
#1
Feb1711, 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 



#2
Feb1711, 04:49 AM

Sci Advisor
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Thanks
P: 26,167

hi autodidude!
(try using the X^{2} icon just above the Reply box ) 



#3
Feb1711, 04:52 AM

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P: 1,347

[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, [tex]\frac{4}{2k} = \frac{2}{k}[/tex] (k + 1)^{2} ≠ k^{2} + 1 ... which is wrong. Use the square of a binomial pattern, or FOIL: (k + 1)^{2} = k^{2} + 2k + 1 Fix these errors and repost, please. EDIT: Beaten to it. 



#4
Feb2111, 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 +132k)] / 4k right? The book says 30k instead of 32k Thanks a lot and sorry about my imprecision lol 



#5
Feb2211, 01:51 AM

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P: 3,436

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. 



#6
Feb2311, 07:55 PM

P: 333

[k+1±√((k^2+130k)/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+130k)]/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 Thanks 



#7
Feb2311, 08:14 PM

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P: 3,436

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^24ac}}{2a}[/tex] we get [tex]\frac{k+1\pm\sqrt{(k+1)^24(2k)(4)}}{2(2k)}[/tex] Simplifying, we get [tex]x=\frac{k+1\pm\sqrt{k^230k+1}}{4k}[/tex] If you try to use the quadratic formula by first dividing through by the coefficient of x^{2}, 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. 



#8
Feb2411, 12:27 AM

P: 333

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 



#9
Feb2411, 12:40 AM

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P: 3,436

[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)^232k}{4k^2}[/tex] [tex]=\frac{k^2+2k+132k}{4k^2}[/tex] [tex]=\frac{k^230k+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^230k+1}}{\sqrt{4k^2}}}{2} [/tex] [tex] x=\frac{\frac{k+1}{2k}\pm\frac{\sqrt{k^230k+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. 



#10
Feb2411, 02:22 AM

P: 333

^ Wow that makes perfect sense! Many thanks!




#11
Feb2411, 03:34 AM

HW Helper
P: 3,436

No worries




#12
Feb2411, 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? 



#13
Feb2511, 06:38 PM

HW Helper
P: 3,436

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^24ac}}{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? 



#14
Feb2611, 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^230k+1}}{2k}}{2} [/tex] ...wait, s that what you were saying was wrong in post #7? 



#15
Feb2611, 01:26 AM

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P: 3,436

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] 



#16
Feb2611, 03:30 AM

P: 333

Gotcha, cheers for putting up with me! :D




#17
Feb2611, 04:04 AM

HW Helper
P: 3,436




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