Question

$I.x^2-x-12=0$
$II.y^2+5y+6=0$

• A. x > y
• B. x $\ge$ y
• C. x < y
• D. x $\le$ y
• E. x = y or the relationship cannot be established

Answer 1

The correct option is B x $\ge$ y

Equation I
$x^2-x-12=0$
$\Rightarrow x^2-4x+3x-12=0$
$\Rightarrow x(x-4)+3(x-4)=0$
$\Rightarrow (x+3)(x-4)=0$
$x=-3,4$
Equation II
$y^2+5y+6=0$
$\Rightarrow y^2+2y+3y+6=0$
$\Rightarrow y(y+2)+3(y+2)=0$
$\therefore y=-2,-3$
Hence, x $\ge$ y