Question

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

• A. $\left\{2\right\}$
• B. $\left\{1\right\}$
• C. $\varphi$
• D. $\{-1\}$

Answer 1

The correct option is B $\left\{1\right\}$

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, $x^2=t$
$\Rightarrow t^2-(1-2m)t+(m^2-1)=0$
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
$x_1=+\sqrt{t_1},x_2=+\sqrt{t_2},x_3=-\sqrt{t_1},x_4=-\sqrt{t_2}$.

Applicable Conditions :
i) $D>0$
ii) $t_2=0$
iii) For $t_1>0,-\frac{b}{2a}>0$
Now, solve all conditions,
i) $D>0$
$\Rightarrow (1-2m{)}^2-\mathrm{4.1.}(m^2-1)$
$\Rightarrow -4m+5>0$
$\Rightarrow 4m<5\Rightarrow m<\frac{5}{4}$
$\therefore m\in (-\infty ,\frac{5}{4})=A$

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

iii) For $t_1>0,-\frac{b}{2a}>0$
$\Rightarrow -\frac{1-2m}{2}>0$
$\Rightarrow 1-2m<0\Rightarrow 2m>1\Rightarrow m>\frac{1}{2}$
$\therefore m\in (\frac{1}{2},\infty )=C$

$\therefore$ Set of possible value of m, is $A\cap B\cap C=(-\infty ,\frac{5}{4})\cap \{-1,1\}\cap (\frac{1}{2},\infty )=\left\{1\right\}$

$i.e,m=\left\{1\right\}$