Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.2

Question 1.

Find the roots of the following quadratic equations by factorisation:

1. x^{2} – 3x – 10 = 0

2. 2x^{2} + x – 6 = 0

3. \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0

4. 2x^{2} – x + \(\frac{1}{8}\) = 0

5. 100x^{2} – 20x + 1 = 0

Solution:

1. x^{2} – 3x – 10 = 0

∴ x^{2} – 5x + 2x – 10 = 0

∴ x(x – 5) + 2(x – 5) = 0

∴ (x – 5)(x + 2) = 0

Hence, x – 5 = 0 or x + 2 = 0

∴ x = 5 or x = -2

Thus, the roots of the given equation are 5 and -2.

Verification:

For x = 5,

LHS = (5)^{2} – 3(5) – 10

= 25 – 15 – 10

= 0

= RHS

For x = -2,

LHS = (2)^{2} – 3(-2) – 10

= 4 + 6 – 10

= 0

= RHS

Hence, both the roots are verified.

Note that verification is not a part of the solution. It is meant only for your confirmation of receiving correct solution.

2. 2x^{2} + x – 6 = 0

∴ 2x^{2} + 4x – 3x – 6 = 0

∴ 2x (x + 2) – 3(x + 2) = 0

∴ (x + 2) (2x – 3) = 0

∴ x + 2 = 0 or 2x – 3 = 0

∴ x = -2 or x = \(\frac{3}{2}\)

Thus, the roots of the given equation are -2 and \(\frac{3}{2}\)

3. \(\sqrt{2}\)x^{2} + 7x + 5\(\sqrt{2}\) = 0

∴ \(\sqrt{2}\)x^{2} + 2x + 5x + 5\(\sqrt{2}\) = 0

∴ \(\sqrt{2}\)x(x + \(\sqrt{2}\)) + 5(x + \(\sqrt{2}\)) = 0

∴ (x + \(\sqrt{2}\))(\(\sqrt{2}\)x + 5) = 0

∴ x + \(\sqrt{2}\) = 0 or \(\sqrt{2}\)x + 5 = 0

∴ x = –\(\sqrt{2}\) or x = \(-\frac{5}{\sqrt{2}}\)

Thus, the roots of the given equation are –\(\sqrt{2}\) and \(-\frac{5}{\sqrt{2}}\)

4. 2x^{2} – x + \(\frac{1}{8}\) = 0

∴ 16x^{2} – 8x + 1 = 0 (Multiplying by 8)

∴ 16x^{2} – 4x – 4x + 1 = 0

∴ 4x(4x – 1) -1 (4x – 1) = 0

∴ (4x – 1) (4x – 1) = 0

∴ 4x – 1 = 0 or 4x – 1 = 0

∴ x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)

Thus, the repeated roots of the given equation are \(\frac{1}{4}\) and \(\frac{1}{4}\)

5. 100x^{2} – 20x + 1 = 0

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1) -1 (10x – 1) = 0

(10x – 1)(10x – 1) = 0

10x – 1 = 0 or 10x – 1 = 0

x = \(\frac{1}{10}\) or x = \(\frac{1}{10}\)

Thus, the repeated roots of the given equation are \(\frac{1}{10}\) and \(\frac{1}{10}\).

Question 2.

Solve the problems given in Textual Examples:

1. 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. Find out how many marbles they had to start with.

2. 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. Find out the number of toys. produced on that day.

Solution:

1. Let the number of marbles that John had be x.

Then, the number of marbles that Jivanti had is (45 – x).

After losing 5 marbles, the number of marbles left with John = x – 5.

After losing 5 marbles, the number of marbles left with Jivanti = (45 – x) – 5 = 40 – x.

Therefore, the product of marbles with them is (x – 5) (40 – x), which is given to be 124.

Hence, we get the following equation:

(x – 5)(40 – x) = 124.

∴ 40x – x^{2} – 200 + 5x = 124

∴ -x^{2} + 45x – 324 = 0

∴ x^{2} – 45x + 324 = 0

∴ x^{2} – 36x – 9x + 324 = 0

∴ x(x – 36) – 9(x – 36) = 0

∴ (x – 36)(x – 9) = 0

∴ x – 36 = 0 or x – 9 = 0

∴ x = 36 or x = 9

Here, both the answers are admissible.

∴ 45 – x = 45 – 36 = 9 or

45 – x = 45 – 9 = 36

Thus, the number of marbles with John and Jivanti to start with are 36 and 9 respectively or 9 and 36 respectively.

2. Let the number of toys produced on that day be x.

Therefore, the cost of production (in rupees) of each toy on that day = 55 – x.

So, the total cost of production (in rupees) on that day = x (55 – x).

Hence, x(55 – x) = 750

∴ 55x – x^{2} – 750 = 0

∴ x^{2} – 55x + 750 = 0

∴ x^{2} – 30x – 25x + 750 = 0

∴ x(x – 30) – 25(x – 30) = 0

∴ (x – 30)(x – 25) = 0

∴ x – 30 = 0 or x – 25 = 0

∴ x = 30 or x = 25

Here, both the answers are admissible.

Hence, the number of toys produced on that day is 30 or 25.

Question 3.

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

Solution:

Let the first of the two numbers whose sum is 27 be x.

Then, the second number is 27 – x and the product of those two numbers is x (27 – x).

Their product is given to be 182.

∴ x (27 – x) = 182

∴ 27x – x^{2} – 182 = 0

∴ x^{2} – 27x + 182 = 0

∴ x^{2} – 14x – 13x + 182 = 0

∴ x(x – 14) – 13(x – 14) = 0

∴ (x – 14)(x – 13) = 0

∴ x – 14 = 0 or x – 13 = 0

∴ x = 14 or x = 13

Here, both the answers are admissible.

Hence, if x = 14, it gives that the first number = x = 14 and the second number = 27 – x = 27 – 14 = 13.

And if x = 13, it gives that the first number = x = 13 and the second number = 27 – x = 27 – 13 = 14.

Thus, in either case, the required numbers are 13 and 14.

Question 4.

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

Solution:

Let two consecutive positive integers be x and x + 1.

Then, the sum of their squares = (x)^{2} + (x + 1)^{2}

= x^{2} + x^{2} + 2x + 1

= 2x^{2} + 2x + 1

This sum is given to be 365.

∴ 2x^{2} + 2x + 1 = 365

∴ 2x^{2} + 2x – 364 = 0

∴ x^{2} + x – 182 = 0

∴ x^{2} + 14x – 13x – 182 = 0

∴ x(x + 14) – 13(x + 14) = 0

∴ (x + 14)( x- 13) = 0

∴ x + 14 = 0 or x – 13 = 0

∴ x = -14 or x = 13

Since x is a positive integer, x = -14 is inadmissible.

∴ x = 13 and x + 1 = 13 + 1 = 14

Thus, the required consecutive positive integers are 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 the base of the right triangle be x cm.

Then, its altitude is (x – 7) cm.

The hypotenuse of the right triangle is given to be 13 cm.

Now, by Pythagoras theorem.

(Base)^{2} + (Altitude)^{2} = (Hypotenuse)^{2}

∴ (x)^{2} + (x – 7)^{2} = (13)^{2}

∴ x^{2} + x^{2} – 14x + 49 = 169

∴ 2x^{2} – 14x – 120 = 0

∴ x^{2} – 7x – 60 = 0

∴ x^{2} – 12x + 5x – 60 = 0

∴ x(x – 12) + 5(x – 12) = 0

∴ (x – 12)(x + 5) = 0

∴ x – 12 = 0 or x + 5 = 0

∴ x = 12 or x = -5

As the base of a triangle cannot be negative. x = -5 is inadmissible.

Hence, x = 12.

Then, the base of the triangle = x = 12 cm and the altitude of the triangle = x – 7 = 12 – 7

= 5 cm.

Thus, the base and the altitude of the given triangle are 12 cm and 5 cm respectively.

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

Then, according to the given, the cost of production (in rupees) of each article = 2x + 3.

Hence, total cost of production (in rupees) on that day = x(2x + 3) = 2x^{2} + 3x.

This total cost of production is given to be ₹ 90.

∴ 2x^{2} + 3x = 90

∴ 2x^{2} + 3x – 90 = 0

∴ 2x^{2} – 12x + 15x – 90 = 0

∴ 2x(x – 6) + 15(x – 6) = 0

∴ (x – 6) (2x + 15) = 0

∴ x – 6 = 0 or 2x + 15 = 0

∴ x = 6 or x = –\(\frac{15}{2}\)

Here, x = –\(\frac{15}{2}\) is inadmissible as x represents the number of articles produced.

Hence, x = 6 and 2x + 3 = 2(6) + 3 = 15.

Thus, the number of pottery articles produced on that day is 6 and the cost of production of each article is ₹ 15.