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. $\varphi$

Answer 1

Answer: (D)

$x^4-\lambda x^2+9=0$
Let $x^2=t\ge 0$
$\therefore f(t)=t^2-\lambda t+9=0$ ... (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)
$\frac{-b}{2a}>0$
$\Rightarrow \frac{\lambda }{2}>0$ or $\lambda >\mathrm{0...}$ (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
$\frac{-b}{2a}<0\Rightarrow \lambda <\mathrm{0...}$ (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 $\lambda =6$.