Question

If the quadratic equation $a{x}^2+bx+c=0(a>0)$ has ${\sec }^2\theta$ and ${\text{cosec}}^2\theta$ as its roots, then which of the following must hold good?

• A. $b+c=0$
• B. $b^2-4ac\ge 0$
• C. $c\ge 4a$
• D. $4a+b\ge 0$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $b+c=0$
B $b^2-4ac\ge 0$
C $c\ge 4a$

Given, the quadratic equation $a{x}^2+bx+c=0(a>0)$.......... (1)

has ${\sec }^2\theta$ and ${\text{cosec}}^2\theta$ as its roots,

$a(x-se{c}^2\theta )(x-cose{c}^2\theta )=0$

$a{x}^2-a({\sec }^2\theta +{\text{cosec}}^2\theta )x+a{\sec }^2\theta \cdot {\text{cosec}}^2\theta =0$

$b=-a({\sec }^2\theta +{\text{cosec}}^2\theta )$ and $c=a{\sec }^2\theta \cdot {\text{cosec}}^2\theta$

But ${\sec }^2\theta +{\text{cosec}}^2\theta ={\sec }^2\theta \cdot {\text{cosec}}^2\theta$

Sum of the roots is equal to their product and the roots are real.

Hence,

$-\frac{b}{a}=\frac{c}{a}$

$\Rightarrow b+c=0$

Also,

$b^2-4ac\ge 0$

$\Rightarrow c^2-4ac\ge 0$

$\Rightarrow c(c-4a)\ge 0$

$\Rightarrow c-4a\ge 0(\because c>0)$

$\Rightarrow c\ge 4a$

Further

$b^2+4ab\ge 0$

$\Rightarrow b+4a\le 0(\because b<0)$

Options A,B and C are correct.