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If p+q = j and p-q = k, then 2pq =

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martie11
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If p+q = j and p-q = k, then 2pq = [#permalink] New post   26 Nov 2010, 13:11
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If p+q = j and p-q = k, then 2pq =

A. (j^2 - k^2)/2
B. (k^2 - j^2)/2
C. jk/2
D. (j^2 + k^2)/2
E. (j - k)/2

Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

Can someone post the algebraic solution? Thanks in advance.
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Bunuel
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Re: Solving for 2pq [#permalink] New post   26 Nov 2010, 13:21
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martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.


Given: \(p+q = j\) and \(p-q = k\) --> square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^2-2pq+q^2=k^2\) --> now, subtract (2) form (1): \(4pq=j^2-k^2\) --> \(2pq=\frac{j^2-k^2}{2}\).

Answer: A.
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Re: Solving for 2pq [#permalink] New post   26 Nov 2010, 13:43
Thanks very much Bunuel.

+1
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Re: Solving for 2pq [#permalink] New post   26 Nov 2010, 14:03
Bunuel wrote:
martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.


Given: \(p+q = j\) and \(p-q = k\) --> square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^2-2pq+q^2=k^2\) --> now, subtract (2) form (1): \(4pq=j^2-k^2\) --> \(2pq=\frac{j^2-k^2}{2}\).

Answer: A.


Bunuel did you know to square all of the terms once you looked at the answers - since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach.
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Re: Solving for 2pq [#permalink] New post   26 Nov 2010, 14:23
Good question...my thought exactly...I didn't 'see' that I should square both sides...
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Re: Solving for 2pq [#permalink] New post   27 Nov 2010, 10:49
gettinit wrote:
Bunuel wrote:
martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.


Given: \(p+q = j\) and \(p-q = k\) --> square both expressions: \(p^2+2pq+q^2=j^2\) and \(p^2-2pq+q^2=k^2\) --> now, subtract (2) form (1): \(4pq=j^2-k^2\) --> \(2pq=\frac{j^2-k^2}{2}\).

Answer: A.


Bunuel did you know to square all of the terms once you looked at the answers - since they were all in J^2 or K^2 form? I went the route of understanding this was a play on the a^2 +2ab+b^2 form but wouldn't think to subtract the two unless I looked at the answer choices. Just wanted to understand your approach.


These kind of questions wud b easier if we make a note that....

(a+b)^2 = a^2 + b^2 + 2ab
(a-b)^2 = a^2 + b^2 -2ab

=> a^2 + b^2 = [(a+b)^2 + (a-b)^2]/2...Sum of the squares *0.5

also

=> 4ab = [(a+b)^2 - (a-b)^2]/2 ...difference of the squares *0.5
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Re: Solving for 2pq [#permalink] New post   Updated on: 29 Nov 2010, 16:22
Look right at 2pq which hints at use of the formula- (x+y)2 and (x-y)2

subtracting (x-y)2 from (x+y)2 above gives 4pq which when divided by 2 gives the result.


Answer:- A

Originally posted by Sarang on 27 Nov 2010, 18:46.
Last edited by Sarang on 29 Nov 2010, 16:22, edited 1 time in total.
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Re: Solving for 2pq [#permalink] New post   29 Nov 2010, 07:27
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martie11 wrote:
Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

If p+q = j and p-q = k, then 2pq =

a) (j^2 - k^2)/2
b) (k^2 - j^2)/2
c) jk/2
d) (j^2 + k^2)/2
e) (j - k)/2

Can someone post the algebraic solution? Thanks in advance.


MAKE IT SIMPLE.

ADD 1st equation to second
==> 2p=j+k
SUBTRACT 2nd equation from 1st
==> 2q=j-k

multiply the above 2 equations
4pq=j^2-K^2
==> 2pq = j^2-K^2 / 2
ANSWER "A"

Regards,
Murali.

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Re: If p+q = j and p-q = k, then 2pq = [#permalink] New post   17 Jul 2014, 02:53
p+q = j ........... (1)
p-q = k ............ (2)

(1) + (2)

\(p = \frac{j+k}{2}\)

(1) - (2)

\(q = \frac{j-k}{2}\)

\(2pq = 2 * \frac{j+k}{2} *\frac{j-k}{2}\)

\(= \frac{j^2 - k^2}{2}\)

Answer = A
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Re: If p+q = j and p-q = k, then 2pq = [#permalink] New post   27 Oct 2016, 00:06
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martie11 wrote:
If p+q = j and p-q = k, then 2pq =

A. (j^2 - k^2)/2
B. (k^2 - j^2)/2
C. jk/2
D. (j^2 + k^2)/2
E. (j - k)/2

Hello, I know that I can solve this by plugging in, but I'm can't seem to solve this algebraically...

Can someone post the algebraic solution? Thanks in advance.




here you go,

(p+q)^2 = p^2 + q^2 +2pq = j^2 ................1
(p-q)^2 = p^2 + q^2 -2pq = k^2 ..................2

subtract 2 from 1

4pq=j^2 - k^2 ........................3

divide eqaution 3 by 2 on both sides

2pq = (j^2 - k^2)/2

which is option A.
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If p+q = j and p-q = k, then 2pq = [#permalink] New post   15 Feb 2022, 11:44
Hi,

I solved this question using a method different from the ones in the posts above.

We need to find 2pq in terms of j and k, let's isolate p and q:

p+q=j
+
p-q=k

2p=j+k

Now we need to find q:
p+q=j
-
p-q=k

2q= j-k
q=(j-k)/2

therefore 2pq = (j+k)*(j-k)/2 = (j^2-k^2)/2

Answer A
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If p+q = j and p-q = k, then 2pq = [#permalink]
15 Feb 2022, 11:44
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