\( \Large x^{2}+x-20=0; y^{2}-y-30=0 \)

A) If x>y

B) \( \Large If \ x \ge y \)

C) lfx

D) \( \Large If \ x \le y \)

Correct answer:
D) \( \Large If \ x \le y \)

Description for Correct answer:

\( \Large x^{2}+x-20=0 \) [by factorisation method]

= \( \Large x^{2}+5x-4x-20=0 \)

= \( \Large x \left(x+5\right)-4 \left(x+5\right)=0 \)

= \( \Large \left(x+5\right) \left(x-4\right)=0 \)

Therefore, x = -5 or 4

and \( \Large y^{2}-y-30=0 \)

= \( \Large y^{2}-6y+5y-30=0 \)

= \( \Large y \left(y-6\right)+5 \left(y-6\right)=0 \)

= \( \Large \left(y-6\right) \left(y+5\right)=0 \)

Therefore, y = 6 or -5

Hence, \( \Large y \ge x \   or \   x \le y \)



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