KeerthanaPosted on Directions(Q.21 to Q.24 ): In each of these questions, two equations (i) and (ii) are given, you have to solve both the equations and give answer accordingly. II. 15y2+16y+4=015{y^2} + 16y + 4 = 015y2+16y+4=0 a x > y b x < y c x ≥ y d x ≤ y e x = y or no relation can be established between x & y. Answer : Option BExplanation : I. 2x2+9x+9=02{x^2} + 9x + 9 = 02x2+9x+9=0 2x2+(6+3)x+9=02{x^2} + left( {6 + 3} ight)x + 9 = 02x2+(6+3)x+9=0 2x(x+3)+3(x+3)=02xleft( {x + 3} ight) + 3left( {x + 3} ight) = 02x(x+3)+3(x+3)=0 x=−32,−3x = frac{{ - 3}}{2}, - 3x=2−3,−3 II. 15y2+16y+4=015{y^2} + 16y + 4 = 015y2+16y+4=0 15y2+10y+6y+4=015{y^2} + 10y + 6y + 4 = 015y2+10y+6y+4=0 5y(3y+2)+2(3y+2)=05yleft( {3y + 2} ight) + 2left( {3y + 2} ight) = 05y(3y+2)+2(3y+2)=0 y=−25,−23y = frac{{ - 2}}{5},frac{{ - 2}}{3}y=5−2,3−2 x<yx < yx<y Rate This:NaN / 5 - 1 votesAdd comment