Question

The value of $m$ for which the equation $x^4+(1-2m)x^2+(m^2-1)=0$ has three real distinct root is

• A. $2$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

Given: $x^4+(1-2m)x^2+(m^2-1)=0$ has three real distinct root.
To find: Value of $m.$

Step-$1:$ Assume $x^2=t$
Step-$2:$ Find the roots in term of $t_1,t_2$
Step-$3:$ Write the applicable conditions.
Step-$4:$ Solve all the conditions and combine all the results for $m.$

Let $x^2=t,$ then our equation becomes:
$\Rightarrow t^2+(1-2m)t+(m^2-1)=0$
Roots $x_1,x_2,x_3,x_4$ are transformed to roots $t_1,t_2.$
$x^2=t\Rightarrow x=\pm \sqrt{t}$
$\therefore$ Four roots will be
$x_1=+\sqrt{t_1},x_2=+\sqrt{t_2},x_3=-\sqrt{t_1},x_4=-\sqrt{t_2}.$

For the equation to have three roots the applicable conditions are:
$\left(i\right)D>0$
$\left(ii\right)t_2=0$
$\left(iii\right)\text{For}{t}_1>0,-\frac{b}{2a}>0$
Now, solve all conditions,
$\left(i\right)D>0$
$\Rightarrow (1-2m{)}^2-4\cdot 1\cdot (m^2-1)>0$
$\Rightarrow -4m+5>0$
$\Rightarrow 4m<5\Rightarrow m<\frac{5}{4}$
$\Rightarrow m\in (-\infty ,\frac{5}{4})\cdots \left(1\right)$

$\left(ii\right)t_2=0$
For three real distinct roots, either $t_1$ or $t_2$ should be zero. So, here we are taking $t_2=0$
$\Rightarrow m^2-1=0$
$\Rightarrow m=\pm 1\cdots \left(2\right)$

$\left(iii\right)\text{For}{t}_1>0,-\frac{b}{a}>0$
$\Rightarrow -(1-2m)>0$
$\Rightarrow 1-2m<0\Rightarrow 2m>1\Rightarrow m>\frac{1}{2}$
$\Rightarrow m\in (\frac{1}{2},\infty )\cdots \left(3\right)$

$\therefore \text{Set of possible value of}m,\text{is}A\cap B\cap C$
$m\in (-\infty ,\frac{5}{4})\cap \{-1,1\}\cap (\frac{1}{2},\infty )=\left\{1\right\}$
$m\in \left\{1\right\}$
Thus, for $m=1,$ the equation will have three distinct roots.