exercise 4.2 class 10 solution answer key

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MATHEMATICS CLASS 10TH – QUADRATIC EQUATION SOLUTIONS

  1. (i)   x2 – 3x – 10 = 0
    x2 – 5x + 2x – 10 = 0
    x(x – 5) + 2(x – 5) = 0
    (x – 5)(x + 2) = 0
    Roots: x = 5 or x = –2
  2. (ii)   2x2 + x – 6 = 0
    2x2 + 4x – 3x – 6 = 0
    2x(x + 2) – 3(x + 2) = 0
    (x + 2)(2x – 3) = 0
    Roots: x = –2 or x = 3/2
  3. (iii)   √2x2 + 7x + 5√2 = 0
    √2x2 + 5x + 2x + 5√2 = 0
    x(√2x + 5) + √2(√2x + 5) = 0
    (√2x + 5)(x + √2) = 0
    Roots: x = –5/√2 or x = –√2
  4. (iv)   2x2 – x + 1/8 = 0
    Multiply both sides by 8:
    16x2 – 8x + 1 = 0
    16x2 – 4x – 4x + 1 = 0
    4x(4x – 1) – 1(4x – 1) = 0
    (4x – 1)2 = 0
    Roots: x = 1/4 (double root)
  5. (v)   100x2 – 20x + 1 = 0
    100x2 – 10x – 10x + 1 = 0
    10x(10x – 1) – 1(10x – 1) = 0
    (10x – 1)2 = 0
    Roots: x = 1/10 (double root)

2. Solve the problems given in Example 1:

  1. (i) John and Jivanti’s marbles
    Let John’s marbles = x.
    Jivanti’s marbles = 45 – x.
    After losing 5 each:
    John = x – 5, Jivanti = 40 – x.
    Product given: (x – 5)(40 – x) = 124
    Expanding: x2 – 45x + 324 = 0
    Factorising: (x – 36)(x – 9) = 0
    Solutions: x = 36 or x = 9
    So John had 36 or 9 marbles, and Jivanti had 9 or 36 marbles respectively.
  2. (ii) Cottage industry toys production
    Let number of toys produced = x.
    Cost per toy = Rs. (55 – x).
    Total cost = Rs. 750:
    x(55 – x) = 750
    Expand: x2 – 55x + 750 = 0
    Factorise: (x – 25)(x – 30) = 0
    Solutions: x = 25 or x = 30

3. Find two numbers whose sum is 27 and product is 182:

Let first number = x, second = 27 – x.
Product equation: x(27 – x) = 182
Expand: x2 – 27x + 182 = 0
Factorise: (x – 13)(x – 14) = 0
Solutions: x = 13 or x = 14
So the numbers are 13 and 14.

4. Find two consecutive positive integers, sum of whose squares is 365:

Let the integers be x and x + 1.
Equation: x2 + (x + 1)2 = 365
Simplify: 2x2 + 2x – 364 = 0 or x2 + x – 182 = 0
Factorise: (x + 14)(x – 13) = 0
Positive root: x = 13
Integers: 13 and 14

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides:

Let base = x, altitude = x – 7.
Using Pythagoras theorem:
x2 + (x – 7)2 = 132
Expand: 2x2 – 14x – 120 = 0
Simplify: x2 – 7x – 60 = 0
Factor: (x – 12)(x + 5) = 0
Positive root: x = 12
Altitude = 5 cm
Base: 12 cm, Altitude: 5 cm

6. A cottage industry’s pottery production cost problem:

Let number of articles produced = x.
Cost per article = Rs. (2x + 3).
Total cost = Rs. 90:
x(2x + 3) = 90
Expand: 2x2 + 3x – 90 = 0
Factor: (2x + 15)(x – 6) = 0
Positive root: x = 6
Cost per article = Rs. 15
Number of articles produced: 6, Cost per article: Rs. 15