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Quadratic equations are very common in all the competitive exams and require a lot of practice. It is very different from the linear equation. The main difference between the quadratic equation and linear equation is that the latter will have a term x² in it. When the power of a term is 2 it is called quadratic term or a term with a degree 2.

**Quadratic equations – These are the equations which look like**

**ax ^{2} + bx + c = 0**

These equations are asked in the competitive examinations in a set of 5 questions. Enough with the theory, the question in exams look like –

**In each of the following questions, two equations are given. Solve these equations and give answer:**

**A) If x >= y, i.e. x is greater than or equal to y**

**B) If x > y, i.e. x is greater than y**

**C) If x <= y, i.e. x is less than or equal to y**

**D) If x < y, i.e. x less than y**

**E) If x = y, or no relation can be established between x and y**

**Step – wise tips for practicing a quadratic equation:**

- Firstly the candidate needs to note down twenty sums related to the quadratic equation on a page.
- Then ten sums are to be solved using basic formula.
- To solve these sums using a stopwatch will help in time management.
- The next step is to evaluate the time taken and analyse the performance.
- The remaining ten questions are to be solved next by applying the shortcut tricks.
- The time taken should be noted carefully.
- This will surely have a difference from the time taken while solving the first ten questions.
- Practice is the key to master this topic of the Quantitative Aptitude section.

**Q1. I. 3x² + 10x + 8 = 0 **

** II. 3y² + 7y + 4 = 0 **

**S1. Ans.(b)**

Sol.

I. 3x² + 10x + 8 = 0

⇒ 3x² + 6x + 4x + 8 = 0

⇒ (x + 2) (3x + 4) = 0

⇒ x = –2, –4/3

II. 3y² + 7y + 4 = 0

⇒ 3y² + 3y + 4y + 4 = 0

⇒ (y + 1) (3y + 4) = 0

⇒ y = –1, –4/3

**y ≥ x**

**Q3. I. 8x² + 18x + 9 = 0 **

**II. 4y² + 19y + 21 = 0 **

S3. Ans.(e)

Sol.

I. 8x² + 18x + 9 = 0

⇒ 8x² + 12x + 6x + 9 = 0

⇒ (2x + 3) (4x + 3) = 0

⇒ x = –3/2, –3/4

II. 4y² + 19y + 21 = 0

⇒ 4y² + 12y + 7y + 21 = 0

⇒ (y + 3) (4y + 7) = 0

⇒ x = –3, –7/4

**x >y**

**Q4. I. 3x² + 16x + 21 = 0**

**II. 6y² + 17y + 12 = 0 **

S4. Ans.(a)

Sol.

I. 3x² + 16x + 21 = 0

⇒ 3x² + 9x + 7x + 21 = 0

⇒ (x + 3) (3x + 7) = 0

⇒ x = –3, –7/3

II. 6y² + 17y + 12 = 0

⇒ 6y² + 9y + 8y + 12 = 0

⇒ 3y (2y + 3) + 4 (2y + 3) = 0

⇒ y = – 3/2, –4/3

**y > x**

**Q5. I. 8x² + 6x = 5 **

**II. 12y² – 22y + 8 = 0 **

S5. Ans.(b)

Sol.

I. 8x² + 6x – 5 = 0

⇒ 8x² + 10x – 4x – 5 = 0

⇒ (4x + 5) (2x – 1) = 0

⇒ x = ½, –5/4

II. 12y² – 22y + 8 = 0

⇒ 6y² – 11y + 4 = 0

⇒ 6y² – 3y – 8y + 4 = 0

⇒ (2y – 1) (3y – 4) = 0

⇒ y = 1/2, 4/3

**y ≥ x**