Welcome Students!

Welcome to our comprehensive guide for Class 10 Mathematics Chapter 4: Quadratic Equations. This chapter is fundamental for building algebraic problem-solving skills and is essential for school exams and competitive tests. Below, you will find step-by-step solutions for all questions from Exercise 4.1, Exercise 4.2, and Exercise 4.3.

Exercise 4.1 Solutions

1. Check whether the following are quadratic equations:

(i) (x + 1)² = 2(x - 3)
Solution:
Expanding both sides: x² + 2x + 1 = 2x - 6
x² + 2x - 2x + 1 + 6 = 0
x² + 7 = 0
This is of the form ax² + bx + c = 0 where a = 1 ≠ 0.
Hence, it is a quadratic equation.

(ii) x² - 2x = (-2)(3 - x)
Solution:
x² - 2x = -6 + 2x
x² - 4x + 6 = 0
Hence, it is a quadratic equation.

(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
Solution:
x² - x - 2 = x² + 2x - 3
-3x + 1 = 0
Since the coefficient of x² is 0 (degree is 1), it is not a quadratic equation.

(iv) (x - 3)(2x + 1) = x(x + 5)
Solution:
2x² - 5x - 3 = x² + 5x
x² - 10x - 3 = 0
Hence, it is a quadratic equation.

(v) (2x - 1)(x - 3) = (x + 5)(x - 1)
Solution:
2x² - 7x + 3 = x² + 4x - 5
x² - 11x + 8 = 0
Hence, it is a quadratic equation.

(vi) x² + 3x + 1 = (x - 2)²
Solution:
x² + 3x + 1 = x² - 4x + 4
7x - 3 = 0
Hence, it is not a quadratic equation.

(vii) (x + 2)³ = 2x(x² - 1)
Solution:
x³ + 6x² + 12x + 8 = 2x³ - 2x
x³ - 6x² - 14x - 8 = 0
Degree is 3. Hence, it is not a quadratic equation.

(viii) x³ - 4x² - x + 1 = (x - 2)³
Solution:
x³ - 4x² - x + 1 = x³ - 6x² + 12x - 8
2x² - 13x + 9 = 0
Hence, it is a quadratic equation.

2. Represent the following situations in the form of quadratic equations:

(i) Area of rectangular plot is 528 m². Length is 1 m more than twice its breadth.
Solution:
Let breadth = x m. Then length = (2x + 1) m.
Area = length × breadth = x(2x + 1) = 528
2x² + x - 528 = 0

(ii) Product of two consecutive positive integers is 306.
Solution:
Let the integers be x and x + 1.
x(x + 1) = 306
x² + x - 306 = 0

(iii) Rohan's mother is 26 years older than him. Product of ages 3 years from now will be 360.
Solution:
Let Rohan's age = x years. Mother's age = (x + 26) years.
In 3 years: (x + 3)(x + 29) = 360
x² + 32x + 87 = 360
x² + 32x - 273 = 0

(iv) Train travels 480 km at uniform speed. If speed was 8 km/h less, it takes 3 hours more.
Solution:
Let speed = x km/h.
Time taken = 480/x hours.
New speed = (x - 8) km/h.
480/(x - 8) - 480/x = 3
480[ (x - (x - 8)) / (x(x - 8)) ] = 3
480 × 8 = 3(x² - 8x)
1280 = x² - 8x
x² - 8x - 1280 = 0

Exercise 4.2 Solutions

1. Find roots by factorisation:

(i) x² - 3x - 10 = 0
x² - 5x + 2x - 10 = 0
x(x - 5) + 2(x - 5) = 0
(x - 5)(x + 2) = 0
x = 5, -2

(ii) 2x² + x - 6 = 0
2x² + 4x - 3x - 6 = 0
2x(x + 2) - 3(x + 2) = 0
(2x - 3)(x + 2) = 0
x = 3/2, -2

(iii) √2 x² + 7x + 5√2 = 0
√2 x² + 2x + 5x + 5√2 = 0
√2 x(x + √2) + 5(x + √2) = 0
(√2 x + 5)(x + √2) = 0
x = -5/√2, -√2

(iv) 2x² - x + 1/8 = 0
Multiply by 8: 16x² - 8x + 1 = 0
(4x - 1)² = 0
x = 1/4, 1/4

(v) 100x² - 20x + 1 = 0
(10x - 1)² = 0
x = 1/10, 1/10

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

Solution:
Let numbers be x and 27 - x.
x(27 - x) = 182 → x² - 27x + 182 = 0
(x - 13)(x - 14) = 0
The numbers are 13 and 14.

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

Solution:
x² + (x + 1)² = 365
2x² + 2x + 1 = 365 → 2x² + 2x - 364 = 0
x² + x - 182 = 0 → (x + 14)(x - 13) = 0
Since integers are positive, x = 13.
The integers are 13 and 14.

5. Altitude of right triangle is 7 cm less than base. Hypotenuse is 13 cm. Find other sides.

Solution:
Base = x, Altitude = x - 7.
x² + (x - 7)² = 13² = 169
2x² - 14x + 49 - 169 = 0 → 2x² - 14x - 120 = 0
x² - 7x - 60 = 0 → (x - 12)(x + 5) = 0
x = 12 cm.
Base = 12 cm, Altitude = 5 cm.

6. Cottage industry pottery problem:

Solution:
Let number of articles = x. Cost of each = 2x + 3.
Total cost: x(2x + 3) = 90 → 2x² + 3x - 90 = 0
(2x + 15)(x - 6) = 0 → x = 6.
Articles produced = 6, Cost per article = Rs 15.

Exercise 4.3 Solutions

1. Find nature of roots and solve if real:

(i) 2x² - 3x + 5 = 0
D = (-3)² - 4(2)(5) = 9 - 40 = -31 < 0.
No real roots.

(ii) 3x² - 4√3 x + 4 = 0
D = (-4√3)² - 4(3)(4) = 48 - 48 = 0.
Two equal real roots: x = -b/2a = 4√3 / 6 = 2√3/3 (or 2/√3).

(iii) 2x² - 6x + 3 = 0
D = 36 - 4(2)(3) = 12 > 0.
x = (6 ± √12)/4 = (3 ± √3)/2.

2. Find k for equal roots:

(i) 2x² + kx + 3 = 0
D = k² - 4(2)(3) = 0 → k² = 24 → k = ±2√6.

(ii) kx(x - 2) + 6 = 0 → kx² - 2kx + 6 = 0
D = (-2k)² - 4(k)(6) = 0 → 4k² - 24k = 0 → 4k(k - 6) = 0.
Since k ≠ 0, k = 6.

3. Rectangular mango grove (length = 2 × breadth, Area = 800 m²):

x(2x) = 800 → 2x² = 800 → x² = 400 → x = 20 m.
Yes, it is possible. Breadth = 20 m, Length = 40 m.

4. Age situation: Sum = 20, 4 years ago product = 48.

(x - 4)(16 - x) = 48 → 16x - x² - 64 + 4x = 48 → x² - 20x + 112 = 0.
D = 400 - 448 = -48 < 0.
This situation is NOT possible.

5. Rectangular park (Perimeter = 80 m, Area = 400 m²):

2(l + b) = 80 → l + b = 40 → b = 40 - l.
l(40 - l) = 400 → l² - 40l + 400 = 0 → (l - 20)² = 0 → l = 20 m.
Yes, possible. Length = 20 m, Breadth = 20 m (a square).