NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) x² – 3x – 10 = 0
Step 1: Factorize the equation.
x² – 3x – 10 = 0
x² – 5x + 2x – 10 = 0
x(x – 5) + 2(x – 5) = 0
(x – 5)(x + 2) = 0
Step 2: Find the roots.
x – 5 = 0 or x + 2 = 0
x = 5 or x = -2
(ii) 2x² + x – 6 = 0
Step 1: Factorize the equation.
2x² + x – 6 = 0
2x² + 4x – 3x – 6 = 0
2x(x + 2) – 3(x + 2) = 0
(2x – 3)(x + 2) = 0
Step 2: Find the roots.
2x – 3 = 0 or x + 2 = 0
x = 3/2 or x = -2
(iii) √2x² + 7x + 5√2 = 0
Step 1: Factorize the equation.
√2x² + 7x + 5√2 = 0
√2x² + 5x + 2x + 5√2 = 0
x(√2x + 5) + √2(√2x + 5) = 0
(√2x + 5)(x + √2) = 0
Step 2: Find the roots.
√2x + 5 = 0 or x + √2 = 0
x = -5/√2 or x = -√2
(iv) 2x² – x + 1/8 = 0
Step 1: Factorize the equation.
2x² – x + 1/8 = 0
16x² – 8x + 1 = 0 (Multiplying by 8 to clear the fraction)
16x² – 4x – 4x + 1 = 0
4x(4x – 1) – 1(4x – 1) = 0
(4x – 1)(4x – 1) = 0
Step 2: Find the roots.
4x – 1 = 0
x = 1/4
(v) 100x² – 20x + 1 = 0
Step 2: Factorize the equation.
100x² – 20x + 1 = 0
(10x – 1)² = 0
(10x – 1)(10x – 1) = 0
Step 3: Find the roots.
10x – 1 = 0 or 10x – 1 = 0
x = 1/10 or x = 1/10
2. Solve the problems given in Example 1.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Step 2: Let the number of marbles John had be x. Therefore, Jivanti had (45 – x) marbles.
After losing 5 marbles each, John has (x – 5) marbles, and Jivanti has (40 – x) marbles.
Step 3: Set up the equation using the given product of their marbles.
(x – 5)(40 – x) = 124
Step 4: Expand and rearrange the equation to standard quadratic form.
x(40 – x) – 5(40 – x) = 124
40x – x² – 200 + 5x = 124
-x² + 45x – 324 = 0
Step 5: Factorize the quadratic equation.
-x² + 45x – 324 = 0
x² – 45x + 324 = 0
(x – 9)(x – 36) = 0
Step 6: Find the roots.
x – 9 = 0 or x – 36 = 0
x = 9 or x = 36
John had either 9 or 36 marbles, and Jivanti had the remaining amount out of 45.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day.
Step 2: Let the number of toys produced be x.
Step 3: The cost of production of each toy is (55 – x) rupees.
Step 4: Set up the equation for the total cost.
x(55 – x) = 750
Step 5: Expand and rearrange the equation to standard quadratic form.
55x – x² = 750
-x² + 55x – 750 = 0
Step 6: Factorize the quadratic equation.
-x² + 55x – 750 = 0
x² – 55x + 750 = 0
(x – 25)(x – 30) = 0
Step 7: Find the roots.
x – 25 = 0 or x – 30 = 0
x = 25 or x = 30
The number of toys produced on that day was either 25 or 30.
3. Find two numbers whose sum is 27 and product is 182.
Step 2: Let one number be x. Therefore, the other number is (27 – x).
Step 3: Set up the equation using the given product.
x(27 – x) = 182
Step 4: Expand and rearrange the equation to standard quadratic form.
27x – x² = 182
-x² + 27x – 182 = 0
Step 5: Factorize the quadratic equation.
-x² + 27x – 182 = 0
x² – 27x + 182 = 0
(x – 14)(x – 13) = 0
Step 6: Find the roots.
x – 14 = 0 or x – 13 = 0
x = 14 or x = 13
The two numbers are 14 and 13.
4. Find two consecutive positive integers, sum of whose squares is 365.
Step 2: Let the first integer be x. The next consecutive integer is x + 1.
Step 3: Set up the equation using the sum of squares.
x² + (x + 1)² = 365
Step 4: Expand and rearrange the equation to standard quadratic form.
x² + x² + 2x + 1 = 365
2x² + 2x – 364 = 0
Step 5: Factorize the quadratic equation.
2x² + 2x – 364 = 0
x² + x – 182 = 0
(x – 13)(x + 14) = 0
Step 6: Find the roots.
x – 13 = 0 or x + 14 = 0
x = 13 or x = -14
The two consecutive positive integers are 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.
Step 2: Let the base be x cm. Therefore, the altitude is (x – 7) cm.
Step 3: Apply Pythagoras theorem.
Base² + Altitude² = Hypotenuse²
x² + (x – 7)² = 13²
Step 4: Expand and rearrange the equation to standard quadratic form.
x² + x² – 14x + 49 = 169
2x² – 14x – 120 = 0
Step 5: Factorize the quadratic equation.
2x² – 14x – 120 = 0
x² – 7x – 60 = 0
(x – 12)(x + 5) = 0
Step 6: Find the roots.
x – 12 = 0 or x + 5 = 0
x = 12 or x = -5
Since the base cannot be negative, the base is 12 cm and the altitude is 5 cm.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹ 90, find the number of articles produced and the cost of each article.
Step 2: Let the number of articles produced be x.
Step 3: The cost of production of each article is (2x + 3) rupees.
Step 4: Set up the equation for the total cost.
x(2x + 3) = 90
Step 5: Expand and rearrange the equation to standard quadratic form.
2x² + 3x = 90
2x² + 3x – 90 = 0
Step 6: Factorize the quadratic equation.
2x² + 3x – 90 = 0
x² + 1.5x – 45 = 0
(x – 5)(x + 9) = 0
Step 7: Find the roots.
x – 5 = 0 or x + 9 = 0
x = 5 or x = -9
Since the number of articles cannot be negative, the number of articles produced is 5.
Step 8: Calculate the cost of each article.
Cost of each article = 2x + 3 = 2(5) + 3 = 10 + 3 = 13 rupees
The number of articles produced is 5, and the cost of each article is 13 rupees.