Question

$I.3x^2+29x+56=0II.2y^2+15y+25=0$

• A. $x<y$
• B. $x>y$
• C. $x\ge y$
• D. $x\le y$
• E. Relationship between x and y cannot be established.

Answer 1

The correct option is E Relationship between x and y cannot be established.

Equation I
$3x^2+29x+56=0$
$3x^3+21x+8x+56=0$
$3x(x+7)+8(x+7)=0$
$(3x+8)(x+7)=0$
$x=-7,x=-\frac{8}{3}$
Equation II
$2y^2+15y+25=0$
$2y^2+10y+5y+25=0$
$2y(y+5)+5(y+5)=0$
$(y+5)(2y+5)=0$
$y=-5,y=-\frac{5}{2}$
Hence, relationship between x and y cannot be established