not sure if that's allowed and exactly what I'm doing really

Quote Quote by Mentallic View Post
By the way, the question asked for you to solve it by completing the square. You've used the quadratic formula, which you could've used right at the start anyway before dividing through by 2k.
Yeah I know, but since I'm trying to teach myself, I want to be familiar with different situations and so I'm tackling the more difficult looking questions (for me) using different methods

Thanks
Mentallic
#7
Feb23-11, 08:14 PM
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Quote Quote by autodidude View Post
[k+1√((k^2+1-30k)/4k^2)]/4k
I'm not really sure how you got that... It's not right first of all, but then you get the right answer later down the track?

From [tex]2kx^2-(k+1)x+4=0[/tex] we have
[tex]a=2k[/tex]
[tex]b=-(k+1)[/tex]
[tex]c=4[/tex]

Now, plugging into the quadratic [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

we get [tex]\frac{k+1\pm\sqrt{(k+1)^2-4(2k)(4)}}{2(2k)}[/tex]

Simplifying, we get [tex]x=\frac{k+1\pm\sqrt{k^2-30k+1}}{4k}[/tex]

If you try to use the quadratic formula by first dividing through by the coefficient of x2, then
[tex]a=1[/tex]
[tex]b=\frac{-(k+1)}{2k}[/tex]
[tex]c=\frac{2}{k}[/tex]

[tex]x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}[/tex]

So now you should be able to take out a factor of [tex]\frac{1}{2k}[/tex] from the surd, and bring that down to the denominator of the fraction to get the same result as before.
autodidude
#8
Feb24-11, 12:27 AM
P: 333
Quote Quote by Mentallic View Post
I'm not really sure how you got that... It's not right first of all, but then you get the right answer later down the track?
Well I did get this: [tex]x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}[/tex]


but then I expanded ((k+1)/2k)^2 to get (k+1)^2/4k^2 and multiplied 8/k by 4k/4k to get 32k/4k^2



Quote Quote by Mentallic View Post
So now you should be able to take out a factor of [tex]\frac{1}{2k}[/tex] from the surd, and bring that down to the denominator of the fraction to get the same result as before.
Could you elaborate please? (baby steps lol)
Mentallic
#9
Feb24-11, 12:40 AM
HW Helper
P: 3,531
Quote Quote by autodidude View Post
Well I did get this: [tex]x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}[/tex]


but then I expanded ((k+1)/2k)^2 to get (k+1)^2/4k^2 and multiplied 8/k by 4k/4k to get 32k/4k^2





Could you elaborate please? (baby steps lol)
Yep that's right, so what you have so far is

[tex]
x=\frac{\frac{k+1}{2k}\pm\sqrt{\frac{\left(k+1\right)^2}{4k^2}-\frac{32k}{4k^2}}}{2}
[/tex]

Looking at the discriminant only:

[tex]\Delta=\frac{\left(k+1\right)^2-32k}{4k^2}[/tex]

[tex]=\frac{k^2+2k+1-32k}{4k^2}[/tex]

[tex]=\frac{k^2-30k+1}{4k^2}[/tex]

Now what I mean by taking out a factor of [tex]\frac{1}{2k}[/tex] is that [tex]\sqrt{4k^2}=2k[/tex] so we have:

[tex]
x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{\sqrt{4k^2}}}{2}
[/tex]

[tex]
x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{2k}}{2}
[/tex]

Now we can factor out [tex]\frac{1}{2k}[/tex] from the numerator, and we can move this to the denominator to get our desired result.
autodidude
#10
Feb24-11, 02:22 AM
P: 333
^ Wow that makes perfect sense! Many thanks!
Mentallic
#11
Feb24-11, 03:34 AM
HW Helper
P: 3,531
No worries
autodidude
#12
Feb24-11, 11:46 AM
P: 333
I edited my post to add this 4 minutes after you replied so I'm just gonna post as a new reply :p

Just one last thing, say I didn't have the answer, how do I know not to move the 2k under the 2 before working with the discriminant?
Mentallic
#13
Feb25-11, 06:38 PM
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P: 3,531
Quote Quote by autodidude View Post
how do I know not to move the 2k under the 2 before working with the discriminant?
Eh? I'm at a loss at what you're asking.

You don't need to know the answer, we're just simplifying things. For example, it is customary to rearrange an expression such as

[tex]\frac{\left(\frac{a}{b}\right)}{c}[/tex] into [tex]\frac{a}{bc}[/tex].

We leave the quadratic formula as [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

but we can equivalently (but less customarily) write it as

[tex]x=\frac{\frac{-b}{2}\pm\sqrt{\frac{b^2}{4}-ac}}{a}[/tex]

or

[tex]x=\frac{\frac{-b}{a}\pm\sqrt{\frac{b^2}{a^2}-\frac{4c}{a}}}{2}[/tex]

etc.

Does this answer your question?
autodidude
#14
Feb26-11, 12:31 AM
P: 333
[tex]x=\frac{\frac{k+1}{2k}\pm\sqrt{\left(\frac{k+1}{2k}\right)^2-\frac{8}{k}}}{2}[/tex]

Basically this, what I did wrong before was moving the 2k denominator under k+1 (on the left) down to get 4k which prevented me from being able to do the factoring here:

[tex]
x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^2-30k+1}}{2k}}{2}
[/tex]

...wait, s that what you were saying was wrong in post #7?
Mentallic
#15
Feb26-11, 01:26 AM
HW Helper
P: 3,531
Yes, that's what you did wrong previously.

If you have [tex]\frac{\frac{a}{b}+c}{d}[/tex] it's not valid to convert this into [tex]\frac{a+c}{bd}[/tex] which is what you've essentially done. You first need to drag out of a factor of [tex]\frac{1}{b}[/tex] from the numerator by first multiplying c by b as such,

[tex]\frac{\frac{a}{b}+\frac{cb}{b}}{d}=\frac{\frac{1}{b}\left(a+cb\right)}{ d}=\frac{a+cb}{bd}[/tex]
autodidude
#16
Feb26-11, 03:30 AM
P: 333
Gotcha, cheers for putting up with me! :D
Mentallic
#17
Feb26-11, 04:04 AM
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P: 3,531
Quote Quote by autodidude View Post
Gotcha, cheers for putting up with me! :D
Glad I was able to help!


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