KeerthanaPosted on I. 7x2−19x+10=07{x^2} - 19x + 10 = 07x2−19x+10=0 II. 8y2+2y−3=08{y^2} + 2y - 3 = 08y2+2y−3=0 a if x>y b if x≥y c if xd if x ≤y e if x = y or no relation can be established between x and y. Answer : Option AExplanation : I. 7x2−19x+10=07{x^2} - 19x + 10 = 07x2−19x+10=0 7x2−14x−5x+10=07{x^2} - 14x - 5x + 10 = 07x2−14x−5x+10=0 7x(x−2)−5(x−2)=07xleft( {x - 2} ight) - 5left( {x - 2} ight) = 07x(x−2)−5(x−2)=0 x=2,57x = 2,frac{5}{7}x=2,75 II. 8y2+2y−3=08{y^2} + 2y - 3 = 08y2+2y−3=0 8y2+6y−4y−3=08{y^2} + 6y - 4y - 3 = 08y2+6y−4y−3=0 2y(4y+3)−1(4y+3)=02yleft( {4y + 3} ight) - 1left( {4y + 3} ight) = 02y(4y+3)−1(4y+3)=0 y=−34,12y = frac{{ - 3}}{4},frac{1}{2}y=4−3,21 x>yx > yx>y Rate This:NaN / 5 - 1 votesAdd comment