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ILLUSTRATION 2.13 Let a, b eN and a > 1. Also p is a prime number. If ax² + bx +c=p for two distinct integral values of x, then prove that ax²+bx+c #2p for any integral value of x.

ILLUSTRATION 2.13 Let a, b eN and a > 1. Also p is a prime number. If ax² + bx +c=p for two distinct integral values of x, then prove that ax²+bx+c #2p for any integral value of x.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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f(x)=0. Then find the value of P(4). U are the squares of the roots of Sol. According to the question, and P(x) = k(x- o) (x − B²) (x - y²) ⇒-ko² B²y² = -1 → -αβγ=1 Given P(0) = -1 Also, f(0) = 1 :. --k(-1)² = -1 ⇒ k=1 ²) (x − B²) (x - y²) .. f(x) = x³ + x + 1 = (x - α) (x − B) (x−y) P(x) = (x - Now, P(4) = (2-a) (2-B) (2-1) (2 + a)(2+B) (2 + y) :. =ƒ(2)(-ƒ(-2)) = (2³ + 2 + 1) (−(−2)³ − (−2) – 1) = 11 x 9 = 99 ILLUSTRATION 2.12 Let f(x) be a polynomial with integral coefficients. Iff (1) and f(2) both are odd integers, prove that f(x) = 0 can't have any integral root. Sol. Let us assume that f(x) = 0 for some integer x = k. Then, x-k divides f(x). coefficients f(x)=(x-k) g(x), where g (x) is a polynomial with integral f(1) = (1-k) g(1) and f(2)= (2-k) g(2) ƒ(1) f(2)=(1-k)(2-k) g(1) g(2), (1) which is clearly an even number, which contradicts the given information that both f(1) and f(2) are odd integers. Hence f(x) = 0 can't have any integral root. ILLUSTRATION 2.13 Let a, b eN and a > 1. Also p is a prime number. If ax² + bx +c=p for two distinct integral values of x, then prove that ax²+bx+c #2p for any integral value of x. Sol. Given ax²+bx+c-p=0 has integral roots. Let a, ß be the roots. Then ax²+bx+c-p= a(x-a) (x-3) Now from ax² + bx+c=2p, we have 2 (1)
Transcribed Image Text:f(x)=0. Then find the value of P(4). U are the squares of the roots of Sol. According to the question, and P(x) = k(x- o) (x − B²) (x - y²) ⇒-ko² B²y² = -1 → -αβγ=1 Given P(0) = -1 Also, f(0) = 1 :. --k(-1)² = -1 ⇒ k=1 ²) (x − B²) (x - y²) .. f(x) = x³ + x + 1 = (x - α) (x − B) (x−y) P(x) = (x - Now, P(4) = (2-a) (2-B) (2-1) (2 + a)(2+B) (2 + y) :. =ƒ(2)(-ƒ(-2)) = (2³ + 2 + 1) (−(−2)³ − (−2) – 1) = 11 x 9 = 99 ILLUSTRATION 2.12 Let f(x) be a polynomial with integral coefficients. Iff (1) and f(2) both are odd integers, prove that f(x) = 0 can't have any integral root. Sol. Let us assume that f(x) = 0 for some integer x = k. Then, x-k divides f(x). coefficients f(x)=(x-k) g(x), where g (x) is a polynomial with integral f(1) = (1-k) g(1) and f(2)= (2-k) g(2) ƒ(1) f(2)=(1-k)(2-k) g(1) g(2), (1) which is clearly an even number, which contradicts the given information that both f(1) and f(2) are odd integers. Hence f(x) = 0 can't have any integral root. ILLUSTRATION 2.13 Let a, b eN and a > 1. Also p is a prime number. If ax² + bx +c=p for two distinct integral values of x, then prove that ax²+bx+c #2p for any integral value of x. Sol. Given ax²+bx+c-p=0 has integral roots. Let a, ß be the roots. Then ax²+bx+c-p= a(x-a) (x-3) Now from ax² + bx+c=2p, we have 2 (1)
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