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If p + q = 7 and pq = 12, then what is the value of
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04 Apr 2019, 04:21
04 Apr 2019, 04:21
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If p + q = 7 and pq = 12, then what is the value of \(\frac{1}{p^2} + \frac{1}{q^2}\) ?
(A) \(\frac{1}{6}\)
(B) \(\frac{25}{144}\)
(C) \(\frac{49}{144}\)
(D) \(\frac{7}{12}\)
(E) \(\frac{73}{144}\)
(A) \(\frac{1}{6}\)
(B) \(\frac{25}{144}\)
(C) \(\frac{49}{144}\)
(D) \(\frac{7}{12}\)
(E) \(\frac{73}{144}\)
Re: If p + q = 7 and pq = 12, then what is the value of
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04 Apr 2019, 20:45
04 Apr 2019, 20:45
1
Carcass wrote:
If p + q = 7 and pq = 12, then what is the value of \(\frac{1}{p^2} + \frac{1}{q^2}\) ?
(A) \(\frac{1}{6}\)
(B) \(\frac{25}{144}\)
(C) \(\frac{49}{144}\)
(D) \(\frac{7}{12}\)
(E) \(\frac{73}{144}\)
(A) \(\frac{1}{6}\)
(B) \(\frac{25}{144}\)
(C) \(\frac{49}{144}\)
(D) \(\frac{7}{12}\)
(E) \(\frac{73}{144}\)
Here,
\(p + q = 7\)
or \((p + q)^2 = 49\)
or\(p^2 + q^2 + 2pq = 49\)
or \(p^2 +q^2 = 25\)
Now,
\(\frac{1}{p^2} + \frac{1}{q^2}\) = \(\frac{{p^2 + Q^2}}{(pq)^2} = \frac{25}{{12^2}}= \frac{25}{144}\)
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Re: If p + q = 7 and pq = 12, then what is the value of
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03 Sep 2022, 04:21
03 Sep 2022, 04:21
Given that p + q = 7 and pq = 12 and we need to find the value of \(\frac{1}{p^2} + \frac{1}{q^2}\)
p + q = 7
Squaring both the sides we get
\((p + q)^2 = 7^2\) = 49
=> \(p^2 + 2pq + q^2\) = 49
=> \(p^2 + 2*12 + q^2\) = 49
=> \(p^2 + q^2\) = 49 - 24 = 25 ...(1)
\(\frac{1}{p^2} + \frac{1}{q^2}\) = \(\frac{q^2 + p^2 }{ p^2 * q^2}\) = \(\frac{p^2 + q^2 }{ (pq )^2}\) = \(\frac{25}{(12)^2}\) = \(\frac{25}{144}\)
So, Answer will be B
Hope it helps!
p + q = 7
Squaring both the sides we get
\((p + q)^2 = 7^2\) = 49
=> \(p^2 + 2pq + q^2\) = 49
=> \(p^2 + 2*12 + q^2\) = 49
=> \(p^2 + q^2\) = 49 - 24 = 25 ...(1)
\(\frac{1}{p^2} + \frac{1}{q^2}\) = \(\frac{q^2 + p^2 }{ p^2 * q^2}\) = \(\frac{p^2 + q^2 }{ (pq )^2}\) = \(\frac{25}{(12)^2}\) = \(\frac{25}{144}\)
So, Answer will be B
Hope it helps!
