Class 10 Mathematics Notes Chapter 4 (Chapter 4) – Examplar Problems (English) Book

Alright class, let's get straight into Chapter 4: Quadratic Equations. This is a crucial chapter, not just for your board exams but also for various government competitive exams. We'll focus on the concepts as presented in the NCERT Exemplar, which often tests deeper understanding.

Chapter 4: Quadratic Equations - Detailed Notes for Exam Preparation

1. What is a Quadratic Equation?

• An equation of the form ax² + bx + c = 0, where 'a', 'b', and 'c' are real numbers and a ≠ 0, is called a quadratic equation in the variable 'x'.
• Standard Form: Always try to write the equation in this form (ax² + bx + c = 0) before solving.
• Why a ≠ 0? If 'a' were 0, the x² term would vanish, and it would become a linear equation (bx + c = 0), not quadratic.
• Degree: The highest power of the variable in a quadratic equation is 2.

2. Roots or Solutions of a Quadratic Equation

• A real number 'α' is called a root (or solution) of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0.
• In simpler terms, the values of 'x' that satisfy the equation are its roots.
• A quadratic equation can have at most two roots.

3. Methods for Solving Quadratic Equations

• (a) Factorisation:

  • Concept: Express the quadratic polynomial (ax² + bx + c) as a product of two linear factors (px + q)(rx + s). Then, set each factor to zero to find the roots.
  • Method (Splitting the Middle Term):
    1. Find two numbers whose product is equal to (a × c) and whose sum is equal to 'b'.
    2. Rewrite the middle term (bx) using these two numbers.
    3. Factor by grouping the terms.
    4. Equate each factor to zero to find the roots.
  • Exemplar Focus: Problems might involve factorisation with square roots or fractions in coefficients, requiring careful manipulation.
  • Example (Exemplar Type): Solve √3x² + 10x + 7√3 = 0
    • Product (a × c) = √3 × 7√3 = 21
    • Sum (b) = 10
    • Numbers are 7 and 3.
    • √3x² + 7x + 3x + 7√3 = 0
    • x(√3x + 7) + √3(√3x + 7) = 0
    • (√3x + 7)(x + √3) = 0
    • Roots: x = -7/√3 and x = -√3

• (b) Quadratic Formula (Sridharacharya's Rule):

  • Concept: A direct formula to find the roots, applicable to all quadratic equations.
  • Formula: For ax² + bx + c = 0, the roots are given by:
    x = [-b ± √(b² - 4ac)] / 2a
  • Discriminant (D): The term D = b² - 4ac is called the discriminant. It determines the nature of the roots.
  • When to Use: Always applicable, especially useful when factorisation is difficult or when asked to find the nature of roots.

• (c) Completing the Square: (Less common for direct solving in MCQs, but concept is important)

  • Concept: Convert the equation ax² + bx + c = 0 into the form (x + k)² = p² or (x - k)² = p².
  • Method:
    1. Make the coefficient of x² unity (divide by 'a').
    2. Shift the constant term to the RHS.
    3. Add (half the coefficient of x)² to both sides.
    4. Write LHS as a perfect square and simplify RHS.
    5. Take the square root on both sides and solve for x.

4. Nature of Roots (Based on Discriminant D = b² - 4ac)

• This is a VERY important section for competitive exams.
• Case 1: D > 0 (b² - 4ac > 0)
  • The equation has two distinct real roots.
  • Roots are given by x = (-b + √D) / 2a and x = (-b - √D) / 2a.
  • Sub-case: If D is a perfect square, the roots are real, distinct, and rational.
  • Sub-case: If D is not a perfect square, the roots are real, distinct, and irrational (often occur in conjugate pairs like p + √q and p - √q).
• Case 2: D = 0 (b² - 4ac = 0)
  • The equation has two equal real roots (or sometimes referred to as one real root).
  • The roots are given by x = -b / 2a.
• Case 3: D < 0 (b² - 4ac < 0)
  • The equation has no real roots. (The roots are complex conjugates, usually outside the scope of Class 10 based exams unless specified).

5. Relationship Between Roots and Coefficients

• While not explicitly detailed in basic NCERT, it's useful for competitive exams and Exemplar problems.
• If α and β are the roots of ax² + bx + c = 0:
  • Sum of roots (α + β) = -b/a (Negative of coefficient of x / coefficient of x²)
  • Product of roots (αβ) = c/a (Constant term / coefficient of x²)
• Forming a Quadratic Equation with given roots α and β:
  • The equation is x² - (α + β)x + αβ = 0
  • Or, k[x² - (Sum of roots)x + (Product of roots)] = 0, where k is any non-zero real number.

6. Word Problems Leading to Quadratic Equations

• Common Types: Problems involving numbers, ages, speed-distance-time, geometry (area, perimeter), work-time, cost/price.
• Strategy:
  1. Read the problem carefully and identify the unknown quantity. Assign a variable (usually 'x').
  2. Translate the statements/conditions given in the problem into mathematical equations using the variable.
  3. Simplify the equation and bring it to the standard quadratic form (ax² + bx + c = 0).
  4. Solve the quadratic equation using an appropriate method (factorisation or quadratic formula).
  5. Check the validity of the solution(s) in the context of the problem (e.g., age, speed, length cannot be negative). Discard extraneous solutions.

Key Points for Exam Preparation:

• Always check if a = 0. If a=0, it's not a quadratic equation.
• Master the discriminant (D = b² - 4ac) and its implications for the nature of roots. Many MCQs are based directly on this.
• Be comfortable with both factorisation and the quadratic formula. Choose the method wisely based on the equation's complexity and the question asked.
• Pay close attention to signs (+/-) while using the quadratic formula.
• In word problems, defining the variable correctly and forming the equation accurately is crucial.
• Remember to check your final answers, especially in word problems, for practical validity.

Multiple Choice Questions (MCQs)

1. Which of the following is a quadratic equation?
   (a) x² + 2x + 1 = (4 - x)² + 3
   (b) -2x² = (5 - x)(2x - 2/5)
   (c) (k+1)x² + (3/2)x = 7, where k = -1
   (d) x³ - x² = (x - 1)³

2. The roots of the quadratic equation 2x² - x - 6 = 0 are:
   (a) -2, 3/2
   (b) 2, -3/2
   (c) -2, -3/2
   (d) 2, 3/2

3. What is the value of the discriminant for the quadratic equation √3x² - 2√2x - 2√3 = 0?
   (a) 20
   (b) 24
   (c) 32
   (d) 8

4. For what value(s) of 'k' does the quadratic equation 9x² + 3kx + 4 = 0 have equal roots?
   (a) k = ±2
   (b) k = ±3
   (c) k = ±4
   (d) k = ±5

5. If one root of the equation x² + kx - 5/4 = 0 is 1/2, then the value of k is:
   (a) 2
   (b) -2
   (c) 1/4
   (d) 1/2

6. The nature of roots of the quadratic equation 3x² - 4√3x + 4 = 0 is:
   (a) Real, distinct roots
   (b) Real, equal roots
   (c) No real roots
   (d) More than two roots

7. If α and β are the roots of x² + x - 1 = 0, then the value of 1/α + 1/β is:
   (a) -1
   (b) 1
   (c) 0
   (d) 1/2

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Which quadratic equation represents this situation, where 'x' is the uniform speed in km/h?
   (a) x² + 5x - 1800 = 0
   (b) x² - 5x + 1800 = 0
   (c) x² + 5x + 1800 = 0
   (d) x² - 5x - 1800 = 0

9. The roots of the equation x² + x - p(p+1) = 0, where p is a constant, are:
   (a) p, p+1
   (b) -p, p+1
   (c) p, -(p+1)
   (d) -p, -(p+1)

10. If the equation x² - bx + 1 = 0 does not possess real roots, then:
    (a) -3 < b < 3
    (b) -2 < b < 2
    (c) b > 2
    (d) b < -2

Answers to MCQs:

1. (d) [Explanation: (a) becomes linear, (b) becomes linear, (c) becomes linear as k=-1 makes x² coeff zero. (d) simplifies to x³ - x² = x³ - 3x² + 3x - 1 => 2x² - 3x + 1 = 0, which is quadratic.]
2. (b) [Explanation: Factorise 2x² - 4x + 3x - 6 = 0 => 2x(x-2) + 3(x-2) = 0 => (2x+3)(x-2)=0. Roots are x=2, x=-3/2.]
3. (c) [Explanation: a=√3, b=-2√2, c=-2√3. D = b² - 4ac = (-2√2)² - 4(√3)(-2√3) = 8 - 4(-6) = 8 + 24 = 32.]
4. (c) [Explanation: For equal roots, D = 0. D = (3k)² - 4(9)(4) = 9k² - 144. 9k² - 144 = 0 => 9k² = 144 => k² = 16 => k = ±4.]
5. (a) [Explanation: Substitute x=1/2 in the equation: (1/2)² + k(1/2) - 5/4 = 0 => 1/4 + k/2 - 5/4 = 0 => k/2 = 4/4 = 1 => k = 2.]
6. (b) [Explanation: a=3, b=-4√3, c=4. D = b² - 4ac = (-4√3)² - 4(3)(4) = (16 * 3) - 48 = 48 - 48 = 0. Since D=0, roots are real and equal.]
7. (b) [Explanation: 1/α + 1/β = (α+β)/αβ. Here a=1, b=1, c=-1. Sum (α+β) = -b/a = -1/1 = -1. Product (αβ) = c/a = -1/1 = -1. So, (α+β)/αβ = -1/-1 = 1.]
8. (a) [Explanation: Time taken initially = 360/x. New speed = x+5. New time = 360/(x+5). Given: (360/x) - (360/(x+5)) = 1. Simplify: 360[(x+5)-x] / [x(x+5)] = 1 => 360(5) = x(x+5) => 1800 = x² + 5x => x² + 5x - 1800 = 0.]
9. (c) [Explanation: Use quadratic formula or factorisation. Factorisation: x² + (p+1)x - px - p(p+1) = 0 => x(x + p+1) - p(x + p+1) = 0 => (x-p)(x+p+1)=0. Roots are x=p, x=-(p+1).]
10. (b) [Explanation: No real roots means D < 0. a=1, b=-b, c=1. D = (-b)² - 4(1)(1) = b² - 4. We need b² - 4 < 0 => b² < 4 => -2 < b < 2.]

Study these notes thoroughly, practice problems from the Exemplar, and pay close attention to the conditions for the nature of roots. Good luck!