0
$\begingroup$

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.

$\endgroup$
4
  • 4
    $\begingroup$ Every degree $3$ polynomial has $3$ roots over $\mathbb{C}$, and the non-real roots come in conjugate pairs. So the only reasonable interpretation for this question is that $p$ should have one single root and one double root. $\endgroup$ May 15, 2015 at 20:22
  • $\begingroup$ It can only have exactly two real roots if one of those roots is a repeated root, for the reason that it is a cubic $\endgroup$ May 15, 2015 at 20:22
  • 1
    $\begingroup$ The only possibilities are one real root or three real roots (counted by multiplicity). $\endgroup$
    – copper.hat
    May 15, 2015 at 20:27
  • $\begingroup$ $x^3-x^2=0$ has 2 roots: 0 and 1 as it factors into $x^2=0$ or $x-1=0$, thus there can exist a cubic with precisely 2 real roots. $\endgroup$
    – JB King
    May 15, 2015 at 20:32

3 Answers 3

4
$\begingroup$

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 $-\sqrt5$.

$\endgroup$
3
  • 1
    $\begingroup$ ... or also you could arrange for $(x-2)$ to factor $(x^2+px+5)$ $\endgroup$
    – Joffan
    May 15, 2015 at 20:55
  • 1
    $\begingroup$ Very nice. But you missed $p=-\frac{9}{2}$, i.e. when $2$ is a root of the quadratic. $\endgroup$ May 15, 2015 at 20:56
  • $\begingroup$ (x-2)^2 (x-b), b not= 2, this case is remaining. $\endgroup$
    – ReekMaths
    May 16, 2015 at 19:41
2
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Why the downvote? $\endgroup$
    – copper.hat
    May 22, 2015 at 8:06
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ The "a lot of algebra" gives $p\in \{-\frac{9}{2},2\sqrt{5},-2\sqrt{5}\}$. I think there should be an elegant approach to this. $\endgroup$ May 15, 2015 at 20:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .