When given the quadratic equation with integer roots: mx^2 - (m-2)x + m - 8 = 0

What's the value of integer parameter m? And what are the integer roots?

Guest Jun 15, 2021

#1**+1 **

I believe the solutions to the given equations cannot both be integers if we assume m is an integer, so the problem is unsolvable as stated, and here is a proof; see if you agree with me.

I. For the quadratic equation \(a{x}^{2}+bx +c=0\), the sum of the roots is \(-\frac{b}{a}\), and the product of the roots is \(\frac{c}{a}\).

II. The given quadratic equation, therefore, has the following as the sum and the product of the roots:

\(product=\frac{m-8}{m}=1-\frac{8}{m}\)

\(sum =\frac{m-2}{m}=1-\frac{2}{m}\)

III. The sum and the product of two integers are always integers. So the expressions given above are both integers,

IV. What integer values for m result in an integer value for \(1-\frac{2}{m}\)? m, being an integer, must clearly be 1, -1, 2, or -2.

V. The following table tabulates the sum and the product for each of the four values for m. None of the four pair of values give integer values for the roots.

Guest Jun 16, 2021