Question

If the equation $x^4-(k-1)x^2+(2-k)=0$ has three distinct real roots, then the possible value(s) of $k$ is/are

• A. $\left\{2\right\}$
• B. $\{\sqrt{2}-1,2\}$
• C. $\{\sqrt{5}-1\}$
• D. $\{2\sqrt{2},\sqrt{3}-\sqrt{2}\}$

Answer 1

The correct option is A $\left\{2\right\}$

$x^4-(k-1)x^2+(2-k)=0\cdots \left(1\right)$
Assuming $x^2=t$,
$t^2-(k-1)t+(2-k)=0\cdots \left(2\right)$
Let roots of the equation $\left(2\right)$ be $t_1,t_2(t_1\le t_2)$
Equation $\left(1\right)$ will have three distinct real roots iff for equation $\left(2\right)$,
$t_1=0,t_2>0$


$\left(i\right)f\left(0\right)=0\Rightarrow 2-k=0\Rightarrow k=2\cdots \left(3\right)\left(ii\right)-\frac{b}{2a}>0\Rightarrow \frac{k-1}{2}>0\Rightarrow k>1\cdots \left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$,
$k=2$