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Instructions
In these questions two equations numbered I and II are given. You have to solve both the equations and give answer
a: if x < y
b: if x > y
c: if x = y
d: if x = y
e: if x= y or relationship between x and y cannot be established
Solution
$$3x^{2} + 23x + 44 = 0$$.
$$x=\frac{-23\pm\sqrt{23^{2}-4\times44\times3}}{2\times3}$$
$$x=\frac{-23\pm1}{6}$$.
$$x=\frac{-11}{3},-4$$.
$$3y^{2} + 20y + 33 =0$$.
$$y=\frac{-20\pm\sqrt{20^{2}-4\times33\times3}}{2\times3}$$.
$$y=\frac{-20\pm2}{6}$$.
$$y=-3,\frac{(-11)}{3}$$.
Clearly,
x <= y
Hence, Option D is correct.
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