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

Question 40

I. $$2x^{2} - 13x + 21 = 0$$
II. $$5y^{2} - 22y + 21 =0$$

Solution

$$2x^{2} - 13x + 21 = 0$$.
$$x=\frac{13\pm\sqrt{13^{2}-4\times21\times2}}{2\times2}$$
$$x=\frac{13\pm1}{4}$$.
$$x=\frac{7}{2},3$$.
$$5y^{2} - 22y + 21 =0$$.
$$y=\frac{22\pm\sqrt{22^{2}-4\times21\times5}}{2\times5}$$
$$y=\frac{22\pm8}{10}$$.
$$y=3,\frac{7}{5}$$.
Clearly,
x => y
Hence, Option C is correct.


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