KeerthanaPosted on x2−8x+15=0{x^2} - 8x + 15 = 0x2−8x+15=0 y2−3y+2=0{y^2} - 3y + 2 = 0y2−3y+2=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. x2−8x+15=0{x^2} - 8x + 15 = 0x2−8x+15=0 ⇒x2−5x−3x+15=0 Rightarrow {x^2} - 5x - 3x + 15 = 0⇒x2−5x−3x+15=0 ⇒x(x−5)−3(x−5)=0 Rightarrow xleft( {x - 5} ight) - 3left( {x - 5} ight) = 0⇒x(x−5)−3(x−5)=0 ⇒(x−3)(x−5)=0 Rightarrow left( {x - 3} ight)left( {x - 5} ight) = 0⇒(x−3)(x−5)=0 ∴x=3or5 herefore x = 3{ m{or}}5∴x=3or5II. y2−3y+2=0{y^2} - 3y + 2 = 0y2−3y+2=0 ⇒y2−2y−y+2=0 Rightarrow {y^2} - 2y - y + 2 = 0⇒y2−2y−y+2=0 ⇒y(y−2)−1(y−2)=0 Rightarrow yleft( {y - 2} ight) - 1left( {y - 2} ight) = 0⇒y(y−2)−1(y−2)=0 ⇒(y−1)(y−2)=0 Rightarrow left( {y - 1} ight)left( {y - 2} ight) = 0⇒(y−1)(y−2)=0 ∴y=1or2 herefore y = 1{ m{or}}2∴y=1or2 ∴x>y herefore x > y∴x>y Rate This:NaN / 5 - 1 votesAdd comment