Keerthana

Posted 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. $15{y^2} + 16y + 4 = 0$

a

x > y

b

x < y

c

x ≥ y

d

x ≤ y

e

x = y or no relation can be established between x & y.

Explanation :

I. $2{x^2} + 9x + 9 = 0$

$2{x^2} + left( {6 + 3} ight)x + 9 = 0$

$2xleft( {x + 3} ight) + 3left( {x + 3} ight) = 0$

$x = frac{{ - 3}}{2}, - 3$

II. $15{y^2} + 16y + 4 = 0$

$15{y^2} + 10y + 6y + 4 = 0$

$5yleft( {3y + 2} ight) + 2left( {3y + 2} ight) = 0$

$y = frac{{ - 2}}{5},frac{{ - 2}}{3}$

$x < y$

Rate This:
NaN / 5 - 1 votes