Question

Consider the equation $x^{4}-\lambda x^{2}+9=0$.
If the equation has only two real roots, then the set of values of $\lambda$ is :

• A. $(-\infty,-6)$
• B. $(-6,6)$
• C. $\{6\}$
• D. $\phi$

Answer 1

Answer: (D)

$x^{4}-\lambda x^{2}+9=0$
Let $x^{2}=t \geq 0$
$\therefore f(t)=t^{2}-\lambda t+9=0$ ... (1)
If given equation has four real and distinct roots, then
both roots of $(1)$ must be positive.
$\therefore D>0 $
$\Rightarrow \lambda^{2}-36>0$ ...(2)
$ \frac{-b}{2 a}>0 $
$\Rightarrow \frac{\lambda}{2}>0 $ or $ \lambda>0...$ (3)
$ f(0)>0 $ or $ 9>0$
$\therefore \lambda \in(6, \infty)$
$[$ from $(2),(3),(4)]$
If equation has no real roots, then (1) must have both roots negative for which
$\frac{-b}{2 a} < 0 \Rightarrow \lambda < 0 ...$ (5)
$ f (0)>0 \Rightarrow 9>0$
$\therefore \lambda \in(-\infty,-6)$
$[$ form $(2),(3),(5)]$
Also for no real roots we can have $D <0$
$\Rightarrow \lambda^{2}-36<0 \Rightarrow \lambda \in(-6,6)$
Hence, $\lambda \in(-\infty, 6)$
From above discussion we have $\lambda \in(-\infty, 6) \cup(6, \infty)$ equation has either four distinct real roots or no real roots.
For $\lambda=6$, both roots are $\pm \sqrt{3}$
Alternate method :
$x^{4}-\lambda x^{2}+9=0$
or $x^{2}+\frac{9}{x^{2}}=\lambda$
Now $\left(x-\frac{3}{x}\right)^{2}=\lambda-6$
For above equation
(i) four distinct real roots if $\lambda-6>0$ or $\lambda>6$.
(ii) no real roots if $\lambda-6<0$ or $\lambda>6$
(iii) real and equal repeating roots if $\lambda=6$.