If p and n are integers such that p>n>0 and p^2-n^2=12, which of the following can be the value of p - n? I. 1 II. 2 III. 4

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If p and n are integers such that p>n>0 and p^2-n^2=12, which of the following can be the value of p - n? I. 1 II. 2 III. 4

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Answer 1 · 2015-06-15T11:35:22

First rewrite $p^{2} - n^{2} = (p + n) (p - n) = 12$ $p^2-n^2=(p+n)(p-n)=12$

Explanation:

$12$ $12$ can only be factorised in a few ways:
$12 = 12 \cdot 1 = 6 \cdot 2 = 4 \cdot 3$ $12=12*1=6*2=4*3$ which would be the values of $p + n and p - n$ $p+nandp-n$ respectively.
The difference of these two factors is then $= 2 n$ $=2n$

This leaves only the $6 \cdot 2$ $6*2$ combination, as the other two would give a non-integer solution for $n$ $n$

Answer : $p - n = 2$ $p-n=2$

Extra : $p = 4 and n = 2$ $p=4andn=2$

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If p and n are integers such that p>n>0 and p^2-n^2=12, which of the following can be the value of p - n? I. 1 II. 2 III. 4

1 Answer

Explanation:

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