0

Guest

Courses New

a,b belong to N and a>1. p is prime .ax^2+bx+c=p for 2 distict integral values of x then the number of integral roots of ax^2+bx+c=2p?

a,b belong toN and a>1. p is prime .ax^2+bx+c=p for 2 distict integral values of x then the number of integral roots of ax^2+bx+c=2p?

Grade:11

2 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
9 years ago
Hello student,
Please find the answer to your question below
If a = b = N, then the roots of ax^2 + ax + c - p are -1/2\pmsqrt(1 - 4(c - p)/a)/2 which implies 1 - 4(c - p)/a is an odd square = n^2, and c <= p.
The roots of ax^2 + ax + c - 2p are given by
x = - 1/2\pmsqrt(1 - 4(c - 2p)/a)/2 = -1/2 \pmsqrt(1 - 4(c - p)/a + 4p/a)/2. If these roots are integral then 1 - 4(c - p)/a + 4p/a = m^2 is also an odd square.
So n^2 + 4p/a = m^2 ==> 4p/a = m^2 - n^2 = (m + n)(m - n).
Since p is prime and a > 1 then 4p/a is an integer only if a = 2, 4, p, 2p, 4p. Now just go through the cases:
A) a = 4p ==> 1 = (m + n)(m - n) ==> m + n = m - n = 1 ==> m = 1, n = 0 ==> c = 2p, which is impossible since c must be <= p. So No.
B) a = 2p ==> 2 = (m + n)(m - n) ==> m + n = 2, m - n = 1 ==> m = 3/2, which is impossible since m is an integer. So No
C) a = p ==> 4 = (m + n)(m - n) ==> two cases
Case 1) m + n = 4, m - n = 1 ==> m = 5/2, impossible since m is an integer.
Case 2) m + n = 2, m - n = 2 ==> m = 2, n = 0 ==> c = 7p/4, which is impossible since c must be <= p.
D) a = 2 ==> 2p = (m+n)(m -n) ==> two cases
Case 1) m+ n = 2p and m - n = 1, impossible since m = p + 1/2 must be an integer.
Case 2) m + n = p, m - n = 2 impossible since m = p - 1/2 is not an integer.
mycroft holmes
272 Points
9 years ago
Let m, n be the two integer valued roots of the quadratic ax2+bx+c-p = 0, so that
 
ax2+bx+(c-p) = a (x-m)(x-n) .
 
Let x=N be an integer such that aN2+bN+(c-p) = p, then
 
a(N-m)(N-n) = p. Since a>1, we need a =p. that means (N-m)(N-n)=1.
 
However, since m and n are distinct, we have (N-m) and (N-n) are distinct integers. However there are no distinct integers that are factors of 1.
 
Hence no integers exist satisfying the quadratic ax2+bx+(c-2p) = 0.

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More