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II. 2y2 – y – 21 = 0
I. 2x2+ 3x – 20 = 0
⇒ 2x2+ 8x – 5x – 20 = 0
⇒ 2x(x + 4) – 5(x + 4) = 0
⇒ (2x – 5)(x + 4) = 0
⇒ x = –4 and 2.5
⇒ 2y2 + 6y – 7y – 21 = 0
⇒ 2y(y + 3) – 7(y + 3) = 0
⇒ (y + 3)(2y – 7) = 0
⇒ y = –3 and 3.5
∴ No relationship can be established between x and y.
II. y2 − 13y + 40 = 0
I. x2 − 12x + 35 = 0
⇒ (x2 − 5x) – (7x − 35) = 0
⇒ x(x − 5) – 7(x − 5) = 0
⇒ (x − 7)(x − 5) = 0
⇒ x = 5 and 7
⇒ (y2 – 5y) – (8y – 40) = 0
⇒ y(y – 5) – 8(y – 5) = 0
⇒ (y − 5)(y – 8) = 0
⇒ y = 5 and 8Now, (x = 5) < (y = 8)And, (x = 7) > (y = 5)
Thus, no relationship can be established between x and y.
II. y2− 5y – 36 = 0
x2 + 6x − 55 = 0
⇒ (x2− 5x) + (11x − 55) = 0
⇒ x(x − 5) + 11(x − 5) = 0
⇒ (x + 11)(x − 5) = 0
⇒ x = 5 and −11
y2− 5y – 36 = 0
⇒ (y2+ 4y) – (9y + 36) = 0
⇒ y(y + 4) – 9(y + 4) = 0
⇒ (y + 4)(y – 9) = 0
⇒ y = 9 and −4
II. y2 + y - 20 = 0
x2 + 12x + 35 = 0
(x2 + 7x) + (5x + 35) = 0
x(x + 7) + 5(x + 7) = 0
(x + 5)(x + 7) = 0
x = –7, –5
y2 + y - 20 = 0
(y2 - 4y) + (5y - 20) = 0
y(y- 4) + 5(y - 4) = 0
(y + 5)(y - 4) = 0
y = 4 and –5
∴ x ≤ y
II. y2 + 14y + 48 = 0
x2 – 14x + 48 = 0
(x2- 6x) - (8x - 48) = 0
x(x - 6) - 8(x - 6) = 0
(x - 6)(x - 8) = 0
x = 6 and 8
y2 + 14y + 48 = 0
(y2 + 6y) + (8y + 48 ) = 0
y(y+6) + 8(y + 6) = 0
(y + 6)(y + 8) = 0
y = –6 and –8
∴ x > y
Correct (-)
Wrong (-)
Skipped (-)