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?
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.
