Question

If all the roots of the equation $p{x}^4+q{x}^2+r=0,p\ne 0,q^2\ge 9pr$ are real, then which of the following option(s) is/are correct?

• A. $q>0,p>0,r>0$
• B. $q>0,p<0,r>0$
• C. $q>0,p<0,r<0$
• D. $q<0,p>0,r>0$

Answer 1

Answer: (C), (D)

$q<0,p>0,r>0$

$p{x}^4+q{x}^2+r=0\cdots \left(1\right)$
Assuming $x^2=t,t\in [0,\infty )$
$p{t}^2+qt+r=0\Rightarrow t^2+\frac{q}{p}t+\frac{r}{p}=0,p\ne 0\cdots \left(2\right)$
Let the roots of equation $\left(2\right)$ be $t_1,t_2$
Equation $\left(1\right)$ will have all real roots iff roots of equation $2$, $t_1,t_2\ge 0$,
Required conditions are,
$\left(i\right)-\frac{b}{2a}>0\Rightarrow -\frac{q}{2p}>0\Rightarrow \frac{q}{p}<0$
$q$ and $p$ are of opposite sign.
$\left(ii\right)f\left(0\right)>0\Rightarrow \frac{r}{p}>0$
$r$ and $p$ are of same sign.
$\left(iii\right)D\ge 0\Rightarrow \frac{q^2-4pr}{p^2}\ge 0$
Which is always true as $q^2\ge 9pr$

Therefore,
When $p>0\Rightarrow q<0,r>0$
When $p<0\Rightarrow q>0,r<0$