(i) x² − 3x − 10 = 0

Solution:

The given quadratic equation is x² − 3x − 10 = 0.

Splitting the middle term, we get:

x² − 5x + 2x − 10 = 0

x(x − 5) + 2(x − 5) = 0

(x − 5)(x + 2) = 0

Either x − 5 = 0 or x + 2 = 0

x = 5 or x = −2

Hence, the roots of the quadratic equation are 5 and −2.

(ii) 2x² + x − 6 = 0

Solution:

The given quadratic equation is 2x² + x − 6 = 0.

Splitting the middle term, we get:

2x² + 4x − 3x − 6 = 0

2x(x + 2) − 3(x + 2) = 0

(2x − 3)(x + 2) = 0

Either 2x − 3 = 0 or x + 2 = 0

2x = 3 or x = −2

x = 3/2 or x = −2

Hence, the roots of the quadratic equation are 3/2 and −2.

(iii) √2x² + 7x + 5√2 = 0

Solution:

The given quadratic equation is √2x² + 7x + 5√2 = 0.

Splitting the middle term, we get:

√2x² + 2x + 5x + 5√2 = 0

√2x(x + √2) + 5(x + √2) = 0

(√2x + 5)(x + √2) = 0

Either √2x + 5 = 0 or x + √2 = 0

√2x = −5 or x = −√2

x = −5/√2 or x = −√2

Hence, the roots of the quadratic equation are −5/√2 and −√2.

(iv) 2x² − x + 1/8 = 0

Solution:

The given quadratic equation is 2x² − x + 1/8 = 0.

Multiplying the entire equation by 8 to clear the fraction:

16x² − 8x + 1 = 0

Splitting the middle term, we get:

16x² − 4x − 4x + 1 = 0

4x(4x − 1) − 1(4x − 1) = 0

(4x − 1)(4x − 1) = 0

Either 4x − 1 = 0 or 4x − 1 = 0

4x = 1

x = 1/4

Hence, the roots of the quadratic equation are 1/4 and 1/4.

(v) 100x² − 20x + 1 = 0

Solution:

The given quadratic equation is 100x² − 20x + 1 = 0.

Splitting the middle term, we get:

100x² − 10x − 10x + 1 = 0

10x(10x − 1) − 1(10x − 1) = 0

(10x − 1)(10x − 1) = 0

Either 10x − 1 = 0 or 10x − 1 = 0

10x = 1

x = 1/10

Hence, the roots of the quadratic equation are 1/10 and 1/10.

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

Solution:

Let the first number be x.

Since the sum of the two numbers is 27, the second number will be 27 − x.

According to the given condition, their product is 182:

x(27 − x) = 182

27x − x² = 182

Rearranging the terms into standard quadratic form:

x² − 27x + 182 = 0

Splitting the middle term to factorise the equation:

x² − 13x − 14x + 182 = 0

x(x − 13) − 14(x − 13) = 0

(x − 13)(x − 14) = 0

Either x − 13 = 0 or x − 14 = 0

x = 13 or x = 14

If the first number is 13, the second number is 27 − 13 = 14. If the first number is 14, the second number is 27 − 14 = 13.

Hence, the required two numbers are 13 and 14.

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

Solution:

Let the first positive integer be x. The next consecutive positive integer will be x + 1.

According to the given condition, the sum of their squares is 365:

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x + 1 − 365 = 0

2x² + 2x − 364 = 0

Dividing the entire equation by 2:

x² + x − 182 = 0

Splitting the middle term to factorise:

x² + 14x − 13x − 182 = 0

x(x + 14) − 13(x + 14) = 0

(x − 13)(x + 14) = 0

Either x − 13 = 0 or x + 14 = 0

x = 13 or x = −14

Since the problem specifies positive integers, x cannot be −14. Therefore, x = 13.

The second consecutive integer is 13 + 1 = 14.

Hence, the required 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.

Solution:

Let the base of the right-angled triangle be x cm.

Since the altitude is 7 cm less than its base, the altitude will be (x − 7) cm.

Using Pythagoras theorem (Base² + Altitude² = Hypotenuse²):

x² + (x − 7)² = 13²

x² + x² − 14x + 49 = 169

2x² − 14x + 49 − 169 = 0

2x² − 14x − 120 = 0

Dividing the entire equation by 2:

x² − 7x − 60 = 0

Splitting the middle term to factorise:

x² − 12x + 5x − 60 = 0

x(x − 12) + 5(x − 12) = 0

(x − 12)(x + 5) = 0

Either x − 12 = 0 or x + 5 = 0

x = 12 or x = −5

Since a side length cannot be negative, we reject x = −5. Therefore, the base is 12 cm.

The altitude is 12 − 7 = 5 cm.

Hence, the other two sides of the triangle are 12 cm and 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.

(Note- Important Question for board exam)

Solution:

Let the number of pottery articles produced on that day be x.

According to the given condition, the cost of production of each article is 3 more than twice the number of articles, which is (2x + 3) rupees.

The total cost of production is given by multiplying the number of articles by the cost of each article:

x(2x + 3) = 90

2x² + 3x = 90

2x² + 3x − 90 = 0

Splitting the middle term to factorise:

2x² − 12x + 15x − 90 = 0

2x(x − 6) + 15(x − 6) = 0

(2x + 15)(x − 6) = 0

Either 2x + 15 = 0 or x − 6 = 0

2x = −15 or x = 6

x = −15/2 or x = 6

Since the number of articles produced cannot be a negative value or a fraction, we reject x = −15/2. Therefore, x = 6.

The number of articles produced is 6, and the cost of each article is 2(6) + 3 = 15 rupees.

Hence, the number of articles produced is 6 and the cost of each article is ₹ 15.