Question

$I.2x^2+18x+40=0II.2y^2+15y+27=0$

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

Answer 1

The correct option is E $x=y$or relationship cannot be established

Equation I
$2x^2+18x+40=0$
$\Rightarrow 2x^2+10x+8x+40=0$
$\Rightarrow 2x(x+5)+8(x+5)=0$
$\Rightarrow (2x+8)(x+5)=0$
$\therefore x=-4,x=-5$
Equation II
$2y^2+15y+27=0$
$\Rightarrow 2y^2+9y+6y+27=0$
$\Rightarrow y(2y+9)+3(2y+9)=0$
$\therefore y=-3,y=-\frac{9}{2}$
Hence, relationship cannot be established.