RRB NTPC & Group D Maths: Master Quadratic Equations – Concepts, Shortcuts & Solved Questions

Quadratic Equations form a fundamental building block of algebra and quantitative aptitude in competitive exams conducted by the Railway Recruitment Board (RRB). Whether you are appearing for RRB NTPC (CBT 1 & CBT 2), RRB Group D, or RRB Technician Grade I & III, having a solid grasp of quadratic equations can help you secure easy, high-accuracy marks.

Introduction to Quadratic Equations for RRB Exams

An algebraic equation of second degree is called a Quadratic Equation. In simple terms, it is an equation where the highest power (degree) of the variable is 2. The standard form of a quadratic equation in a single variable \(x\) is expressed as:

\(ax^2 + bx + c = 0\)

Where:

• \(a\), \(b\), and \(c\) are real numbers known as coefficients.
• \(a \neq 0\) (if \(a = 0\), the equation reduces to a linear equation \(bx + c = 0\)).
• \(x\) is the unknown variable whose values (called roots or solutions) satisfy the equation.

In railway examinations, questions range from finding the nature of roots, forming equations from given roots, to solving word problems that translate into quadratic equations.

Topic Weightage and Importance

In Railway Recruitment Board exams, algebra and quadratic equations hold substantial weightage:

Because quadratic equation questions follow predictable patterns and formulas, mastering this topic ensures 100% accuracy in very little time.

Key Concepts and Formulas

1. Methods of Solving Quadratic Equations

There are three primary methods used to find the roots (\(\alpha\) and \(\beta\)) of a quadratic equation:

• Factorization Method (Splitting the Middle Term): Rewrite the term \(bx\) as the sum of two terms whose product is equal to \(a \cdot c\).
• Quadratic Formula (Sridharacharya Formula):
  

  \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

• Completing the Square Method: Transform the equation into the form \((x + p)^2 = q\).

2. Nature of Roots (Discriminant Method)

The term \(D = b^2 - 4ac\) is called the Discriminant. The value of \(D\) determines the nature of the roots:

3. Relation Between Roots and Coefficients

Let \(\alpha\) and \(\beta\) be the two roots of \(ax^2 + bx + c = 0\):

• Sum of Roots (\(S\)): \(\alpha + \beta = -\frac{b}{a}\)
• Product of Roots (\(P\)): \(\alpha \cdot \beta = \frac{c}{a}\)

4. Formation of a Quadratic Equation

If the roots \(\alpha\) and \(\beta\) are known, the quadratic equation is given by:

\(x^2 - (\text{Sum of Roots})x + (\text{Product of Roots}) = 0\)

\(x^2 - (\alpha + \beta)x + (\alpha \beta) = 0\)

5. Important Algebraic Identities Frequently Used

• \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
• \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
• \(\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)\)
• \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}\)

6. Sign Shortcut Trick for Fast Root Determination

For quick calculations without writing out full steps, observe the signs of \(b\) and \(c\):

Solved Examples (Step-by-Step)

Example 1: Finding Roots via Splitting the Middle Term

Question: Find the roots of the quadratic equation \(2x^2 - 7x + 6 = 0\).

Solution:

• Step 1: Identify coefficients: \(a = 2\), \(b = -7\), \(c = 6\).
• Step 2: Compute \(a \cdot c = 2 \times 6 = 12\).
• Step 3: Find two numbers whose product is 12 and sum is \(-7\). These numbers are \(-3\) and \(-4\).
• Step 4: Split the middle term:
  \(2x^2 - 4x - 3x + 6 = 0\)
  \(2x(x - 2) - 3(x - 2) = 0\)
  \((2x - 3)(x - 2) = 0\)
• Step 5: Solve for \(x\):
  \(2x - 3 = 0 \implies x = \frac{3}{2}\)
  \(x - 2 = 0 \implies x = 2\)

Answer: The roots are \(x = 2\) and \(x = \frac{3}{2}\).

Example 2: Determining the Nature of Roots

Question: For what value of \(k\) does the equation \(kx^2 - 6x + 1 = 0\) have equal and real roots?

Solution:

• Step 1: For real and equal roots, the discriminant must be zero: \(D = b^2 - 4ac = 0\).
• Step 2: Identify coefficients: \(a = k\), \(b = -6\), \(c = 1\).
• Step 3: Substitute into discriminant formula:
  \((-6)^2 - 4(k)(1) = 0\)
  \(36 - 4k = 0\)
  \(4k = 36 \implies k = 9\)

Answer: \(k = 9\).

Example 3: Working with Symmetric Functions of Roots

Question: If \(\alpha\) and \(\beta\) are the roots of \(x^2 - 5x + 4 = 0\), find the value of \(\alpha^2 + \beta^2\).

Solution:

• Step 1: Identify sum and product of roots:
  Sum \((\alpha + \beta) = -\frac{b}{a} = -\frac{-5}{1} = 5\)
  Product \((\alpha\beta) = \frac{c}{a} = \frac{4}{1} = 4\)
• Step 2: Use identity \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\):
  \(\alpha^2 + \beta^2 = (5)^2 - 2(4) = 25 - 8 = 17\)

Answer: \(17\).

Example 4: Forming an Equation from Given Roots

Question: Form the quadratic equation whose roots are \(3 + \sqrt{5}\) and \(3 - \sqrt{5}\).

Solution:

• Step 1: Find the sum of roots \(S\):
  \(S = (3 + \sqrt{5}) + (3 - \sqrt{5}) = 6\)
• Step 2: Find the product of roots \(P\):
  \(P = (3 + \sqrt{5})(3 - \sqrt{5}) = 3^2 - (\sqrt{5})^2 = 9 - 5 = 4\)
• Step 3: Substitute into equation formula \(x^2 - Sx + P = 0\):
  \(x^2 - 6x + 4 = 0\)

Answer: \(x^2 - 6x + 4 = 0\).

Common Mistakes to Avoid

• Sign Errors in Sum of Roots: Forgetting the negative sign in \(\alpha + \beta = -\frac{b}{a}\). If \(b = -6\), \(-\frac{b}{a} = -(-6) = +6\).
• Ignoring the Leading Coefficient \(a\): When dividing by \(2a\) in the quadratic formula, candidates often divide only by \(2\), forgetting \(a\) when \(a \neq 1\).
• Discriminant Sign Confusion: Calculating \(b^2 - 4ac\) with negative coefficients incorrectly (e.g., \((-4)^2 = 16\), not \(-16\)).
• Disregarding Domain Constraints in Word Problems: In speed, distance, or age-based quadratic word problems, roots can be negative, but physical quantities cannot be negative. Always reject negative solutions where physically impossible.

Practice Questions with Solutions

Practice Questions

1. Find the roots of the equation \(x^2 - 11x + 30 = 0\).
2. If one root of \(2x^2 + kx - 6 = 0\) is \(2\), find the value of \(k\) and the other root.
3. Find the discriminant and state the nature of roots for \(3x^2 - 5x + 3 = 0\).
4. If \(\alpha\) and \(\beta\) are roots of \(x^2 - 7x + 12 = 0\), calculate the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\).
5. Find the quadratic equation whose roots are twice the roots of \(x^2 - 4x + 3 = 0\).
6. Solve for \(x\): \(\sqrt{6 + \sqrt{6 + \sqrt{6 + \dots}}} = x\).

Detailed Solutions

1. Solution:
   \(x^2 - 11x + 30 = 0\)
   Splitting middle term: \(-5\) and \(-6\) (since \((-5) \times (-6) = 30\) and \(-5 + (-6) = -11\)).
   \((x - 5)(x - 6) = 0 \implies x = 5, 6\).
2. Solution:
   Substitute \(x = 2\) into the equation:
   \(2(2)^2 + k(2) - 6 = 0 \implies 8 + 2k - 6 = 0 \implies 2k + 2 = 0 \implies k = -1\).
   Equation becomes \(2x^2 - x - 6 = 0\).
   Product of roots \(= \alpha \cdot \beta = \frac{-6}{2} = -3\).
   Since \(\alpha = 2\), \(2 \cdot \beta = -3 \implies \beta = -\frac{3}{2}\).
3. Solution:
   \(a = 3\), \(b = -5\), \(c = 3\).
   \(D = b^2 - 4ac = (-5)^2 - 4(3)(3) = 25 - 36 = -11\).
   Since \(D < 0\), the roots are imaginary / complex (no real roots).
4. Solution:
   For \(x^2 - 7x + 12 = 0\):
   \(\alpha + \beta = 7\) and \(\alpha\beta = 12\).
   \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{7}{12}\).
5. Solution:
   For original equation \(x^2 - 4x + 3 = 0\), sum \(S = 4\), product \(P = 3\).
   Let new roots be \(\alpha' = 2\alpha\) and \(\beta' = 2\beta\).
   New Sum \(S' = 2\alpha + 2\beta = 2(\alpha + \beta) = 2(4) = 8\).
   New Product \(P' = (2\alpha)(2\beta) = 4\alpha\beta = 4(3) = 12\).
   New equation: \(x^2 - S'x + P' = 0 \implies x^2 - 8x + 12 = 0\).
6. Solution:
   Given \(x = \sqrt{6 + x}\).
   Squaring both sides: \(x^2 = 6 + x \implies x^2 - x - 6 = 0\).
   \((x - 3)(x + 2) = 0 \implies x = 3\) or \(x = -2\).
   Since the square root principal value is always positive, \(x = 3\).

Frequently Asked Questions (FAQs)

1. Can a quadratic equation have more than two roots?

No. According to the Fundamental Theorem of Algebra, a polynomial equation of degree \(n\) has exactly \(n\) roots. Therefore, a quadratic equation (degree 2) has exactly two roots (which can be real and distinct, real and equal, or complex).

2. How can I quickly find roots in CBT exams without full factorisation?

Use the Sign Rule trick and factors of the product \(a \cdot c\). First determine the signs of the roots, then find two factors of \(a \cdot c\) that add up to \(b\), and divide both numbers by \(a\) with their signs reversed.

3. What type of word problems on quadratic equations appear in RRB exams?

The most frequent word problem types include:
• Number relation problems (e.g., product of two consecutive integers is 182).
• Speed, distance, and time (e.g., an increase in train speed reduces journey time).
• Work and time relations involving two workers with dependent rates.

Conclusion and Final Tips

Quadratic Equations are a scoring topic across all RRB exams. Ensure you have the standard formula, discriminant conditions, and symmetric root properties memorized. Practice 15-20 problems daily using middle-term splitting and sign tricks to build speed and accuracy. Consistent practice will give you full marks in the algebraic section of your upcoming Railway exam!