What are the values of $p$ so that equation $x^3+(p-2)x^2+(5-2p)x-10=0$ has exactly $2$ real roots........

Question

I found this question in a maths-group in Facebook-

What are the values of $p$ so that equation $x^3+(p-2)x^2+(5-2p)x-10=0$ has exactly $2$ real roots........

I think we do not count repeated roots as one root. I'm confused about the meaning of the question. Please explain.

Answer 1

The polynomial has a root $x=2$ as can be seen; this can be verified by replacing x by 2 in the polynomial. By division, the polynomial can be seen to factor into: $(x-2)(x^2+px+5)$. For the polynomial to have only 2 real roots, $x^2+px+5$ must have only one root. This happens if its discriminant $p^2-20=0$. The two values for $p$ are then $2 \sqrt5$ and $-2 \sqrt5$. The two roots of the polynomial are the 2 and either $\sqrt5$ or $-\sqrt5$.

Answer 2

If you call the above expression $y$, then one of the solutions of the equation $\frac {dy}{dx}=0$ must also be a solution of the equation $y=0$. This will give an equation for $p$, but that's a lot of algebra.

Answer 3

Assuming that the form is $(x-a)^2(x-b)$, we can equate coefficients and solve for $p$.

The equations are \begin{eqnarray} p-2 &=& -(b+2a) \\ 5-2p &=& 2ab+a^2 \\ -10 &=& -a^2b \end{eqnarray} This gives $b = {10 \over a^2}$ and substituting in and eliminating $p$ gives the quartic $a^4-4 a^3-a^2+20a -20$ which has solutions $\{2, \pm \sqrt{5} \}$ ($2$ is repeated).

We have $p= 2 - ({10 \over a^2} + 2a)$, substituting the roots of the quartic give $p \in \{ -{9\over 2}, \pm 2 \sqrt{5} \}$. It is easy to verify that these values given rise to two real roots (ignoring multiplicity).