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If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal

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faifai0714
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If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal [#permalink] New post   Updated on: 30 Jan 2014, 22:54
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If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal to p^2 for all values of p?

A. -2
B. -1
C. 0
D. 1
E. 2
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C

Originally posted by faifai0714 on 19 Mar 2007, 21:39.
Last edited by Bunuel on 30 Jan 2014, 22:54, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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KarishmaB
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Re: PS: Ampersand Function [#permalink] New post   30 Jan 2014, 21:08
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joshlevin90 wrote:
techjanson wrote:
faifai0714 wrote:
If p&q = p2 + q2 – 2pq, for what value of q is p&q equal to p2 for all values of p?

a)-2
b)-1
c)0
d)1
e)2

(p2=p to the power 2)


C

p2 + q2 – 2pq = p2
=> q2 - 2pq = 0
=> q (q - 2p) = 0
=> q = 0 or q = 2p


im not following the logic you used here.. how did you replace p&q with 0 could someone help explain it in MORE DETAIL


Given \(p&q = p^2 + q^2 - 2pq\) (I am assuming the sign before 2pq is '-')

We want the value of q such that \(p&q = p^2\)
i.e. \(p^2 + q^2 - 2pq = p^2\)
This means that \(q^2 - 2pq\) should be 0 so that left side becomes equal to right side.
If q = 0, both q^2 and 2pq will become 0.
or you can do it algebraically too: \(q^2 - 2pq = 0\)
\(q(q - 2p) = 0\)
So if q = 0 or if q = 2p, p&q will be equal to \(p^2\).

We have q = 0 in the options hence answer (C)
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ricokevin
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[#permalink] New post   20 Mar 2007, 00:49
p&q = p^2 + q^2 – 2pq = (p-q)^2
Therefore, if q = 0, p&q = p^2.

The answer is C.

:)
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[#permalink] New post   20 Mar 2007, 18:49
C for me...

though I saw it could be factored, I just plugged 0 in for q to see what happened... lucky guess...

but yes, the approach above is what you would normally do
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Re: PS: Ampersand Function [#permalink] New post   21 Mar 2007, 16:18
faifai0714 wrote:
If p&q = p2 + q2 – 2pq, for what value of q is p&q equal to p2 for all values of p?

a)-2
b)-1
c)0
d)1
e)2

(p2=p to the power 2)


C

p2 + q2 – 2pq = p2
=> q2 - 2pq = 0
=> q (q - 2p) = 0
=> q = 0 or q = 2p
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Re: PS: Ampersand Function [#permalink] New post   30 Jan 2014, 18:05
techjanson wrote:
faifai0714 wrote:
If p&q = p2 + q2 – 2pq, for what value of q is p&q equal to p2 for all values of p?

a)-2
b)-1
c)0
d)1
e)2

(p2=p to the power 2)


C

p2 + q2 – 2pq = p2
=> q2 - 2pq = 0
=> q (q - 2p) = 0
=> q = 0 or q = 2p


im not following the logic you used here.. how did you replace p&q with 0 could someone help explain it in MORE DETAIL
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Re: If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal [#permalink] New post   22 Nov 2016, 17:27
I have a question, can i convert p^2 + 2pq + q^2 => (p + q)^2

If I use (p + q)^2, the I can solve the question much easier because at the end I see the below:

(square root(14) + square root(13) + square root(14) - square root(13))^2 => (square root(14) + square root(14))^2 =>
=> 14 + 2square root(14)square root(14) + 14 => 28 + 28 = 56

What are your thoughts?
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Re: If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal [#permalink] New post   23 Nov 2016, 03:59
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lalania1 wrote:
I have a question, can i convert p^2 + 2pq + q^2 => (p + q)^2

If I use (p + q)^2, the I can solve the question much easier because at the end I see the below:

(square root(14) + square root(13) + square root(14) - square root(13))^2 => (square root(14) + square root(14))^2 =>
=> 14 + 2square root(14)square root(14) + 14 => 28 + 28 = 56

What are your thoughts?


I don't understand your logic at all.

You are given that:
p&q = p^2 + q^2 - 2pq = p^2

So q^2 = 2pq
So q is either 0 or 2p.
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Re: If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal [#permalink] New post   24 Nov 2016, 07:44
thank you for your response. My apologies for the confusion.

in essence, my question is... is p^2 + 2pq + q^2 = (p + q)^2. if so, how would it work in making this question easier
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Re: If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal [#permalink] New post   13 Jun 2018, 16:26
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faifai0714 wrote:
If p&q = p^2 + q^2 – 2pq, for what value of q is p&q equal to p^2 for all values of p?

A. -2
B. -1
C. 0
D. 1
E. 2


We see that if q = 0, then p&0 = p^2 - 0^2 - 2p(0) = p^2. So q must be 0.

Alternate solution:

Notice that p^2 + q^2 - 2pq = (p - q)^2, so p&q = (p - q)^2. We see that if q = 0, then (p - q)^2 = p^2.

Answer: C
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Re: If p&q = p^2 + q^2 2pq, for what value of q is p&q equal [#permalink] New post   28 Oct 2023, 09:31
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