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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)

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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 08 Mar 2017, 05:29
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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2) ?

(A) 1/6

(B) 25/144

(C) 49/144

(D) 7/12

(E) 73/144
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 08 Mar 2017, 07:38
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(1/p^2) + (1/q^2) = (p^2 + q^2) / (pq)^2 = ((p+q)^2 - 2pq)/(pq)^2 = (7^2 - 2*12)/(12^2) = 25/144

(B) it is.
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 08 Mar 2017, 08:11
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\(\frac{1}{p^2}\)+\(\frac{1}{q^2}\)
=\(\frac{p^2+q^2}{p^2q^2}\)
=\(\frac{(p+q)^2-2pq}{(pq)^2}\)
=\(\frac{7^2-2*12}{12^2}\)
=\(\frac{49-24}{144}\)
=\(\frac{25}{144}\)

Answer B
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If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 08 Mar 2017, 08:14
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vikasp99 wrote:
If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2) ?

(A) 1/6

(B) 25/144

(C) 49/144

(D) 7/12

(E) 73/144


\(p + q = 12\)

So, we can say \(p = 4\) and \(q =3\)

Hence, the value of \((\frac{1}{p^2}) + (\frac{1}{q^2})\) -

\(= (\frac{1}{3^2}) + (\frac{1}{4^2})\)

\(= (\frac{1}{9}) + (\frac{1}{16})\)

\(= \frac{25}{144}\)

Thus, answer must be (B) \(\frac{25}{144}\)
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 10 Mar 2017, 10:38
vikasp99 wrote:
If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2) ?

(A) 1/6

(B) 25/144

(C) 49/144

(D) 7/12

(E) 73/144



We can start by adding together (1/p^2) + (1/q^2) by first obtaining a common denominator:

(q^2/q^2)(1/p^2) + (p^2/p^2)(1/q^2)

q^2/[(q^2)(p^2)] + p^2/[(q^2)(p^2)]

(q^2 + p^2)/(pq)^2

Since pq =12, we have:

(q^2 + p^2)/(12)^2

(q^2 + p^2)/144

To determine the value of q^2 + p^2, we can do the following:

(p + q)^2 = (7)^2

p^2 + q^2 + 2pq = 49

Since pq = 12, we know:

p^2 + q^2 + 24 = 49

p^2 + q^2 = 25

Thus, (q^2 + p^2)/144 = 25/144

Answer: B
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 23 Mar 2017, 11:05
I used a different method, which was a bit over 2 minutes, but basically:

plug in p = 7 - p into th eequation pq = 12, (or we can plug in q = 7 - p).

Then we foil the formula which will end up being p^2 - 7p +12 = 0,
(p - 4)(p - 3),
p = 3 or = 4. And we know that we will get the same values for q,

Therefore there are 3 possible answer:

(1/3^3) + (1/3^2) = 2/9 - not available
(1/4^2) + (1/4^2) = 2/16 = 1/8 - not available
(1/4^2) + (1/3^2) = 1/16 + 1/9 = 25/144,

Thus, the answer is B.

I am still practicing hoping I can use more time efficient. Tapabrata's ways surely seems way faster!
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 22 Oct 2018, 22:11
tapabrata wrote:
\(\frac{1}{p^2}\)+\(\frac{1}{q^2}\)
=\(\frac{p^2+q^2}{p^2q^2}\)
=\(\frac{(p+q)^2-2pq}{(pq)^2}\)
=\(\frac{7^2-2*12}{12^2}\)
=\(\frac{49-24}{144}\)
=\(\frac{25}{144}\)

Answer B


Can someone please elaborate on how the \(-2pq\) is obtained in going from \(\frac{p^2+q^2}{p^2q^2}\) to \(\frac{(p+q)^2-2pq}{(pq)^2}\)

I can solve this formula by using the available statements to know that p and q = 3 and 4, but would like to understand solving this algebraically. Thanks!
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)  [#permalink]

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New post 22 Oct 2018, 23:39
1
LeoPTY wrote:
tapabrata wrote:
\(\frac{1}{p^2}\)+\(\frac{1}{q^2}\)
=\(\frac{p^2+q^2}{p^2q^2}\)
=\(\frac{(p+q)^2-2pq}{(pq)^2}\)
=\(\frac{7^2-2*12}{12^2}\)
=\(\frac{49-24}{144}\)
=\(\frac{25}{144}\)

Answer B


Can someone please elaborate on how the \(-2pq\) is obtained in going from \(\frac{p^2+q^2}{p^2q^2}\) to \(\frac{(p+q)^2-2pq}{(pq)^2}\)

I can solve this formula by using the available statements to know that p and q = 3 and 4, but would like to understand solving this algebraically. Thanks!



to make the p^2 + q^2 to (p+q)^2 = p^2+q^2 + 2pq, we would need to add and subtract (to cancel the terms) 2pq. The added 2pq is in the (p+q)^2 term, while the "-2pq" remains outside it
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Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)   [#permalink] 22 Oct 2018, 23:39
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