Question

The equation $9y^2(m+3)+6(m-3)y+(m+3)=0$, where $m$ is real has real roots then

• A. $m<0$
• B. $m>0$
• C. $m\le 0$
• D. $m\ge 0$

Answer 1

Answer: (C)

$m\le 0$

For an equation $a{x}^2+bx+c=0$, the discriminant $△=b^2-4ac$ helps us understand the nature of the roots.
When $△\ge 0$ then roots are real and equal.
${\left[6\right(m-3\left)\right]}^2-4\times 9(m+3)\times (m+3)\ge 0$
$36(m^2+9+-6m)-36(m^2+9+6m)\ge 0$
$36m^2+324-216m-36m^2-324-216m\ge 0$
$-432m\ge 0$
$\therefore m\le 0$