KeerthanaPosted on I. 2x2−7x−60=02{x^2} - 7x - 60 = 02x2−7x−60=0 II. 3y2+13y+4=03{y^2} + 13y + 4 = 03y2+13y+4=0 a If x > y b If x ≥ y c If y > x d If y ≥ x e If x = y or no relation can be established Answer : Option EExplanation : I. 2x2−7x−60=02{{ m{x}}^2} - 7{ m{x}} - 60 = 02x2−7x−60=0 2x2−15x+8x−60=02{{ m{x}}^2} - 15{ m{x}} + 8{ m{x}} - 60 = 02x2−15x+8x−60=0 x(2x−15)+4(2x−15)=0{ m{x}}left( {2{ m{x}} - 15} ight) + 4left( {2{ m{x}} - 15} ight) = 0x(2x−15)+4(2x−15)=0 (x+4)(2x−15)=0left( {{ m{x}} + 4} ight)left( {2x - 15} ight) = 0(x+4)(2x−15)=0 x=−4,152x = - 4,frac{{15}}{2}x=−4,215II. 3y2+13y+4=03{{ m{y}}^2} + 13{ m{y}} + 4 = 03y2+13y+4=0 3y2+12y+y+4=03{{ m{y}}^2} + 12{ m{y}} + { m{y}} + 4 = 03y2+12y+y+4=0 3y(y+4)+1(y+4)=03{ m{y}}left( {{ m{y}} + 4} ight) + 1left( {{ m{y}} + 4} ight) = 03y(y+4)+1(y+4)=0 (3y+1)(y+4)=0left( {3{ m{y}} + 1} ight)left( {{ m{y}} + 4} ight) = 0(3y+1)(y+4)=0 y=−13,−4y = - frac{1}{3}, - 4y=−31,−4 no relation can be established Rate This:NaN / 5 - 1 votesAdd comment