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If (5(pq)^3 + 35(pq)^2 - 40pq)/((p - 1)(p + 4)) = 0 and p, q are both

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If (5(pq)^3 + 35(pq)^2 - 40pq)/((p - 1)(p + 4)) = 0 and p, q are both  [#permalink]

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New post 12 Nov 2019, 04:15
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A
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C
D
E

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33% (02:25) correct 67% (02:30) wrong based on 21 sessions

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Re: If (5(pq)^3 + 35(pq)^2 - 40pq)/((p - 1)(p + 4)) = 0 and p, q are both  [#permalink]

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New post 12 Nov 2019, 04:47
if you divide both sides of the equation by "pq" you have 5(pq)2+35pq−40/((p−1)(p+4))=0 and then dividing everything by 5 we have (pq)2+7pq−8/((p−1)(p+4))=0

we know that p cant be 1 and it cant be -4 because the denominator can't be 0

the numerator of the function can be written as (pq-1)(pq+8) which means that pq should be equal to either 1 or -8

if q=-4 we can have a 0 in the numerator by using p=2, which doesn't lead to a 0 in the denominator
if q=1 we can have a 0 in the numerator by using p=1 or p=-8, p=1 can't be used but p=-8 can be used
if q=2 we can have a 0 in the numerator by using p=-4 which can't be used

in conclusion, answer C both I and II can be used
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Re: If (5(pq)^3 + 35(pq)^2 - 40pq)/((p - 1)(p + 4)) = 0 and p, q are both  [#permalink]

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New post 12 Nov 2019, 04:59
If \(\frac{5(pq)^3+35(pq)^2-40pq}{(p-1)(p+4)}=0\) and p, q are both non-zero integers, which of the following would be the value of q?

\(\frac{5(pq)^3+35(pq)^2-40pq}{(p-1)(p+4)}=0\) means \(5(pq)^3+35(pq)^2-40pq=0......(pq)^2+7pq-8=0....(pq+8)(pq-1)=0....pq=-8..or..pq=1\)
But p cannot be 1 or -4 as (p-1)(p+4) cannot be 0

I. -4 ..........pq=-8..p*-4=8...p=2 Possible
II. 1 .........pq=-8.....p=-8.......Possible
III. 2 ........pq=-8....p=-4...or..pq=-1...p=-1/2...Both NOT possible

I and II possible

C
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Re: If (5(pq)^3 + 35(pq)^2 - 40pq)/((p - 1)(p + 4)) = 0 and p, q are both   [#permalink] 12 Nov 2019, 04:59
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