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# Class 10 NCERT Solutions- Chapter 4 Quadratic Equations – Exercise 4.2

### (i) x2– 3x – 10 = 0

Solution:

Here, LHS = x2– 3x – 10

= x2 – 5x + 2x – 10

= x(x – 5) + 2(x – 5)

= (x – 5)(x + 2)

The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which

(x – 5)(x + 2) = 0

Hence, x – 5 = 0 or x + 2 = 0

x = 5 or x = -2

### (ii) 2x2 + x – 6 = 0

Solution:

Here, LHS = 2x2 + x – 6

= 2x2 + 4x – 3x – 6

= 2x(x + 2) – 3(x + 2)

= (2x– 3)(x + 2)

The roots of this equation, 2x2 + x – 6 = 0 are the values of x for which

(2x– 3)(x + 2) = 0

Hence, 2x– 3 = 0 or x + 2 = 0

⇒ x = 3/2 or x = –2

### (iii) √2x2 + 7x + 5√2 = 0

Solution:

Here, LHS = √2x2 + 7x + 5√2

= √2x2 + 5x + 2x + 5√2

= x(√2x + 5) + √2(√2x + 5)

= (√2x + 5) (x +√2)

The roots of this equation, √2x2 + 7x + 5√2 = 0 are the values of x for which

(√2x + 5) (x +√2) = 0

Hence, √2x + 5 = 0 or x +√2 = 0

x = –5/√2 or x = –√2

### (iv) 2x2 – x + 1/8 = 0

Solution:

Here, LHS = 2x2 – x + 1/8

= 1/8(16x2 – 8x + 1)

= 1/8(16x2 – 4x -4x + 1)

= 1/8(4x(4x-1) -1 (4x-1))

= 1/8 (4x-1) (4x-1)

The roots of this equation, 2x2 – x + 1/8 = 0 are the values of x for which

1/8 (4x-1) (4x-1)  = 0

(4x-1)2 = 0

Hence, 4x-1 = 0 or 4x-1 = 0

⇒ x = 1/4 or x = 1/4

### (v) 100x2 – 20x + 1 = 0

Solution:

Here, LHS = 100x2 – 20x + 1

= 100x2 – 10x – 10x + 1

= 10x(10x – 1) – 1(10x – 1)

= (10x – 1) (10x – 1)

The roots of this equation, 100x2 – 20x + 1 = 0 are the values of x for which

(10x – 1) (10x – 1) = 0

(10x – 1)2 = 0

Hence, 10x – 1 = 0 or 10x – 1 = 0

x = 1/10 or x = 1/10

### (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.

Solution:

Let’s say,

The number of marbles John have = x.

So, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

Number of marble Jivanti have = 45 – x – 5 = 40 – x

Here, According to the given condition

(x – 5)(40 – x) = 124

x2 – 45x + 324 = 0

x2 – 36x – 9x + 324 = 0

x(x – 36) -9(x – 36) = 0

(x – 36)(x – 9) = 0

Hence, x – 36 = 0 or x – 9 = 0

x = 36 or x = 9

Therefore,

If, John’s marbles = 36, then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9, then, Jivanti’s marbles = 45 – 9 = 36

### (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.

Solution:

Let us say,

Number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 – x)

Given, total cost of production of the toys = Rs 750

So, x(55 – x) = 750

x2 – 55x + 750 = 0

x2 – 25x – 30x + 750 = 0

x(x – 25) -30(x – 25) = 0

(x – 25)(x – 30) = 0

Hence, x – 25 = 0 or x – 30 = 0

x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

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

Solution:

Let’s say,

First number be x and the second number is 27 – x.

Therefore, the product of two numbers will be:

x(27 – x) = 182

x2 – 27x – 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) -14(x – 13) = 0

(x – 13)(x -14) = 0

Hence, x – 13 = 0 or x – 14= 0

x = 13 or x = 14

Hence, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

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

Solution:

Let’s say,

Two consecutive positive integers be x and x + 1.

Here, According to the given condition,

x2 + (x + 1)2 = 365

x2 + x2 + 1 + 2x = 365

2x2 + 2x – 364 = 0

x2 + x – 182 = 0

x2 + 14x – 13x – 182 = 0

x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Hence, x – 13 = 0 or x + 14= 0

x = 13 or x = – 14

As, here it is said positive integers, so x can be 13, only.

So,

x = 13

and, x + 1 = 13 + 1 = 14

Hence, two consecutive positive integers will be 13 and 14.

### Question 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’s say,

Base of the right triangle be x cm.

So, the altitude of right triangle = (x – 7) cm

Base2 + Altitude2 = Hypotenuse2                       (Pythagoras theorem)

x2 + (x – 7)2 = 132

x2 + x2 + 49 – 14x = 169    (using identity (a-b)2 = a2 – 2ab + b2)

2x2 – 14x – 120 = 0

x2 – 7x – 60 = 0               (Dividing by 2)

x2 – 12x + 5x – 60 = 0

x(x – 12) + 5(x – 12) = 0

(x – 12)(x + 5) = 0

Hence, x – 12 = 0 or x + 5= 0

x = 12 or x = – 5

As, here side will be a  positive integers, so x can be 12, only.

Therefore, the base of the given triangle is 12 cm and,

the altitude of this triangle will be (12 – 7) cm = 5 cm.

### Question 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.

Solution:

Let’s say,

Number of articles produced be x.

So, cost of production of each article = ₹ (2x + 3)

Here, According to the given condition

x(2x + 3) = 90

2x2 + 3x – 90 = 0

2x2 + 15x -12x – 90 = 0

x(2x + 15) -6(2x + 15) = 0

(2x + 15)(x – 6) = 0

Hence, 2x +15 = 0 or x – 6= 0

x = –15/2 or x = 6

As the number of articles produced can only be a positive integer,

So, x = 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = ₹ 15.

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