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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
p and m are positive integers, and p is a prime number. If x [#permalink]
05 Jul 2014, 01:10
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
05 Jul 2014, 01:10