8.3 Determine two values of \( p \) for which the equation \( x^{2}+6 x-8 p=0 \) will have rational roots.

To find the values of \( p \) for which the quadratic equation \( x^{2}+6x-8p=0 \) has rational roots, we can use the discriminant method. A quadratic equation \( ax^2 + bx + c = 0 \) has rational roots if its discriminant \( D = b^2 - 4ac \) is a perfect square. In our equation, \( a = 1 \), \( b = 6 \), and \( c = -8p \). Thus, the discriminant is: \[ D = 6^2 - 4(1)(-8p) = 36 + 32p = 36 + 32p. \] For the equation to have rational roots, \( D \) should be a perfect square, say \( k^2 \). Therefore, we set up the equation: \[ 36 + 32p = k^2. \] Rearranging the equation gives: \[ 32p = k^2 - 36 \implies p = \frac{k^2 - 36}{32}. \] Now, \( k^2 - 36 \) must be divisible by 32 for \( p \) to be an integer. We can explore various integer values of \( k \) that will make \( k^2 - 36 \) a multiple of 32. Trying \( k = 10 \): \[ k^2 - 36 = 100 - 36 = 64 \implies p = \frac{64}{32} = 2. \] Trying \( k = 2 \): \[ k^2 - 36 = 4 - 36 = -32 \implies p = \frac{-32}{32} = -1. \] Thus, the two values of \( p \) for which the equation \( x^{2}+6x-8p=0 \) will have rational roots are \( p = 2 \) and \( p = -1 \).