Question

Let $A=\{x\in R:x^2+(m-1)x-2(m+1)=0\}$ and $B=\{x\in R:(m-1)x^2+mx+1=0\},$ where $m\in R.$ If $A\cup B$ has exactly three distinct elements, then the number of possible values of $m$ is

Answer 1

Given : $A=\{x\in R:x^2+(m-1)x-2(m+1)=0\}$
and $B=\{x\in R:(m-1)x^2+mx+1=0\}$

When $m=1$, then
$A=\{x:x^2-4=0\}\Rightarrow A=\{-2,2\}B=\{x:x+1=0\}\Rightarrow B=\{-1\}\therefore A\cup B=\{-2,-1,2\}$

When $m\ne 1$, then
$x^2+(m-1)x-2(m+1)=0\Rightarrow x=\frac{-(m-1)\pm \sqrt{(m-1{)}^2+8(m+1)}}{2}\Rightarrow x=\frac{-(m-1)\pm \sqrt{(m+3{)}^2}}{2}\Rightarrow x=2,-(m+1)\cdots \left(1\right)$

$(m-1)x^2+mx+1=0\Rightarrow x=\frac{-m\pm \sqrt{m^2-4(m-1)}}{2(m-1)}\Rightarrow x=\frac{-m\pm \sqrt{(m-2{)}^2}}{2(m-1)}\Rightarrow x=-1,\frac{1}{1-m}\cdots \left(2\right)$

For exactly three elements in $A\cup B$, two roots of both the quadratic equation must be same, so the possibilities are
$\left(i\right)2=-(m+1)\Rightarrow m=-3\Rightarrow A\cup B=\{-1,\frac{1}{4},2\}\left(ii\right)2=\frac{1}{1-m}\Rightarrow m=\frac{1}{2}\Rightarrow A\cup B=\{-1,-\frac{3}{2},2\}\left(iii\right)-(m+1)=-1\Rightarrow m=0\Rightarrow A\cup B=\{-1,1,2\}\left(iv\right)-(m+1)=\frac{1}{1-m}\Rightarrow m^2=2\Rightarrow m=\pm \sqrt{2}\left(v\right)\frac{1}{1-m}=-1\Rightarrow m=2\Rightarrow A\cup B=\{-3,-1,2\}$

$\therefore m=\{1,-3,\frac{1}{2},0,-\sqrt{2},\sqrt{2},2\}$
Hence, the number of possible values of $m$ is $7.$