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Quadratic equations is one of the most important topic for banking as well as insurance exams as 5 questions are expected from this topic. In these types of questions, you will be given two quadratic equations roots of which you have to find & compare the values of the roots.
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Quadratic Equations Quadratic equations is one of the most important topic for banking as well as insurance exams as 5 questions are expected from this topic. In these types of questions, you will be given two quadratic equations roots of which you have to find & compare the values of the roots.
Quadratic Equations • Directions (1-5): In each of the following questions two equations are given. You have to solve the equations and give answer as: (a) if x < y(b) if x ≤ y (c) relationship between x and y cannot be determined (d) if x ≥ y(e) if x > y
Quadratic Equations Q1. I. 3x² + 10x + 8 = 0 II. 3y² + 7y + 4 = 0 Sol.I. 3x² + 10x + 8 = 0⇒ 3x² + 6x + 4x + 8 = 0⇒ (x + 2) (3x + 4) = 0⇒ x = –2, –4/3II. 3y² + 7y + 4 = 0⇒ 3y² + 3y + 4y + 4 = 0⇒ (y + 1) (3y + 4) = 0⇒ y = –1, –4/3y ≥ x Ans.(b)
Quadratic Equations Q2. I. 2x² + 21x + 10 = 0 II. 3y² + 13y + 14 = 0 Sol.I. 2x² + 21x + 10 = 0⇒ 2x² + 20 + x + 10 = 0⇒ (x + 10) (2x + 1) = 0⇒ x = –10, –1/2II. 3y² + 13y + 14 = 0⇒ 3y² + 6y + 7y + 14 = 0⇒ (y + 2) (3y + 7) = 0⇒ y = –2, –7/3No relation Ans. (c)
Quadratic Equations Q3. I. 8x² + 18x + 9 = 0 II. 4y² + 19y + 21 = 0 Sol.I. 8x² + 18x + 9 = 0⇒ 8x² + 12x + 6x + 9 = 0⇒ (2x + 3) (4x + 3) = 0⇒ x = –3/2, –3/4II. 4y² + 19y + 21 = 0⇒ 4y² + 12y + 7y + 21 = 0⇒ (y + 3) (4y + 7) = 0⇒ x = –3, –7/4x >y Ans.(e)
Quadratic Equations Q4. I. 3x² + 16x + 21 = 0II. 6y² + 17y + 12 = 0 Sol.I. 3x² + 16x + 21 = 0⇒ 3x² + 9x + 7x + 21 = 0⇒ (x + 3) (3x + 7) = 0⇒ x = –3, –7/3II. 6y² + 17y + 12 = 0⇒ 6y² + 9y + 8y + 12 = 0⇒ 3y (2y + 3) + 4 (2y + 3) = 0⇒ y = – 3/2, –4/3y > x Ans.(a)
Quadratic Equations Q5. I. 8x² + 6x = 5 II. 12y² – 22y + 8 = 0 Sol.I. 8x² + 6x – 5 = 0⇒ 8x² + 10x – 4x – 5 = 0⇒ (4x + 5) (2x – 1) = 0⇒ x = ½, –5/4II. 12y² – 22y + 8 = 0⇒ 6y² – 11y + 4 = 0⇒ 6y² – 3y – 8y + 4 = 0⇒ (2y – 1) (3y – 4) = 0⇒ y = 1/2, 4/3y ≥ x Ans.(b)