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

Question

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

GyanOne · Answer

(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.

tapabrata · Answer

\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

Abhishek009 · Answer

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}

JeffTargetTestPrep · Answer

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/  + 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

vmelgargalan · Answer

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!

LeoPTY · Answer

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!

rahulkashyap · Answer

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

BrushMyQuant · Answer

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!

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)

Prep Toolkit

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My Rewards