For what values of $m$ is there a common root to $mx^2+2x+1=0$ and $x^2+2x+m=0$?

Question

If the equation $mx^2+2x+1 = 0$ and $x^2+2x+m = 0$ have a common root, find the possible values of $m$ and the value of the common root in each case.

Answer 1

Let $a$ be the common root, then $ma^2 - a^2 + 1 - m=0\implies (m-1)(a^2-1) = 0$. Thus if $m \neq 1$, then $a = 1$, and this means $m = -3$. If $m = 1\implies a = -1$ . Thus we have $m = -3, 1$ and the common roots are $1,-1$ respectively.

Answer 2

$mx^2 + 2x + 1 = 0$ will have 2 roots $\frac {-2 \pm \sqrt{4 - 4m}}{2m}$.

(We must consider that $m = 0$. If $m = 0$ then $2x+1 = 0$ has root $-\frac 12$ and $x^2 +2x +m = 0$ has roots $x = 0$ and $x = -2$. So if $m =0$ there are no common roots. So we may assume $m \ne 0$.

$x^2 + 2x + m= 0$ will have 2 roots $\frac {-2 \pm \sqrt{4-4m}}{2}$.

So there are four cases that this will have a common root:

1) $\frac {-2 + \sqrt{4 - 4m}}{2m}=\frac {-2 + \sqrt{4-4m}}{2}$

$2m = 2$ and $m = 1$.

And $x^2 + 2x + 1=0$ and $x^2 + 2x + 1=0$ are the same equation with the same roots.

2) $\frac {-2 + \sqrt{4 - 4m}}{2m} = \frac {-2 - \sqrt{4-4m}}{2}$

$\frac{1 -\sqrt{1-m}}m = 1 + \sqrt{1-m}$

$1 -\sqrt{1-m} = m + m\sqrt{1-m}$

$1-m = \sqrt{1-m}(m + 1)$

$\frac {1-m}{m+1} = \sqrt {1 -m}$

$\frac {1-2m + m^2}{1 + 2m + m^2} = 1 -m$

$1 - \frac {4m}{1+2m + m^2} = 1-m$

$4 = 1 + 2m + m^2$

$m^2 + 2m -3 = 0$

$(m+3)(m - 1) = 0$

So $m = -3$ and $m =1$ are solutions.

If $m = -3$ then $-3x^2 + 2x +1 =0$ has a root $x = \frac {-2-\sqrt{4+4*3}}{-6} = 1$ and $x^2 + 2x -3=0$ has a root $x = 1$.

3) $\frac {-2 - \sqrt{4 - 4m}}{2m} = \frac {-2 + \sqrt{4-4m}}{2}$

$\frac {1+\sqrt{1-m}}{m} = 1-\sqrt{1-m}$

$1 + \sqrt{1-m} = m - m\sqrt{1-m}$

$\sqrt{1-m}(1+m) = m -1$

Which is the same as 2) above.

4)$\frac {-2 - \sqrt{4 - 4m}}{2m} = \frac {-2 - \sqrt{4-4m}}{2}$

Whih means $m = 1$ (the same as 1) above.

So $m = 1,-3$ are the two possible values of $m$.

Answer 3

For reference:

If the equation $mx^2+2x+1=0$ and $x^2+2x+m=0$ have a common root, find the possible values of $m$ and the value of the common root in each case.

Let $a$ be the root. Multiply the second by $m^2$ and subtract the first: $$2ma+m^2-2a-1=0 \Rightarrow (m-1)(m+1+2a)=0 \Rightarrow \\ \color{red}{m_1=1} \Rightarrow \color{red}{a_1=-1};\\ m+1+2a=0 \Rightarrow m=-1-2a.$$ Plug this into the first equation: $$(-1-2a)a^2+2a+1=0 \Rightarrow (2a+1)(1-a^2)=0 \Rightarrow \\ \color{red}{a_2=-\frac12} \Rightarrow \color{red}{m_2\in \emptyset};\\ a_3=-1 \Rightarrow m_3=1;\\ \color{red}{a_4=1} \Rightarrow \color{red}{m_4=-3}.$$ Hence, the possible answers are: $(m,a)=(1,-1), (-3,1)$.