91Ó°ÊÓ

The equation given is: \(x^2 + Mx + N\) Where M and N are odd integers. We need to prove that the equation has no integer zeros.

For the equation to have an integer zero, the possible values of x can be either odd or even. 1. If x is even (x = 2a, where a is also an integer), the terms of the equation become: \(x^2 = (2a)^2 = 4a^2\) \(Mx = M(2a) = 2(Ma)\) \(N = odd\) So the equation becomes: \(4a^2 + 2(Ma) + odd\) Notice that each term is either a multiple of 2 or an odd number (which is not divisible by 2). The sum of the first two terms will be a multiple of 2, since both terms are even. (2 * even = even) This means that the equation overall has an odd value, since adding an odd number to an even number gives an odd number. 2. If x is odd (x = 2b + 1, where b is also an integer), the terms of the equation become: \(x^2 = (2b + 1)^2 = 4b^2 + 4b + 1\) \(Mx = M(2b + 1) = 2(Mb) + M\) \(N = odd\) So the equation becomes: \((4b^2 + 4b + 1) + (2(Mb) + M) + odd\) Notice that this equation also has an odd value, since the sum of the last term and the constant term within the first bracket is the sum of two odd values. (odd + odd = even) Therefore, in both cases, the quadratic equation has an odd value, which means it cannot be equal to zero. Hence, we can conclude that the given quadratic equation \(x^2+Mx+N\) has no integer zeros.