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p and m are positive integers, and p is a prime number. If x

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p and m are positive integers, and p is a prime number. If x [#permalink]

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New post 07 Sep 2013, 10:57
Hi,

Need your guidance for below qq.

p and m are positive integers, and p is a prime number. If x^2-mx+p=0 has a positive interger solution what is the value of (m-p)?


Thanks in advance !!

Last edited by Narenn on 08 Sep 2013, 19:31, edited 1 time in total.
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Re: p and m are positive integers, and p is a prime number. If x [#permalink]

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New post 08 Sep 2013, 19:30
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alokgupta1009 wrote:
Hi,

Need your guidance for below qq.

p and m are positive integers, and p is a prime number. If x^2-mx+p=0 has a positive interger solution what is the value of (m-p)?


Thanks in advance !!


In any quadratic equation \(ax^2 + bx + c = 0\), the relationship between b and c is that b is the sum of the factors of c*a.

In \(x^2 - mx + p = 0\), we know -m is the sum of the factors of p, and we also know that p is the prime number that means it has only two factors, p and 1

So -m = -p -1 -------> m - p = 1

Hope that helps :)
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Re: p and m are positive integers, and p is a prime number. If x [#permalink]

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New post 03 Jul 2014, 05:19
Narenn wrote:
alokgupta1009 wrote:
Hi,

Need your guidance for below qq.

p and m are positive integers, and p is a prime number. If x^2-mx+p=0 has a positive interger solution what is the value of (m-p)?


Thanks in advance !!


In any quadratic equation \(ax^2 + bx + c = 0\), the relationship between b and c is that b is the sum of the factors of c*a.

In \(x^2 - mx + p = 0\), we know -m is the sum of the factors of p, and we also know that p is the prime number that means it has only two factors, p and 1

So -m = -p -1 -------> m - p = 1

Hope that helps :)



I didn't understand the explanation. Could you elaborate and give a numerical example?

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Re: p and m are positive integers, and p is a prime number. If x [#permalink]

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New post 04 Jul 2014, 06:49
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iftach1 wrote:
Narenn wrote:
alokgupta1009 wrote:
Hi,

Need your guidance for below qq.

p and m are positive integers, and p is a prime number. If x^2-mx+p=0 has a positive interger solution what is the value of (m-p)?


Thanks in advance !!


In any quadratic equation \(ax^2 + bx + c = 0\), the relationship between b and c is that b is the sum of the factors of c*a.

In \(x^2 - mx + p = 0\), we know -m is the sum of the factors of p, and we also know that p is the prime number that means it has only two factors, p and 1

So -m = -p -1 -------> m - p = 1

Hope that helps :)



I didn't understand the explanation. Could you elaborate and give a numerical example?


In quadratic equation \(x^2 + bx + c = 0\), -b is the sum of the factors and c is the product. Suppose the factors of the equation are p and q then we can express \(x^2 + bx + c = 0\) as \(x^2 + (p+q)x + pq = 0\) where pq = c and p+q = -b
Consider the equation \(x^2 - 5x + 4 = 0\) The factors of this equation are 4 & 1 so this equation can be expressed as \(x^2 - (4+1)x + (4*1) = 0\)

In the equation \(x^2 - mx + p = 0\) we know that P is the prime number, so it can have only two factors: p and 1, so we can express the equation as \(x^2 - (p+1)x + (p*1) = 0\). SO we have that -m = -p -1 and P = p ------> so m - p = p + 1 - p -----> 1

Hope that helps!!
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p and m are positive integers, and p is a prime number. If x [#permalink]

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New post 05 Jul 2014, 02:10
Hi Narren,

Why you did not consider:-p+1=-m
Along with -p-1=-m I think above mentioned relation is also valid because m is sum or difference of factors.

Then -1 is also a possible value.
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p and m are positive integers, and p is a prime number. If x   [#permalink] 05 Jul 2014, 02:10
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