Answer:
[tex]\displaystyle p = \frac{16}{q + 1}[/tex].
Step-by-step explanation:
Factor out the coefficient of [tex]x^{2}[/tex]:
[tex]\displaystyle \frac{1}{q + 1}\, [(q + 1)\, x^2 - 8\, x + p] = 0[/tex].
[tex]\displaystyle x^2 - \frac{8}{q + 1} \, x + \frac{p}{q + 1} = 0[/tex].
Let [tex]m[/tex] denote the real root of this equation. By the Factor Theorem, this equation would have the factor [tex](x - m)[/tex] repeated for two times in total:
[tex](x - m)^{2} = 0[/tex].
Expand to obtain: [tex]x^2 - 2\, m\cdot x + m^{2} =0[/tex].
Compare this expression with [tex]\displaystyle x^2 - \frac{8}{q + 1} \, x + \frac{p}{q + 1} = 0[/tex]:
[tex]\displaystyle -\frac{8}{q + 1} = -2\, m[/tex] (for the coefficient of [tex]x[/tex].)
[tex]\displaystyle \frac{p}{q + 1} = m^2[/tex] (for the constant.)
Rewrite the first equation to find an expression for [tex]m[/tex] in terms of [tex]q[/tex]:
[tex]\displaystyle m = \frac{4}{q + 1}[/tex].
Substitute this expression for [tex]m[/tex] into the second equation to find an expression for [tex]p[/tex] in terms of [tex]q[/tex]:
[tex]\displaystyle \frac{p}{q + 1} = m^2[/tex].
[tex]\displaystyle \frac{p}{q + 1} = \frac{4^2}{(q+1)^{2}}[/tex].
[tex]\displaystyle p = \frac{16}{q + 1}[/tex].
Verify that this expression for [tex]p[/tex] satisfies the requirements:
[tex]\displaystyle (q + 1) \, x^{2} - 8\, x + \frac{16}{q + 1} = 0[/tex].
[tex]\displaystyle x^2 - \frac{8}{q + 1} + \frac{16}{(q+1)^2} = 0[/tex].
[tex]\displaystyle x = \frac{4}{q + 1}[/tex] is the (repeated) real root of this quadratic equation.
