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
I. $$2x^{2} + 23x + 63 = 0$$
II. $$4y^{2} + 19y + 21 = 0$$
$$2x^{2} + 23x + 63 = 0$$.
$$x=\frac{-23\pm\sqrt{23^{2}-4\times63\times2}}{2\times2}$$
$$x=\frac{-23\pm5}{4}$$.
$$x=\frac{-9}{2},-7$$.
$$4y^{2} + 19y + 21 =0$$.
$$y=\frac{-19\pm\sqrt{19^{2}-4\times21\times4}}{2\times4}$$.
$$y=\frac{-19\pm5}{8}$$.
$$y=-3,\frac{(-7)}{4}$$.
Clearly,
x < y
Hence, Option A is correct.
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