KeerthanaPosted on I. 2x2+x−28=02{x^2} + x - 28 = 02x2+x−28=0 II. 2y2−23y+56=02{y^2} - 23y + 56 = 02y2−23y+56=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 DExplanation : I. 2x2+x−28=02{{ m{x}}^2} + { m{x}} - 28 = 02x2+x−28=0 2x2+8x−7x−28=02{{ m{x}}^2} + 8{ m{x}} - 7{ m{x}} - 28 = 02x2+8x−7x−28=0 2x(x+4)−7(x+4)=02{ m{x}}left( {{ m{x}} + 4} ight) - 7left( {{ m{x}} + 4} ight) = 02x(x+4)−7(x+4)=0 (2x−7)(x+4)=0left( {2{ m{x}} - 7} ight)left( {{ m{x}} + 4} ight) = 0(2x−7)(x+4)=0 x=−4,72x = - 4,frac{7}{2}x=−4,27II. 2y2−23y+56=02{{ m{y}}^2} - 23{ m{y}} + 56 = 02y2−23y+56=0 2y2−16y−7y+56=02{{ m{y}}^2} - 16{ m{y}} - 7{ m{y}} + 56 = 02y2−16y−7y+56=0 2y(y−8)−7(y−8)=02{ m{y}}left( {{ m{y}} - 8} ight) - 7left( {{ m{y}} - 8} ight) = 02y(y−8)−7(y−8)=0 (2y−7)(y−8)=0left( {2{ m{y}} - 7} ight)left( {{ m{y}} - 8} ight) = 0(2y−7)(y−8)=0 y=72,8y = frac{7}{2},8y=27,8 y≥x{ m{y}} ge { m{x}}y≥x Rate This:NaN / 5 - 1 votesAdd comment