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Second-Degree Equations

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by cramya » Fri Mar 27, 2009 7:06 pm
Take LCM on the left hand side which is k

k - (3-2k^2)/ k = x/k

k^2 - (3-2k^2) / k = x/k

Both k's in the denominator of left hand side and right hand side cancel

k^2 - (3-2k^2) = x

k^2 - 3 + 2k^2 = x

3k^2 - 3 = x


C

Hope this helps!

Regards,
CR
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by BlindVision » Fri Mar 27, 2009 7:13 pm
cramya wrote:
Both k's in the denominator of left hand side and right hand side cancel

C

Hope this helps!

Regards,
CR
Thank you for emphasizing this! I understand this problem better now. You are correct :wink:
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Re: Second-Degree Equations

by hrishikesh05 » Sat Mar 28, 2009 7:22 am
BlindVision wrote:if k NOT EQUAL 0 and k - (3 - 2k^2 / k) = x/k, then x =

A) -3 - k^2

B) k^2 - 3

C) 3k^2 - 3

D) k - 3 - 2k^2

E) k - 3 + 2k^2


OA = C



Please put the brackets in the correct place. :roll:
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Re: Second-Degree Equations

by BlindVision » Sat Mar 28, 2009 8:03 am
hrishikesh05 wrote:
BlindVision wrote:if k NOT EQUAL 0 and k - (3 - 2k^2 / k) = x/k, then x =

A) -3 - k^2

B) k^2 - 3

C) 3k^2 - 3

D) k - 3 - 2k^2

E) k - 3 + 2k^2


OA = C



Please put the brackets in the correct place. :roll:
That was how the question was wrote. Could you please show me what you meant, instead of showing an emoticon with rolling eyes? :roll:
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by gmat740 » Sat Mar 28, 2009 5:57 pm
if k NOT EQUAL 0 and k - (3 - 2k^2 / k) = x/k, then x =
it has to be k - [3 - 2k^2]/k

otherwise what you have written it simple means:

k - (3 - 2k^2 / k)
==> k - (3 - 2k)

Because the K in denominator will divide 2k^2 and we will be left with only 2k

Hope now its clear where to put the brackets
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by BlindVision » Sat Mar 28, 2009 6:54 pm
Thank you for the explanation, GMAT740 :)
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by gmat740 » Sat Mar 28, 2009 8:33 pm
You are welcome :D
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by arunsharma » Mon Mar 07, 2011 9:48 am
Hey guys,

I am not understanding one thing in this question and that is why first K becomes K2?

I understand both denominator K's go away and when brackets open - sign changes.

Just can't understand K to k2 part.

I saw Cramya's reply:

"Take LCM on the left hand side which is k

k - (3-2k^2)/ k = x/k

k^2 - (3-2k^2) / k = x/k "

I don't understand why are we taking LCM and how are we finding that LCM. I feel I am missing a small part and will be good to go once get it. Any help would be appreciated... thanks in advance.
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by GMATGuruNY » Mon Mar 07, 2011 10:03 am
BlindVision wrote:if k NOT EQUAL 0 and k - (3 - 2k^2)/k = x/k, then x =

A) -3 - k^2

B) k^2 - 3

C) 3k^2 - 3

D) k - 3 - 2k^2

E) k - 3 + 2k^2


OA = C
Plug in k=2 into k - (3 - 2k^2)/k = x/k and solve for x:

2 - (3 - 2*2^2)/2 = x/2
2 - (3-8)/2 = x/2
2 + 5/2 = x/2
9/2 = x/2
x = 9. This is our target.

Now we plug k=2 into all the answers to see which yields our target of 9.

Only answer choice C works:
3k^2 - 3 = 3*2^2 - 3 = 12-3 = 9.

The correct answer is C.
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