Class 10 Notes

Class 10 Mathematics Notes Chapter 4 (Quadratic equations) – Mathematics Book

Philoid

Philoid

— 8 min read

Mathematics
Below is the detailed notes of Chapter 4: Quadratic Equations. This is a fundamental chapter, not just for your Class 10 exams but also as a building block for higher mathematics and frequently tested in various government exams. Pay close attention to the concepts and methods.

Chapter 4: Quadratic Equations - Detailed Notes

1. Introduction & Definition

  • Polynomial: An expression consisting of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
  • Quadratic Polynomial: A polynomial of degree 2. Its general form is ax² + bx + c, where a, b, c are real numbers and a ≠ 0.
  • Quadratic Equation: When we set a quadratic polynomial equal to zero, we get a quadratic equation.
  • Standard Form: The standard form of a quadratic equation in variable x is:
    ax² + bx + c = 0
    where:
    • a, b, c are real numbers.
    • a ≠ 0 (This condition is crucial; if a=0, the equation becomes linear).
    • a is the coefficient of x² (leading coefficient).
    • b is the coefficient of x.
    • c is the constant term.

Example: 2x² - 5x + 3 = 0 is a quadratic equation where a=2, b=-5, c=3.
x² - 9 = 0 is also a quadratic equation where a=1, b=0, c=-9.

2. Identifying Quadratic Equations

To check if a given equation is quadratic:

  1. Simplify the equation by expanding brackets and bringing all terms to one side (usually the left-hand side), setting the other side to zero.
  2. Check if the highest power (degree) of the variable in the simplified equation is 2.
  3. Ensure the coefficient of the term with power 2 is non-zero.

Example: Check if (x+1)² = 2(x-3) is quadratic.

  • Expand: x² + 2x + 1 = 2x - 6
  • Simplify: x² + 2x - 2x + 1 + 6 = 0
  • Result: x² + 7 = 0 (or x² + 0x + 7 = 0)
  • Conclusion: Yes, it is a quadratic equation (a=1, b=0, c=7).

3. Solutions (Roots) of a Quadratic Equation

  • A root (or solution) of the quadratic equation ax² + bx + c = 0 is a real number α such that aα² + bα + c = 0.
  • In other words, the roots are the values of the variable x that satisfy the equation.
  • A quadratic equation can have at most two roots.

4. Methods for Solving Quadratic Equations

There are three primary methods:

a) Factorisation:

  • Principle: If a quadratic expression ax² + bx + c can be factorised into the product of two linear factors, say (px + q)(rx + s), then the roots are found by setting each factor to zero.
    (px + q)(rx + s) = 0 implies px + q = 0 or rx + s = 0.
    So, x = -q/p or x = -s/r.

  • Method (Splitting the Middle Term):

    1. Write the equation in standard form ax² + bx + c = 0.
    2. Find two numbers, say p and q, such that:
      • p + q = b (sum equals the middle coefficient)
      • p * q = ac (product equals the product of the first and last coefficients)
    3. Rewrite the middle term bx as px + qx.
      ax² + px + qx + c = 0
    4. Group the terms and factor by grouping:
      x(ax + p) + (q/a)(ax + p) = 0 (assuming a divides q here, adjust factors accordingly) or more generally, find common factors in pairs.
      Example: (ax+p) becomes a common factor.
    5. Set each linear factor to zero and solve for x.

  • Example: Solve 2x² - 5x + 3 = 0 by factorisation.

    1. Standard form: Already is. a=2, b=-5, c=3.
    2. Find p, q: Need p+q = -5 and p*q = a*c = 2*3 = 6. The numbers are -2 and -3. (-2 + -3 = -5, -2 * -3 = 6).
    3. Split middle term: 2x² - 2x - 3x + 3 = 0
    4. Factor by grouping: 2x(x - 1) - 3(x - 1) = 0
    5. Factor out common binomial: (2x - 3)(x - 1) = 0
    6. Set factors to zero:
      2x - 3 = 0 => 2x = 3 => x = 3/2
      x - 1 = 0 => x = 1
    • Roots are 1 and 3/2.

b) Completing the Square:

  • Principle: Convert the quadratic equation ax² + bx + c = 0 into the form (x + k)² = m² or (x - k)² = m², from which x can be easily found.

  • Method:

    1. Write the equation in standard form.
    2. Make the coefficient of x² unity (1) by dividing the entire equation by a (assuming a≠1): x² + (b/a)x + (c/a) = 0.
    3. Move the constant term to the right side: x² + (b/a)x = -c/a.
    4. Take half of the coefficient of x, square it, and add it to both sides. Half of b/a is b/(2a). Its square is (b/(2a))² = b²/(4a²).
      x² + (b/a)x + b²/(4a²) = -c/a + b²/(4a²)
    5. The left side is now a perfect square: (x + b/(2a))² = (b² - 4ac) / (4a²)
    6. Take the square root of both sides: x + b/(2a) = ±√(b² - 4ac) / (2a)
    7. Isolate x: x = -b/(2a) ± √(b² - 4ac) / (2a)
    8. Combine: x = [-b ± √(b² - 4ac)] / 2a (This leads directly to the quadratic formula).

  • Example: Solve x² + 4x - 5 = 0 by completing the square.

    1. a=1, b=4, c=-5. Coefficient of x² is already 1.
    2. Move constant: x² + 4x = 5.
    3. Half of coefficient of x is 4/2 = 2. Square it: 2² = 4. Add to both sides:
      x² + 4x + 4 = 5 + 4
    4. Factor LHS: (x + 2)² = 9
    5. Take square root: x + 2 = ±√9 = ±3
    6. Solve for x:
      Case 1: x + 2 = 3 => x = 3 - 2 = 1
      Case 2: x + 2 = -3 => x = -3 - 2 = -5
    • Roots are 1 and -5.

c) Quadratic Formula (Sridharacharya's Formula):

  • Principle: A direct formula derived from the method of completing the square, applicable to any quadratic equation.

  • Formula: For the equation ax² + bx + c = 0, the roots are given by:
    x = [-b ± √(b² - 4ac)] / 2a

  • Discriminant (D): The expression b² - 4ac is called the discriminant (D). It determines the nature of the roots.
    D = b² - 4ac
    The formula becomes: x = [-b ± √D] / 2a

  • Example: Solve 3x² - 5x + 2 = 0 using the quadratic formula.

    1. Identify coefficients: a=3, b=-5, c=2.
    2. Calculate Discriminant: D = b² - 4ac = (-5)² - 4(3)(2) = 25 - 24 = 1.
    3. Apply the formula:
      x = [-(-5) ± √1] / (2 * 3)
      x = [5 ± 1] / 6
    4. Find the two roots:
      x₁ = (5 + 1) / 6 = 6 / 6 = 1
      x₂ = (5 - 1) / 6 = 4 / 6 = 2/3
    • Roots are 1 and 2/3.

5. Nature of Roots (Using Discriminant D = b² - 4ac)

The value of the discriminant D tells us about the type of roots the equation ax² + bx + c = 0 has, without actually solving for them:

  • Case 1: D > 0 (Discriminant is positive)

    • The equation has two distinct real roots.
    • Formula gives x = (-b + √D) / 2a and x = (-b - √D) / 2a.
    • If D is a perfect square, the roots are rational. If D is not a perfect square, the roots are irrational.

  • Case 2: D = 0 (Discriminant is zero)

    • The equation has two equal real roots (or one repeated real root).
    • Formula gives x = (-b ± √0) / 2a = -b / 2a. Both roots are equal to -b / 2a.

  • Case 3: D < 0 (Discriminant is negative)

    • The equation has no real roots.
    • The square root of a negative number (√D) is not a real number. The roots are complex/imaginary (usually studied in Class 11). For Class 10 and most government exams based on it, we say "no real roots exist".

Example: Find the nature of roots for x² - 6x + 9 = 0.

  • a=1, b=-6, c=9.
  • D = b² - 4ac = (-6)² - 4(1)(9) = 36 - 36 = 0.
  • Since D = 0, the equation has two equal real roots.

Example: Find the nature of roots for 2x² + x + 1 = 0.

  • a=2, b=1, c=1.
  • D = b² - 4ac = (1)² - 4(2)(1) = 1 - 8 = -7.
  • Since D < 0, the equation has no real roots.

6. Relationship Between Roots and Coefficients

If α and β are the roots of the quadratic equation ax² + bx + c = 0, then:

  • Sum of Roots: α + β = -b/a
  • Product of Roots: α * β = c/a

These relationships are very useful for:

  • Finding a quadratic equation when roots are known:
    x² - (Sum of roots)x + (Product of roots) = 0
    x² - (α + β)x + αβ = 0
    (Note: This form assumes a=1. For the general form, multiply by a: a[x² - (α + β)x + αβ] = 0 which simplifies back to ax² + bx + c = 0).
  • Solving problems where relationships between roots are given.

Example: If the roots of x² - 7x + 12 = 0 are α and β, find α + β and αβ.

  • a=1, b=-7, c=12.
  • Sum: α + β = -b/a = -(-7)/1 = 7.
  • Product: α * β = c/a = 12/1 = 12.
    (We can check: the roots are 3 and 4. Sum = 3+4=7, Product = 3*4=12).

Example: Form a quadratic equation whose roots are 5 and -2.

  • Sum of roots = 5 + (-2) = 3.
  • Product of roots = 5 * (-2) = -10.
  • Equation: x² - (Sum)x + (Product) = 0
    x² - (3)x + (-10) = 0
    x² - 3x - 10 = 0.

7. Applications of Quadratic Equations (Word Problems)

Quadratic equations are used to model various real-world situations involving:

  • Area and dimensions of geometric shapes.
  • Problems related to speed, distance, and time.
  • Age-related problems.
  • Problems involving numbers.

Steps to Solve Word Problems:

  1. Read the problem carefully and identify the unknown quantity. Assign a variable (e.g., x) to it.
  2. Translate the statements of the problem into mathematical equations using the variable.
  3. Formulate a quadratic equation in standard form (ax² + bx + c = 0).
  4. Solve the quadratic equation using any suitable method (factorisation, quadratic formula).
  5. Interpret the solution(s) in the context of the problem. Reject any solution that doesn't make sense (e.g., negative length, negative speed).
  6. State the final answer clearly.

Example: The product of two consecutive positive integers is 306. Find the integers.

  1. Let the integers be x and x+1.
  2. Product: x(x+1) = 306.
  3. Form equation: x² + x = 306 => x² + x - 306 = 0.
  4. Solve (using formula or factorisation): a=1, b=1, c=-306.
    D = 1² - 4(1)(-306) = 1 + 1224 = 1225. √D = 35.
    x = [-1 ± 35] / 2.
    x₁ = (-1 + 35) / 2 = 34 / 2 = 17.
    x₂ = (-1 - 35) / 2 = -36 / 2 = -18.
  5. Interpret: The problem asks for positive integers, so we reject x = -18.
    If x = 17, the next integer is x+1 = 18.
  6. Answer: The integers are 17 and 18. (Check: 17 * 18 = 306).

Multiple Choice Questions (MCQs)

  1. Which of the following is a quadratic equation?
    (A) x³ - x² + 5 = 0
    (B) x² + 1/x² = 2
    (C) (x+2)² = 2(x-1)
    (D) x + 5 = 0

  2. The roots of the quadratic equation x² - 3x - 10 = 0 are:
    (A) 5, 2
    (B) 5, -2
    (C) -5, 2
    (D) -5, -2

  3. The discriminant (D) of the quadratic equation 2x² - 6x + 3 = 0 is:
    (A) 12
    (B) -12
    (C) 60
    (D) -60

  4. The nature of roots of the equation x² + 4x + 5 = 0 is:
    (A) Two distinct real roots
    (B) Two equal real roots
    (C) No real roots
    (D) More than two real roots

  5. If the roots of ax² + bx + c = 0 are equal, then:
    (A) b² - 4ac < 0
    (B) b² - 4ac > 0
    (C) b² - 4ac = 0
    (D) b² + 4ac = 0

  6. If one root of the equation kx² - 5x + 2 = 0 is 1/2, what is the value of k?
    (A) 3
    (B) 2
    (C) 1/3
    (D) 1/6

  7. If α and β are the roots of x² - 5x + 6 = 0, then the value of α + β is:
    (A) 6
    (B) -6
    (C) 5
    (D) -5

  8. The quadratic equation whose roots are 3 and -1 is:
    (A) x² + 2x - 3 = 0
    (B) x² - 2x - 3 = 0
    (C) x² + 2x + 3 = 0
    (D) x² - 2x + 3 = 0

  9. For what value of k does the equation 9x² + 3kx + 4 = 0 have equal roots?
    (A) ±2
    (B) ±3
    (C) ±4
    (D) ±5

  10. The sum of the squares of two consecutive natural numbers is 61. The numbers are:
    (A) 4 and 5
    (B) 5 and 6
    (C) 6 and 7
    (D) 3 and 4

Answer Key for MCQs:

  1. C (Simplify: x² + 4x + 4 = 2x - 2 => x² + 2x + 6 = 0. Highest power is 2. Option B becomes x⁴ + 1 = 2x² which is biquadratic).
  2. B (Factorise: (x-5)(x+2)=0 => x=5 or x=-2).
  3. A (D = b² - 4ac = (-6)² - 4(2)(3) = 36 - 24 = 12).
  4. C (D = b² - 4ac = 4² - 4(1)(5) = 16 - 20 = -4. Since D < 0, no real roots).
  5. C (Condition for equal roots is D = 0).
  6. B
  7. C (α + β = -b/a = -(-5)/1 = 5).
  8. B (Sum = 3 + (-1) = 2. Product = 3 * (-1) = -3. Equation: x² - (Sum)x + (Product) = 0 => x² - 2x - 3 = 0).
  9. C (Equal roots means D = 0. D = (3k)² - 4(9)(4) = 9k² - 144. 9k² - 144 = 0 => 9k² = 144 => k² = 16 => k = ±4).
  10. B (Let numbers be n and n+1. n² + (n+1)² = 61. n² + n² + 2n + 1 = 61. 2n² + 2n - 60 = 0. n² + n - 30 = 0. Factorise: (n+6)(n-5) = 0. Since numbers are natural, n=5. The numbers are 5 and 5+1=6. Check: 5² + 6² = 25 + 36 = 61).

Make sure you practice solving various types of problems, especially word problems, as they test your understanding and application skills. Good luck!

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