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### Tip of the month

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

ax2 + 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