If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root then which of the following is true

Question

If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root then which of the following is true:

$1$. $p=q$

$2$. $p+q+1=0$

$3$. $p+q=0$

$4$. $\textrm {both} 1,2$.

MY WORK:

$$x=\dfrac {-p\pm \sqrt {p^2-4q}}{2}$$ And, $$x=\dfrac {-q\pm \sqrt {q^2-4p}}{2}$$

HOW DO I GET IT COMPLETED?

Answer 1

Hint:

Common root for $a_1x^2+b_1x+c_1=0 $ and $ a_2x^2+b_2x+c_2=0$ is

$$α = \frac{c_1a_2 - c_2a_1}{a_1b_2 - a_2b_1} \text{ or } α = \frac{b_1c_2 - b_2c_1}{c_1a_2 - c_2a_1}$$

Answer 2

If they have a single root, they have two each, so we can factor. Now let us call these polynomials by $P$ and $Q$, respectively.

Edit: I have assumed throughout here that $\beta \neq \gamma$. If it turned out they were equal, then we could not conclude $\alpha = -1$, since this would be division by $0$.

$P(x) = x^2 + px + q = (x + \alpha)(x + \beta) \implies \alpha + \beta = p, \ \ \ \ \alpha\beta = q $

$Q(x) = x^2 + qx + p = (x + \alpha)(x + \gamma) \implies \alpha + \gamma = q,\ \ \ \ \ \alpha\gamma = p$

Can we do anything with this? Well, from the first and third equations, it follows that $\beta - \gamma = p - q$, and similarly, $\alpha \gamma - \alpha \beta = p - q$

So then $-\alpha(\beta - \gamma) = \beta - \gamma$, which implies that $\alpha = -1$

This implies that $\beta = p+1$ and $\gamma = q + 1$. But also that $-\beta = q$ and $- \gamma = p$. Putting these together gives $p+q + 1 =0$, which is $\# 2 $ on your list.

On the other hand, we can take the difference of the two original polynomials, and get $(p-q)x +(q-p) = 0$, which seems to imply $p=q$, since $x$ is just an indeterminate. This in fact would force $\beta = \gamma$, and so this is contrary to hypothesis above.

This appears to be in contradiction to the other answer here, so feedback is appreciated.

Answer 3

If $$x^2+px+q = 0 = x^2+qx+p$$ then $$px+q=qx+p$$ So either $p=q$ or $x=1$ and thus $1+p+q=0$.

So one of (1) or (2) must be true (it is not necessary for both to be true.) It's not clear what "both" is doing in the question

Answer 4

$$a_1x^2+b_1x+c_1=0$$ $$a_2x^2+b_2x+c_2=0$$.

Condition for one root to be common of two quadratic equations is: $$(c_1a_2-c_2a_1)^2=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1)$$

In your case it is: $$(q-p)^2=(p^2-q^2)(q-p)$$ $$(q-p)=(p+q)(p-1)$$ $$p+q=-1$$ $$p+q+1=0$$

So the correct option would be $2$.

Answer 5

If $k $ is the common root then

$k^2+pk+q=k^2+qk+p=0$

$(p-q)k=(p-q ) $ so either $p=q $ or $k=1$. If $k=1$ then $q+p+1=0$.

If $p=q $ then solutions are $x=\frac {-1\pm\sqrt {p^2-4p}}{2}$ so $ p=q \le4$ is only restriction.

So 1 OR 2 are true but neither have to be.

So I'd say none of the above. Although all could be true (3 can be true if $p=q=0$).

===

It's not possible for 1) and 2) to both be false.

1) alone is true if $p=q\le4$ but $p\ne 0$ and $p\ne -1/2$

2) Alone is true if $p\ne q $.

3) can only be true if $p=q=0$ so 1 would also be true and 2 would be false.

1 and 2) can be mutually true if $p=q=-1/2$

2 and 3 are mutually exclusive and can never both be true.