Quadratic Equations ⇒⇒ Exercise 4.2

Question 1 (i)

Find the roots of the following quadratic equations by factorisation:
(i) $ x ^2 – 3x – 10 = 0 $

Solution :

The given equation is: $ x ^2 – 3x – 10 = 0 $

By expanding middle term. Here -3x can be written as −5x + 2x

$$ ⇒{ x^2 − 5x + 2x – 10 = 0} $$

By taking common,

$$ ⇒{ x( x − 5) + 2( x − 5) = 0} $$

$$ ⇒{ ( x − 5) ( x + 2) = 0} $$

Set each factor equal to zero to find the roots of the equation:

x − 5 = 0 ⇒ x = 5

x + 2 = 0 ⇒ x = – 2

The roots of the equation $ x ^2 – 3x – 10 = 0 $ are ( x = 5 and x = – 2)

Question 1 (ii)

Find the roots of the following quadratic equations by factorisation:
(ii) $ 2x^2 + x – 6 = 0 $

Solution :

The given equation is: $ 2x^2 + x – 6 = 0 $

By expanding middle term. Here x can be written as 4x -3x

$$ ⇒{ 2x^2 + 4x -3x– 6 = 0} $$

By taking common,

$$ ⇒{ 2x( x + 2) -3( x + 2) = 0} $$

$$ ⇒{ ( 2x − 3) ( x + 2) = 0} $$

Set each factor equal to zero to find the roots of the equation:

2x − 3 = 0 ⇒ x = $ {3 \over 2} $

x + 2 = 0 ⇒ x = – 2

The roots of the equation $ 2x^2 + x – 6 = 0 $ are ( x = $ {3 \over 2} $ and x = – 2)

Question 1 (iii)

Find the roots of the following quadratic equations by factorisation:
(iii) $ {\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 } $

Solution :

The given equation is: $ {\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 } $

By expanding middle term. Here 7x can be written as 5x + 2x

$$ ⇒{\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 } $$

By taking common,

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

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

Set each factor equal to zero to find the roots of the equation:

$ ( x + \sqrt{2}) =0 $ ⇒⇒ $ ({ x = - \sqrt{2}})$

$ (\sqrt{2}x + 5) = 0 $ ⇒⇒ $( x ={ - 5 \over \sqrt{2} }) $

The roots of the equation $ {\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 } $ are ( $ { x = - \sqrt{2}}$ and $ x ={ - 5 \over \sqrt{2} } $ )

Question 1 (iv)

Find the roots of the following quadratic equations by factorisation:
(iv) $ {2x^2 - x +{ 1 \over 8} = 0 } $

Solution :

$$ {2x^2 - x +{ 1 \over 8} = 0 } $$

$$ ⇒ {{16x^2 - 8x + 1 } \over 8 } = 0 $$

$$ ⇒ {16x^2 - 8x + 1 } = 0 $$

By expanding middle term. Here - 8x can be written as -4x -4x

$$ ⇒ {16x^2 -4x -4x + 1 } = 0 $$

By taking common,

$$ ⇒ {4x (4x -1 )- 1 (4x -1 ) } = 0 $$

$$ ⇒{ (4x -1 ) (4x -1 ) = 0} $$

Set each factor equal to zero to find the roots of the equation:

4x -1 = 0 ⇒ $ x = { 1 \over 4} $

4x -1 = 0 ⇒ $ x = { 1 \over 4} $

The roots of the equation $ {2x^2 - x +{ 1 \over 8} = 0 } $ are $ x = { 1 \over 4} $ and $ x = { 1 \over 4} $

Question 1 (v)

Find the roots of the following quadratic equations by factorisation:
(v) $ {100x^2 - 20x + 1 = 0 } $

Solution :

$$ {100x^2 - 20x + 1 = 0 } $$

By expanding middle term. Here - 20x can be written as -10x -10x

$$ ⇒ {100x^2 -10x -10x + 1 = 0 } $$

By taking common,

$$ ⇒ {10x (10x - 1 )- 1 (10x - 1 ) } = 0 $$

$$ ⇒{ (10x - 1 ) (10x - 1 ) = 0} $$

Set each factor equal to zero to find the roots of the equation:

10x - 1 = 0 ⇒ $ x = { 1 \over 10} $

10x - 1 = 0 ⇒ $ x = { 1 \over 10} $

The roots of the equation $ {100x^2 - 20x + 1 = 0 } $ are $ x = { 1 \over 10} $ and $ x = { 1 \over 10} $

Question 2 (i)

(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 the number of marbles John had be x.

Given, Number of marbles John and Jivanti together have = 45

The number of marbles Jivanti had will be (Total marbles MINUS the Number of marbles John had) = 45 − x

Thus, after losing of 5 marbles by each of them, Number of total marbles left

Now, let after losing total number of marbles John have =(x – 5)

∴ After losting number of marbles Jivanti have = ( 45 – x – 5) = (40 – x).

Now, given, after lost the product of marbles both of them have = 124

$${ (x−5)(40−x)=124 }$$

$$ ⇒ { 40x - x^2 - 200 + 5x = 124 }$$

$$ ⇒ { - x^2 + 45x - 200 = 124 }$$

$$ ⇒ { - x^2 + 45x - 200 -124 = 0 }$$

$$ ⇒ { - x^2 + 45x - 324 = 0 }$$

$$ ⇒ { x^2 - 45x + 324 = 0 }$$

By expanding middle term. Here - 45x can be written as -36x -9x

$$ ⇒ { x^2 -36x -9x + 324 = 0 }$$

By taking common,

$$ ⇒ { x(x -36) -9(x−36) = 0 }$$

$$ ⇒ { (x -36) (x−9) = 0 }$$

Set each factor equal to zero to find the roots of the equation:

x -36 = 0 ⇒ x = 36

x − 9 = 0 ⇒ 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.

Question 2 (ii)

(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 the number of toys produced on a particular day be x

Given, Cost of production of each toy = Rs (55 − x)

Given, On a particular day total cost of production = Rs 750

Total cost of production = Cost of each toy × Total number of toys

$${ (55 − x ) × (x) = 750}$$

$$ ⇒ { 55x - x^2 = 750 }$$

$$ ⇒ { 55x - x^2 - 750 = 0 }$$

$$ ⇒ { x^2 - 55x + 750 = 0 }$$

By expanding middle term. Here - 55x can be written as -25x -30x

$$ ⇒ { x^2 -25x -30x + 750 = 0 }$$

By taking common,

$$ ⇒ { x(x -25) -30(x -25) = 0 }$$

$$ ⇒ { (x -30) (x -25) = 0 }$$

Set each factor equal to zero to find the roots of the equation:

x -30 = 0 ⇒ x = 30

x − 25 = 0 ⇒ x = 25

This means the number of toys produced on that day could be either 25 or 30.

Question 3

Find two numbers whose sum is 27 and product is 182

Solution :

Let one of the numbers be x

Given, sum of two numbers is equal to 27

∴ second number = 27 − x

Given, It is given that the product of these numbers is 182

Therefore, their product = x × (27 − x)

$$ {x (27 − x) = 182 }$$

$$ ⇒ { 27x - x^2 = 182 }$$

$$ ⇒ { 27x - x^2 - 182 = 0 }$$

$$ ⇒ { x^2 - 27x + 182 = 0 }$$

By expanding middle term. Here - 27x can be written as -13x -14x

$$ ⇒ { x^2 -13x -14x + 182 = 0 }$$

By taking common,

$$ ⇒ { x(x -13) -14(x -13) = 0 }$$

$$ ⇒ { (x -13) (x -14) = 0 }$$

Set each factor equal to zero to find the roots of the equation:

x - 13 = 0 ⇒ x = 13

x − 14 = 0 ⇒ x = 14

Therefore, 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 the first integer be x.

The next consecutive positive integer will be x + 1 .

Now, as given, Sum of their squares = 365

$$ {x^2 ×(x + 1)^2 = 365 }$$

$$ ⇒ {x^2 +(x^2 + 1^2 + 2x) = 365 }$$

$$ ⇒ {2x^2 + 1 + 2x = 365 }$$

$$ ⇒ {2x^2 + 1 + 2x - 365 =0 }$$

$$ ⇒ {2x^2 + 2x - 364 =0 }$$

Divide by 2

$$ ⇒ {x^2 + x - 182 =0 }$$

By expanding middle term. Here x can be written as 14x – 13x

$$ ⇒ { x^2 + 14x – 13x - 182 = 0 }$$

By taking common,

$$ ⇒ { x(x + 14) -13(x + 14) = 0 }$$

$$ ⇒ { (x -13) (x +14) = 0 }$$

Set each factor equal to zero to find the roots of the equation:

x - 13 = 0 ⇒ x = 13

x + 14 = 0 ⇒ x = -14

Since, the integers are positive, so x can be 13, only .Therefore, 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 :

In a right triangle,

Let the base be x cm

The altitude will be (x − 7)cm

Now, we can apply the Pythagoras theorem to the given right triangle.

Given, hypotenuse = 13cm

We know, $ hypotenuse^2 = altitude^2 + base^2 $

$$ ⇒ 13^2 = {(x-7)^2 +(x^2) }$$

$$ ⇒ 169 = {(x^2 + 7^2 -2×7 ×x) +(x^2) }$$

$$ ⇒ 169 = {(x^2 + 49 -14x) +(x^2) }$$

$$ ⇒ 169 = {2x^2 -14x + 49 }$$

$$ ⇒ {2x^2 -14x + 49 - 169 = 0 }$$

$$ ⇒ {2x^2 -14x - 120 = 0 }$$

Divide by 2

$$ ⇒ {x^2 - 7x - 60 = 0 }$$

By expanding middle term. Here - 7x can be written as 5x – 12x

$$ ⇒ { x^2 + 5x – 12x - 60 = 0 }$$

By taking common,

$$ ⇒ { x(x + 5) -12(x + 5) = 0 }$$

$$ ⇒ { (x -12) (x +5) = 0 }$$

Set each factor equal to zero to find the roots of the equation:

x - 12 = 0 ⇒ x = 12

x + 5 = 0 ⇒ x = -5

Value of x cannot be negative (This is not possible).

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 Rs 90, find the number of articles produced and the cost of each article

Solution :

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

∴ Cost of production of each article on that day = Rs (2x + 3)

Given, On a particular day total cost of production = Rs 90

Total cost of production = Cost of each article × Total number of article

$${ (2x + 3 )×(x) = 90}$$

$$ ⇒ { 2x^2 + 3x = 90 }$$

$$ ⇒ {2x^2 + 3x - 90 = 0 }$$

By expanding middle term. Here 3x can be written as 15x -12x

$$ ⇒ {2x^2 + 15x -12x - 90 = 0 }$$

By taking common,

$$ ⇒ { x(2x +15) -6(2x +15) = 0 }$$

$$ ⇒ { (x -6) (2x +15) = 0 }$$

Set each factor equal to zero to find the roots of the equation:

x - 6 = 0 ⇒ x = 6

2x + 15 = 0 ⇒ x = $ { - 15 \over 2} $

Value of x cannot be negative (This is not possible).

Thus, number of articles proudced on the given day = 6

And cost of each article = Rs 15