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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
$$4x^{2} - 29x + 45 = 0$$.
$$x=\frac{29\pm\sqrt{29^{2}-4\times45\times4}}{2\times4}$$
$$x=\frac{29\pm3}{8}$$.
$$x=\frac{13}{4},4$$.
$$3y^{2} - 19y + 28 =0$$.
$$y=\frac{19\pm\sqrt{19^{2}-4\times28\times3}}{2\times3}$$.
$$y=\frac{19\pm5}{6}$$.
$$y=4,\frac{7}{3}$$.
Clearly,
if x= y or relationship between x and y cannot be established
Hence, Option E is correct.
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