Quadratics can be dirty! Really?

Algebra Level 5

Let $A= \{ x|x^2+(m-1)x-2(m+1)=0, x \in R \}$

$B= \{ x|x^2(m-1)+mx+1=0, x \in R \}$

Number of values of $m$ such that $A\cup B$ has exactly 3 distinct elements, is

None of the others. 8 5 9 6 4 3 7

1 solution

Pranjal Jain
Jan 21, 2015

This is not a solution. I will post one soon if no one else will post.

The values of $m$ can be $1,-3,\dfrac{1}{2}, 0, 2, \pm\sqrt{2}$

Edit: Solution

By observation, $2\in A$ and $-1\in B$ .

By using Vieta's, $-(m+1)\in A$ and $\dfrac{-1}{m-1}\in B$

Now since $A\cup B$ has 3 elements, two of the roots must be equal.

$2=-(m+1)\Rightarrow m=-3$
$2=\frac{-1}{m-1}\Rightarrow m=\frac{1}{2}$
$-1=-(m+1)\Rightarrow m=0$
$-1=\frac{-1}{m-1}\Rightarrow m=2$
$-(m+1)=\frac{-1}{m-1}\Rightarrow m^2-2=0\Rightarrow m=\pm\sqrt{2}$
At the end, don't forget that a quadratic can be made a linear if coefficient of $x^2$ becomes 0!! So $m$ can also be 1.

please upload the solution

manish bhargao - 6 years, 4 months ago

Perfect. Didn't forget m=1.

Vineeth Chelur - 6 years, 4 months ago

I submitted it as Level 3 but maybe $x=1$ made it a level 5 worth 320 points!

Pranjal Jain - 6 years, 4 months ago

@Pranjal Jain I am so bad!!!! Dude, I forgot the case of m=1.... :(
But by Vieta's why do we get one of the roots to be undefined at m=1???

Aaghaz Mahajan - 3 years, 3 months ago