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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
$$3x^{2} + 29x + 56 = 0$$.
$$x=\frac{-29\pm\sqrt{29^{2}-4\times56\times3}}{2\times3}$$
$$x=\frac{-29\pm13}{6}$$.
$$x=\frac{-8}{3},-7$$.
$$2y^{2} + 15y + 25 =0$$.
$$y=\frac{-15\pm\sqrt{15^{2}-4\times25\times2}}{2\times2}$$.
$$y=\frac{-15\pm5}{4}$$.
$$y=-5,\frac{(-5)}{2}$$.
Clearly,
x < y
Hence, Option A is correct.
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