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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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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)
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08 Mar 2017, 07:38
(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)
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08 Mar 2017, 08:11
\(\frac{1}{p^2}\)+\(\frac{1}{q^2}\) =\(\frac{p^2+q^2}{p^2q^2}\) =\(\frac{(p+q)^22pq}{(pq)^2}\) =\(\frac{7^22*12}{12^2}\) =\(\frac{4924}{144}\) =\(\frac{25}{144}\)
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)
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08 Mar 2017, 08:14
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)
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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)
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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)
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22 Oct 2018, 22:11
tapabrata wrote: \(\frac{1}{p^2}\)+\(\frac{1}{q^2}\) =\(\frac{p^2+q^2}{p^2q^2}\) =\(\frac{(p+q)^22pq}{(pq)^2}\) =\(\frac{7^22*12}{12^2}\) =\(\frac{4924}{144}\) =\(\frac{25}{144}\)
Answer B Can someone please elaborate on how the \(2pq\) is obtained in going from \(\frac{p^2+q^2}{p^2q^2}\) to \(\frac{(p+q)^22pq}{(pq)^2}\) 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)
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22 Oct 2018, 23:39
LeoPTY wrote: tapabrata wrote: \(\frac{1}{p^2}\)+\(\frac{1}{q^2}\) =\(\frac{p^2+q^2}{p^2q^2}\) =\(\frac{(p+q)^22pq}{(pq)^2}\) =\(\frac{7^22*12}{12^2}\) =\(\frac{4924}{144}\) =\(\frac{25}{144}\)
Answer B Can someone please elaborate on how the \(2pq\) is obtained in going from \(\frac{p^2+q^2}{p^2q^2}\) to \(\frac{(p+q)^22pq}{(pq)^2}\) 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




Re: If p + q = 7 and pq = 12, then what is the value of (1/p^2) + (1/q^2)
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22 Oct 2018, 23:39






