The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.
thanks
Use the quadratic equation.
$\displaystyle x=\frac{-(p+1) \pm \sqrt{(p+1)^2 - 4(8)(2)}}{4}$
$\displaystyle p+1$ is restricted by the radical or discriminant in order for x to have real solutions.
$\displaystyle (p+1)^2 - 4(8)(2) \ge 0$
$\displaystyle p^2 + 2p -63 \ge 0$
$\displaystyle (p+9)(p-7) = 0$
$\displaystyle p = -9, 7$
You have to test the three values that are less than -9, between -9 and 7, and greater than 7 to see which sub domains hold for the inequality.
It turns out that $\displaystyle p \le 9, p \ge 7$ OR $\displaystyle p: (-\infty, -9] \cap [7, \infty)$