Two equations I and II are given below in each question. You have to solve these equations and give answer I. 2x2 + 3x – 20 = 0 II. 2y2 + 19y + 44 = 0

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Two equations I and II are given below in each question. You have to solve these equations and give answer I. 2x2 + 3x – 20 = 0 II. 2y2 + 19y + 44 = 0

Asked on 2019-07-11 21:11:47 by Guest | Votes 0 | Views: 32 | Tags: algebra | numerical ability | Add Bounty

Two equations I and II are given below in each question. You have to solve these equations and give answer

I. 2x² + 3x – 20 = 0

II. 2y² + 19y + 44 = 0
1). if x < y
2). if x > y
3). if x ≤ y
4). if x ≥ y

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Answered by Guest on 2019-07-11 21:12:13 | Votes 2 | #

I. 2x² + 3x – 20 = 0

⇒ 2x² + 8x – 5x – 20 = 0

⇒ 2x(x + 4) – 5(x + 4) = 0

⇒ (2x – 5)(x + 4) = 0

Then, x = (5/2) or x = (-4)

II. 2y² + 19y + 44 = 0

⇒ 2y² + 11y + 8y + 44 = 0

⇒ y(2y + 11) + 4(2y + 11) = 0

⇒ (y + 4)(2y + 11) = 0

Then, y = (-4) or y = (-11/2)

So, when x = (5/2), x > y for y = (-11/2) and x > y for y = (-4)

And when x = (-4), x > y for y = (-11/2) and x = y for y = (-4)

∴ So, we can find x ≥ y.

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