CBSE NCERT Solutions for Class 10 Mathematics Chapter 4€¦ · Thus, the given equation is a quadratic equation as 𝑎𝑎≠0. 2. Represent the following situations in the form

Class- X-CBSE-Mathematics Quardatic Equations

Practice more on Quardatic Equations Page - 1 www.embibe.com

CBSE NCERT Solutions for Class 10 Mathematics Chapter 4 Back of Chapter Questions

1. Check whether the following are quadratic equations:

(i) (𝑥𝑥 + 1)2 = 2(𝑥𝑥 − 3)

(ii) 𝑥𝑥2 − 2𝑥𝑥 = (−2)(3− 𝑥𝑥)

(iii) (𝑥𝑥 − 2)(𝑥𝑥 + 1) = (𝑥𝑥 − 1)(𝑥𝑥 + 3)

(iv) (𝑥𝑥 − 3)(2𝑥𝑥 + 1) = 𝑥𝑥(𝑥𝑥 + 5)

(v) (2𝑥𝑥 − 1)(𝑥𝑥 − 3) = (𝑥𝑥 + 5)(𝑥𝑥 − 1)

(vi) 𝑥𝑥2 + 3𝑥𝑥 + 1 = (𝑥𝑥 − 2)2

(vii) (𝑥𝑥 + 2)3 = 2𝑥𝑥(𝑥𝑥2 − 1)

(viii) 𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 1 = (𝑥𝑥 − 2)3

Solution:

(i) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called aquadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: (𝑥𝑥 + 1)2 = 2(𝑥𝑥 − 3)

Using the formula (𝑎𝑎 + 𝑏𝑏)2 = 𝑎𝑎2 + 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2

⇒ 𝑥𝑥2 + 2𝑥𝑥 + 1 = 2𝑥𝑥 − 6

⇒ 𝑥𝑥2 + 7 = 0

Here, 𝑎𝑎 = 1, 𝑏𝑏 = 0 and 𝑐𝑐 = 7.

Thus, the given equation is a quadratic equation as 𝑎𝑎 ≠ 0.

(ii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called aquadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: 𝑥𝑥2 − 2𝑥𝑥 = (−2)(3− 𝑥𝑥)

⇒ 𝑥𝑥2 − 2𝑥𝑥 = −6 + 2𝑥𝑥

⇒ 𝑥𝑥2 − 4𝑥𝑥 + 6 = 0

Here, 𝑎𝑎 = 1, 𝑏𝑏 = −4 and 𝑐𝑐 = 6.


(iii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called aquadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

https://www.embibe.com/foundation-10/test/foundation-10/chapterwise-test/mathematics/quadratic-equations/paper?utm_source=Exams&utm_medium=internal&utm_campaign=Exams_to_PTL

http://www.embibe.com/



Given equation: (𝑥𝑥 − 2)(𝑥𝑥 + 1) = (𝑥𝑥 − 1)(𝑥𝑥 + 3)

⇒ 𝑥𝑥2 − 𝑥𝑥 − 2 = 𝑥𝑥2 + 2𝑥𝑥 − 3

⇒ 3𝑥𝑥 − 1 = 0

But, here 𝑎𝑎 = 0.

So, the given equation is not a quadratic equation.

(iv) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called a quadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: (𝑥𝑥 − 3)(2𝑥𝑥 + 1) = 𝑥𝑥(𝑥𝑥 + 5)

⇒ 2𝑥𝑥2 − 5𝑥𝑥 − 3 = 𝑥𝑥2 + 5𝑥𝑥

⇒ 𝑥𝑥2 − 10𝑥𝑥 − 3 = 0

Here, 𝑎𝑎 = 1, 𝑏𝑏 = −10 and 𝑐𝑐 = −3.


(v) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called a quadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: (2𝑥𝑥 − 1)(𝑥𝑥 − 3) = (𝑥𝑥 + 5)(𝑥𝑥 − 1)

⇒ 2𝑥𝑥2 − 7𝑥𝑥 + 3 = 𝑥𝑥2 + 4𝑥𝑥 − 5

⇒ 𝑥𝑥2 − 11𝑥𝑥 + 8 = 0

Here, 𝑎𝑎 = 1, 𝑏𝑏 = −11 and 𝑐𝑐 = 8.


(vi) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called a quadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: 𝑥𝑥2 + 3𝑥𝑥 + 1 = (𝑥𝑥 − 2)2

Using the formula (𝑎𝑎 − 𝑏𝑏)2 = 𝑎𝑎2 − 2𝑎𝑎𝑏𝑏 + 𝑏𝑏2

⇒ 𝑥𝑥2 + 3𝑥𝑥 + 1 = 𝑥𝑥2 − 4𝑥𝑥 + 4

⇒ 7𝑥𝑥 − 3 = 0

But, here 𝑎𝑎 = 0.


(vii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called a quadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: (𝑥𝑥 + 2)3 = 2𝑥𝑥(𝑥𝑥2 − 1)

Using the formula (𝑎𝑎 + 𝑏𝑏)3 = 𝑎𝑎3 + 𝑏𝑏3 + 3𝑎𝑎2𝑏𝑏 + 3𝑎𝑎𝑏𝑏2





⇒ 𝑥𝑥3 + 8 + 6𝑥𝑥2 + 12𝑥𝑥 = 2𝑥𝑥3 − 2𝑥𝑥

⇒ 𝑥𝑥3 − 14𝑥𝑥 − 6𝑥𝑥2 − 8 = 0

This equation is not of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0


(viii) We know that any equation of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 is called a quadratic equation, where 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 are real numbers and 𝑎𝑎 ≠ 0.

Given equation: 𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 1 = (𝑥𝑥 − 2)3

Using the formula (𝑎𝑎 − 𝑏𝑏)3 = 𝑎𝑎3 − 3𝑎𝑎2𝑏𝑏 + 3𝑎𝑎𝑏𝑏2 − 𝑏𝑏3

⇒ 𝑥𝑥3 − 4𝑥𝑥2 − 𝑥𝑥 + 1 = 𝑥𝑥3 − 8 − 6𝑥𝑥2 + 12𝑥𝑥

⇒ 2𝑥𝑥2 − 13𝑥𝑥 + 9 = 0

Here, 𝑎𝑎 = 2, 𝑏𝑏 = −13 and 𝑐𝑐 = 9.


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

(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km h⁄ less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:

(i) Let the breadth of the plot be 𝑥𝑥 m.

Hence, the length of the plot is (2𝑥𝑥 + 1) m. (Since, given that length is one more than twice its breadth)

Therefore, area of a rectangle = length × breadth

Given: area of rectangle = 528 m2

∴ 528 = 𝑥𝑥(2𝑥𝑥 + 1)

⇒ 2𝑥𝑥2 + 𝑥𝑥 − 528 = 0 ………(i), which is of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0





Here 𝑎𝑎 = 2(≠ 0), 𝑏𝑏 = 1 and 𝑐𝑐 = −528

Thus, quadratic equation (i) represents the situation given in the question and roots of this equation will represent the breadth of the plot.

(ii) We know that the difference between two consecutive positive integers is 1.

So, let the consecutive positive integers be 𝑥𝑥 and 𝑥𝑥 + 1.

Given that their product is 306.

∴ 𝑥𝑥(𝑥𝑥 + 1) = 306

⇒ 𝑥𝑥2 + 𝑥𝑥 − 306 = 0 ……….. (i), which is of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

Here 𝑎𝑎 = 1(≠ 0),𝑏𝑏 = 1 and 𝑐𝑐 = −306

Thus, quadratic equation (i) represents the situation given in the question and roots of this equation will represent the smaller positive integer.

(iii) Let Rohan's age be 𝑥𝑥,

His mother's age = 𝑥𝑥 + 26 (given that Rohan’s mother is 26 years older than him)

3 years from now:

Rohan's age will be = 𝑥𝑥 + 3

Mother's age will be = 𝑥𝑥 + 26 + 3 = 𝑥𝑥 + 29

Also given that the product of their ages after 3 years is 360.

∴ (𝑥𝑥 + 3)(𝑥𝑥 + 29) = 360

On simplification, we get

𝑥𝑥2 + 32𝑥𝑥 − 273 = 0……… (i), which is of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

Here 𝑎𝑎 = 1(≠ 0),𝑏𝑏 = 32 and 𝑐𝑐 = −273

Thus, quadratic equation (i) represents the situation given in the question and positive root of this equation will represent the Rohan’s present age.

(iv) In first case,

Let the speed of train be 𝑥𝑥 km/h.

Total time taken to travel 480 km = 480𝑥𝑥

hrs

In second case,

Given: speed became 8 km/h less

So, the speed of train = (𝑥𝑥 − 8)km/h





Also given that the train will take 3 more hours to cover the same distance.

Therefore, time take to travel 480 km = �480𝑥𝑥

+ 3�hrs

Speed × Time = Distance

(𝑥𝑥 − 8) �480𝑥𝑥 + 3� = 480

⇒ 480 + 3𝑥𝑥 −3840𝑥𝑥 − 24 = 480

⇒ 3𝑥𝑥 −3840𝑥𝑥 = 24

⇒ 3𝑥𝑥2 − 24𝑥𝑥 − 3840 = 0

⇒ 𝑥𝑥2 − 8𝑥𝑥 − 1280 = 0…….. (i), which is of the form 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

Here 𝑎𝑎 = 1(≠ 0),𝑏𝑏 = −8 and 𝑐𝑐 = −1280

Thus, quadratic equation (i) represents the situation given in the question and positive root of this equation will represent the speed of train.

EXERCISE 4.2

1. Find the roots of the following quadratic equations by factorisation:

(i) 𝑥𝑥2 − 3𝑥𝑥 − 10 = 0

(ii) 2𝑥𝑥2 + 𝑥𝑥 − 6 = 0

(iii) √2 𝑥𝑥2 + 7𝑥𝑥 + 5√2 = 0

(iv) 2𝑥𝑥2 − 𝑥𝑥 + 18

= 0

(v) 100𝑥𝑥2 − 20𝑥𝑥 + 1 = 0

Solution:

(i) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 𝑥𝑥2 − 3𝑥𝑥 − 10. The given quadratic expression can be written as follows:

𝑥𝑥2 − 3𝑥𝑥 − 10

= 𝑥𝑥2 − 5𝑥𝑥 + 2𝑥𝑥 − 10 (we factorise by method of splitting the middle term)

= 𝑥𝑥(𝑥𝑥 − 5) + 2(𝑥𝑥 − 5)

= (𝑥𝑥 − 5)(𝑥𝑥 + 2)





Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (𝑥𝑥 − 5)(𝑥𝑥 + 2) = 0

∴ 𝑥𝑥 − 5 = 0 or 𝑥𝑥 + 2 = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = 5 or 𝑥𝑥 = −2

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

(ii) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 2𝑥𝑥2 + 𝑥𝑥 − 6 . The given quadratic expression can be written as follows:

2𝑥𝑥2 + 𝑥𝑥 − 6

= 2𝑥𝑥2 + 4𝑥𝑥 − 3𝑥𝑥 − 6 (we factorise by method of splitting the middle term)

= 2𝑥𝑥(𝑥𝑥 + 2)− 3(𝑥𝑥 + 2)

= (𝑥𝑥 + 2)(2𝑥𝑥 − 3)

Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (𝑥𝑥 + 2)(2𝑥𝑥 − 3) = 0

∴ 𝑥𝑥 + 2 = 0 or 2𝑥𝑥 − 3 = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = −2 or 𝑥𝑥 = 32

Hence, the roots of this quadratic equation are −2 and 32.

(iii) To find the roots of given quadratic equation, lets first factorise the given quadratic expression √2𝑥𝑥2 + 7𝑥𝑥 + 5√2 . The given quadratic expression can be written as follows:

√2𝑥𝑥2 + 7𝑥𝑥 + 5√2

= √2𝑥𝑥2 + 5𝑥𝑥 + 2𝑥𝑥 + 5√2 (we factorise by method of splitting the middle term)

= 𝑥𝑥�√2𝑥𝑥 + 5� + √2�√2𝑥𝑥 + 5�

= �√2𝑥𝑥 + 5��𝑥𝑥 + √2�

Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which

(√2𝑥𝑥 + 5)�𝑥𝑥 + √2� = 0

∴ √2𝑥𝑥 + 5 = 0 or 𝑥𝑥 + √2 = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = − 5√2

or 𝑥𝑥 = −√2





Hence, the roots of this quadratic equation are − 5√2

and −√2.

(iv) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 2𝑥𝑥2 − 𝑥𝑥 + 1

8 . The given quadratic expression can be

written as follows:

2𝑥𝑥2 − 𝑥𝑥 +18

=18

(16𝑥𝑥2 − 8𝑥𝑥 + 1)

= 18

(16𝑥𝑥2 − 4𝑥𝑥 − 4𝑥𝑥 + 1) (we factorise by method of splitting the middle term)

=18�4𝑥𝑥(4𝑥𝑥 − 1)− 1(4𝑥𝑥 − 1)�

=18

(4𝑥𝑥 − 1)2

Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (4𝑥𝑥 − 1)2 = 0

Thus, (4𝑥𝑥 − 1) = 0 or (4𝑥𝑥 − 1) = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = 14 or 𝑥𝑥 = 1

4

Hence, the roots of this quadratic equation are 14 and 1

4.

(v) To find the roots of given quadratic equation, lets first factorise the given quadratic expression 100𝑥𝑥2 − 20𝑥𝑥 + 1. The given quadratic expression can be written as follows:

100𝑥𝑥2 − 20𝑥𝑥 + 1

= 100𝑥𝑥2 − 10𝑥𝑥 − 10𝑥𝑥 + 1 (we factorise by method of splitting the middle term)

= 10𝑥𝑥(10𝑥𝑥 − 1)− 1(10𝑥𝑥 − 1)

= (10𝑥𝑥 − 1)2

Now, the roots of this quadratic equation are the values of 𝑥𝑥 for which (10𝑥𝑥 − 1)2 = 0

Thus, (10𝑥𝑥 − 1) = 0 or (10𝑥𝑥 − 1) = 0

i. e., 𝑥𝑥 = 110

or 𝑥𝑥 = 110

Hence, the roots of this quadratic equation are 110

and 110

.





2. Solve the problems given below.

Represent the following situations mathematically:

(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.

(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:

(i) Let the number of John's marbles be 𝑥𝑥.

So, the number of Jivanti's marbles = 45− 𝑥𝑥

If both lost 5 marbles each,

Then number of marbles left with John = 𝑥𝑥 − 5

Then number of marbles left with Jivanti = 45− 𝑥𝑥 − 5 = 40 − 𝑥𝑥

Given that the product of their marbles is 124.

∴ (𝑥𝑥 − 5)(40− 𝑥𝑥) = 124

⇒ 𝑥𝑥2 − 45𝑥𝑥 + 324 = 0

⇒ 𝑥𝑥2 − 36𝑥𝑥 − 9𝑥𝑥 + 324 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 36)− 9(𝑥𝑥 − 36) = 0

⇒ (𝑥𝑥 − 36)(𝑥𝑥 − 9) = 0

Either 𝑥𝑥 = 36 = 0 or 𝑥𝑥 − 9 = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = 36 or 𝑥𝑥 = 9

If the number of John's marbles = 36

Then, the number of Jivanti's marbles = 45− 36 = 9

If the number of John's marbles = 9

Then, the number of Jivanti's marbles = 45− 9 = 36.

(ii) Let the number of toys produced on that day be 𝑥𝑥.

∴ The cost of production of each toy that day = ₹ (55− 𝑥𝑥)

So, the total cost of production that day = 𝑥𝑥(55− 𝑥𝑥)

As per the question, the total cost of production of the toys = ₹ 750





∴ (55− 𝑥𝑥)𝑥𝑥 = 750

⇒ 𝑥𝑥2 − 55𝑥𝑥 + 750 = 0

⇒ 𝑥𝑥2 − 25𝑥𝑥 − 30𝑥𝑥 + 750 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 25)− 30(𝑥𝑥 − 25) = 0

⇒ (𝑥𝑥 − 25)(𝑥𝑥 − 30) = 0

Either 𝑥𝑥 − 25 = 0 or 𝑥𝑥 − 30 = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = 25 or 𝑥𝑥 = 30

Thus, the number of toys produced that day will be either 25 or 30.

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

Solution:

Let the first number be 𝑥𝑥.

Then the second number is 27 − 𝑥𝑥. (Given sum of two numbers = 27)

Thus, their product = 𝑥𝑥(27− 𝑥𝑥)

According to the question, the product of these numbers is 182.

Therefore, 𝑥𝑥(27− 𝑥𝑥) = 182

⇒ 𝑥𝑥2 − 27𝑥𝑥 + 182 = 0

⇒ 𝑥𝑥2 − 13𝑥𝑥 − 14𝑥𝑥 + 182 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 13)− 14(𝑥𝑥 − 13) = 0

⇒ (𝑥𝑥 − 13)(𝑥𝑥 − 14) = 0

Either 𝑥𝑥 = 13 = 0 or 𝑥𝑥 − 14 = 0

i. e. , 𝑥𝑥 = 13 or 𝑥𝑥 = 14

If first number = 13, then

Second number = 27− 13 = 14

If first number = 14, then

Second number = 27 − 14 = 13

Hence, the numbers are 13 and 14.

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

Solution:

We know that the difference between two consecutive positive integers is 1.





So, let the consecutive positive integers be 𝑥𝑥 and 𝑥𝑥 + 1.

As per the question, 𝑥𝑥2 + (𝑥𝑥 + 1)2 = 365

⇒ 𝑥𝑥2 + 𝑥𝑥2 + 1 + 2𝑥𝑥 = 365

⇒ 2𝑥𝑥2 + 2𝑥𝑥 − 364 = 0

⇒ 𝑥𝑥2 + 𝑥𝑥 − 182 = 0

⇒ 𝑥𝑥2 + 14𝑥𝑥 − 13𝑥𝑥 − 182 = 0

⇒ 𝑥𝑥(𝑥𝑥 + 14)− 13(𝑥𝑥 + 14) = 0

⇒ (𝑥𝑥 + 14)(𝑥𝑥 − 13) = 0

Either 𝑥𝑥 + 14 = 0 or 𝑥𝑥 − 13 = 0, i. e. , 𝑥𝑥 = −14 or 𝑥𝑥 = 13

Since given that integers are positive, 𝑥𝑥 can only be 13.

∴ 𝑥𝑥 + 1 = 13 + 1 = 14

Hence, the two 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: As per the question, hypotenuse is 13 cm

Let the base of the right triangle be 𝑥𝑥 cm.

Its altitude = (𝑥𝑥 − 7)cm

From Pythagoras theorem,

Base2 + Altitude2 = Hypotenuse2

∴ 𝑥𝑥2 + (𝑥𝑥 − 7)2 = 132

⇒ 𝑥𝑥2 + 𝑥𝑥2 + 49 − 14𝑥𝑥 = 169

⇒ 2𝑥𝑥2 − 14𝑥𝑥 − 120 = 0

⇒ 𝑥𝑥2 − 7𝑥𝑥 − 60 = 0

⇒ 𝑥𝑥2 − 12𝑥𝑥 + 5𝑥𝑥 − 60 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 12) + 5(𝑥𝑥 − 12) = 0

⇒ (𝑥𝑥 − 12)(𝑥𝑥 + 5) = 0

Either 𝑥𝑥 − 12 = 0 or 𝑥𝑥 + 5 = 0, i. e. , 𝑥𝑥 = 12 or 𝑥𝑥 = −5

Since sides of a triangle are positive, 𝑥𝑥 can only take 12.

Hence, the base of the right triangle is 12 cm and the altitude of this triangle is (12− 7)cm = 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.

Solution:

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

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

According to the question, the total cost of production on that day was ₹ 90.

We know that

Total cost of production = Cost of each article × Number of articles produced

∴ 𝑥𝑥(2𝑥𝑥 + 3) = 90

⇒ 2𝑥𝑥2 + 3𝑥𝑥 − 90 = 0

⇒ 2𝑥𝑥2 + 15𝑥𝑥 − 12𝑥𝑥 − 90 = 0

⇒ 𝑥𝑥(2𝑥𝑥 + 15) − 6(2𝑥𝑥 + 15) = 0

⇒ (2𝑥𝑥 + 15)(𝑥𝑥 − 6) = 0

Either 2𝑥𝑥 + 15 = 0 or 𝑥𝑥 − 6 = 0, 𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = −152

or 𝑥𝑥 = 6

It’s clear that number of articles produced can only be a positive integer, so, 𝑥𝑥 can only be 6.

Therefore, number of articles produced on that day = 6

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

EXERCISE 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2𝑥𝑥2 − 7𝑥𝑥 + 3 = 0

(ii) (ii) 2𝑥𝑥2 + 𝑥𝑥 − 4 = 0

(iii) 4𝑥𝑥2 + 4√3𝑥𝑥 + 3 = 0

(iv) 2𝑥𝑥2 + 𝑥𝑥 + 4 = 0

Solution:

(i) Given quadratic equation: 2𝑥𝑥2 − 7𝑥𝑥 + 3 = 0





⇒ 2𝑥𝑥2 − 7𝑥𝑥 = −3

On dividing both sides of the equation by 2, we obtain

⇒ 𝑥𝑥2 −72 𝑥𝑥 = −

32

⇒ 𝑥𝑥2 − 2 × 𝑥𝑥 ×74 = −

32

On adding �74�2 to both sides of equation by, we obtain

⇒ (𝑥𝑥)2 − 2 × 𝑥𝑥 ×74 + �

74�

2

= �74�

2

−32

⇒ �𝑥𝑥 −74�

2

=4916−

32

⇒ �𝑥𝑥 −74�

2

=2516

⇒ �𝑥𝑥 − 74� = ± 5

4 (Cancelling square both the sides)

⇒ 𝑥𝑥 =74 ±

54

⇒ 𝑥𝑥 = 74

+ 54 or 𝑥𝑥 = 7

4− 5

4

⇒ 𝑥𝑥 = 124

or 𝑥𝑥 = 24

⇒ 𝑥𝑥 = 3 or 12

Hence, the roots of this quadratic equation are 3 and 12 .

(ii) 2𝑥𝑥2 + 𝑥𝑥 − 4 = 0

⇒ 2𝑥𝑥2 + 𝑥𝑥 = 4


⇒ 𝑥𝑥2 +12 𝑥𝑥 = 2

On adding �14�2 to both sides of the equation, we obtain

⇒ (𝑥𝑥)2 + 2 × 𝑥𝑥 ×14 + �

14�

2

= 2 + �14�

2

⇒ �𝑥𝑥 +14�

2

=3316





⇒ 𝑥𝑥 + 14

= ± √334

(Cancelling square both the sides)

⇒ 𝑥𝑥 = ±√33

4 −14

⇒ 𝑥𝑥 =±√33− 1

4

⇒ 𝑥𝑥 = √33−14

or −√33−14

Hence, the roots of this quadratic equation are −1+√334

and −1−√334

.

(iii) 4𝑥𝑥2 + 4√3𝑥𝑥 + 3 = 0

⇒ (2𝑥𝑥)2 + 2 × 2𝑥𝑥 × √3 + �√3�2

= 0

⇒ �2𝑥𝑥 + √3�2

= 0

⇒ �2𝑥𝑥 + √3�

= 0 and �2𝑥𝑥 + √3�

= 0

⇒ 𝑥𝑥 = −√32

and 𝑥𝑥 = −√32

Hence, the roots of this quadratic equation are −√32

and − √32

.

(iv) 2𝑥𝑥2 + 𝑥𝑥 + 4 = 0

⇒ 2𝑥𝑥2 + 𝑥𝑥 = −4


⇒ 𝑥𝑥2 +12 𝑥𝑥 = −2

⇒ 𝑥𝑥2 + 2 × 𝑥𝑥 ×14 = −2

On adding �14�2to both sides of the equation, we obtain

⇒ (𝑥𝑥)2 + 2 × 𝑥𝑥 ×14 + �

14�

2

= �14�

2

− 2

⇒ �𝑥𝑥 +14�

2

=1

16− 2

⇒ �𝑥𝑥 +14�

2

= −3116

Since, the square of a number cannot be negative.





Thus, there is no real root for the given equation.

2. Find the roots of the quadratic equations given by applying the quadratic formula.

(i) 2𝑥𝑥2 − 7𝑥𝑥 + 3 = 0

(ii) 2𝑥𝑥2 + 𝑥𝑥 − 4 = 0

(iii) 4𝑥𝑥2 + 4√3𝑥𝑥 + 3 = 0

(iv) 2𝑥𝑥2 + 𝑥𝑥 + 4 = 0

Solution:

(i) 2𝑥𝑥2 − 7𝑥𝑥 + 3 = 0

On comparing this equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we obtain

𝑎𝑎 = 2,𝑏𝑏 = −7, 𝑐𝑐 = 3

By using quadratic formula, we obtain

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐

2𝑎𝑎

⇒ 𝑥𝑥 =7 ± √49− 24

2𝑎𝑎

⇒ 𝑥𝑥 =7 ± √25

4

⇒ 𝑥𝑥 =7 ± 5

4

⇒ 𝑥𝑥 = 7+54

or 7−54

⇒ 𝑥𝑥 = 124

or 24

∴ 𝑥𝑥 = 3 or 12

Hence, the roots of this quadratic equation are 3 and 12 .

(ii) 2𝑥𝑥2 + 𝑥𝑥 − 4 = 0


𝑎𝑎 = 2,𝑏𝑏 = 1, 𝑐𝑐 = −4



2𝑎𝑎





⇒ 𝑥𝑥 =−1 ± √1 + 32

4

⇒ 𝑥𝑥 =−1 ± √33

4

∴ 𝑥𝑥 = −1+√334

or −1−√334

Hence, the roots of this quadratic equation are −1+√334

and −1−√334

.

(iii) 4𝑥𝑥2 + 4√3𝑥𝑥 + 3 = 0

On comparing this equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we obtain 𝑎𝑎 = 4,𝑏𝑏 =4√3, 𝑐𝑐 = 3



2𝑎𝑎

⇒ 𝑥𝑥 =−4√3 ± √48− 48

8

⇒ 𝑥𝑥 =−4√3 ± 0

8

∴ 𝑥𝑥 = −√32

or − √32

Hence, the roots of this quadratic equation are −√32

and − √32

.

(iv) 2𝑥𝑥2 + 𝑥𝑥 + 4 = 0


𝑎𝑎 = 2,𝑏𝑏 = 1, 𝑐𝑐 = 4



2𝑎𝑎

⇒ 𝑥𝑥 =−1 ± √1 − 32

4

⇒ 𝑥𝑥 =−1 ± √−31

4

Hence, roots do not exist for this quadratic equation as 𝐷𝐷 = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = −31 < 0.

3. Find the roots of the following equations:





(i) 𝑥𝑥 − 1𝑥𝑥

= 3, 𝑥𝑥 ≠ 0

(ii) 1𝑥𝑥+4

− 1𝑥𝑥−7

= 1130

, 𝑥𝑥 ≠ −4, 7

Solution:

(i) 𝑥𝑥 − 1𝑥𝑥

= 3 ⇒ 𝑥𝑥2 − 3𝑥𝑥 − 1 = 0


𝑎𝑎 = 1,𝑏𝑏 = −3, 𝑐𝑐 = −1



2𝑎𝑎

⇒ 𝑥𝑥 =3 ± √9 + 4

2

⇒ 𝑥𝑥 =3 ± √13

2

Therefore, 𝑥𝑥 = 3+√132

or 3−√132

(ii) 1𝑥𝑥+4

− 1𝑥𝑥−7

= 1130

⇒𝑥𝑥 − 7− 𝑥𝑥 − 4

(𝑥𝑥 + 4)(𝑥𝑥 − 7) =1130

⇒−11

(𝑥𝑥 + 4)(𝑥𝑥 − 7) =1130

⇒ (𝑥𝑥 + 4)(𝑥𝑥 − 7) = −30

⇒ 𝑥𝑥2 − 3𝑥𝑥 − 28 = −30

⇒ 𝑥𝑥2 − 3𝑥𝑥 + 2 = 0

⇒ 𝑥𝑥2 − 2𝑥𝑥 − 𝑥𝑥 + 2 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 2)− 1(𝑥𝑥 − 2) = 0

⇒ (𝑥𝑥 − 2)(𝑥𝑥 − 1) = 0

⇒ 𝑥𝑥 = 1 or 2

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1

3. Find his present age.

Solution:





Let Rehman’s present age = 𝑥𝑥 years.

His age three years ago = (𝑥𝑥 − 3) years.

His age five years from now = (𝑥𝑥 + 5) years.

As per the question, the sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1

3.

∴1

𝑥𝑥 − 3 +1

𝑥𝑥 + 5 =13

𝑥𝑥 + 5 + 𝑥𝑥 − 3(𝑥𝑥 − 3)(𝑥𝑥 + 5) =

13

2𝑥𝑥 + 2(𝑥𝑥 − 3)(𝑥𝑥 + 5) =

13

⇒ 3(2𝑥𝑥 + 2) = (𝑥𝑥 − 3)(𝑥𝑥 + 5)

⇒ 6𝑥𝑥 + 6 = 𝑥𝑥2 + 2𝑥𝑥 − 15

⇒ 𝑥𝑥2 − 4𝑥𝑥 − 21 = 0

⇒ 𝑥𝑥2 − 7𝑥𝑥 + 3𝑥𝑥 − 21 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 7) + 3(𝑥𝑥 − 7) = 0

⇒ (𝑥𝑥 − 7)( 𝑥𝑥 + 3) = 0

⇒ 𝑥𝑥 = 7 or 𝑥𝑥 = −3

It’s clear that age is always positive.

Thus, Rehman's present age is 7 years.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solution:

Let Shefali’s marks in Mathematics = 𝑥𝑥.

Then, her marks in English = 30− 𝑥𝑥. (Given in question)

As per the question, we get

(𝑥𝑥 + 2)(30− 𝑥𝑥 − 3) = 210

(𝑥𝑥 + 2)(27− 𝑥𝑥) = 210

⇒ −𝑥𝑥2 + 25𝑥𝑥 + 54 = 210

⇒ 𝑥𝑥2 − 25𝑥𝑥 + 156 = 0





⇒ 𝑥𝑥2 − 12𝑥𝑥 − 13𝑥𝑥 + 156 = 0

⇒ 𝑥𝑥(𝑥𝑥 + 2)− 13(𝑥𝑥 − 12) = 0

⇒ (𝑥𝑥 − 12)(𝑥𝑥 − 13) = 0

⇒ 𝑥𝑥 = 12 or 𝑥𝑥 = 13

If the marks in Mathematics is 12, then marks in English will be 30− 12 = 18

If the marks in Mathematics is 13,then marks in English will be 30− 13 = 17

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution:

Let the shorter side of the rectangle be 𝑥𝑥 m.

Then, longer side of the rectangle = (𝑥𝑥 + 30)m

Diagonal of the rectangle = �𝑥𝑥2 + (𝑥𝑥 + 30)2 (By Pythagoras theorem)

But in question, it is given that the diagonal of the rectangular field is 60 𝑚𝑚 more than the shorter side.

∴ �𝑥𝑥2 + (𝑥𝑥 + 30)2 = 𝑥𝑥 + 60

⇒ 𝑥𝑥2 + (𝑥𝑥 + 30)2 = (𝑥𝑥 + 60)2 (By squaring on both the sides)

⇒ 𝑥𝑥2 + 𝑥𝑥2 + 900 + 60𝑥𝑥 = 𝑥𝑥2 + 3600 + 120𝑥𝑥

⇒ 𝑥𝑥2 − 60𝑥𝑥 − 2700 = 0

⇒ 𝑥𝑥2 − 90𝑥𝑥 + 30𝑥𝑥 − 2700 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 90) + 30(𝑥𝑥 − 90)

⇒ (𝑥𝑥 − 90)(𝑥𝑥 + 30) = 0

⇒ 𝑥𝑥 = 90 or 𝑥𝑥 = −30





But, side of a rectangle cannot be negative. So, the length of the shorter side is 90 m. Thus, the length of the longer side will be (90 + 30) m = 120 m

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let the larger and smaller number be 𝑥𝑥 and 𝑦𝑦 respectively.

It is given in the question that,

𝑥𝑥2 − 𝑦𝑦2 = 180 and 𝑦𝑦2 = 8𝑥𝑥

⇒ 𝑥𝑥2 − 8𝑥𝑥 = 180

⇒ 𝑥𝑥2 − 8𝑥𝑥 − 180 = 0

⇒ 𝑥𝑥2 − 18𝑥𝑥 + 10𝑥𝑥 − 180 = 0

⇒ 𝑥𝑥(𝑥𝑥 − 18) + 10(𝑥𝑥 − 18) = 0

⇒ (𝑥𝑥 − 18)(𝑥𝑥 + 10) = 0

⇒ 𝑥𝑥 = 18,−10

If larger number, 𝑥𝑥 = −10

then smaller number, 𝑦𝑦 = ±√8𝑥𝑥

= ±�8(−10)

= ±√−80

Since we cannot have negative number in roots

𝑥𝑥 = −10 is not possible

Therefore, the larger number will be 18 only.

𝑥𝑥 = 18

∴ 𝑦𝑦2 = 8𝑥𝑥 = 8 × 18 = 144

⇒ 𝑦𝑦 = ±√144 = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and −12.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km h⁄ more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let the speed to the train be 𝑥𝑥 km/h.





Total time taken to cover 360 km = 360𝑥𝑥

hr

It is given in question,

(𝑥𝑥 + 5) �360𝑥𝑥− 1� = 360 (Distance = Speed × Time)

⇒ 360− 𝑥𝑥 +1800𝑥𝑥 − 5 = 360

⇒ 𝑥𝑥2 + 5𝑥𝑥 − 1800 = 0

⇒ 𝑥𝑥2 + 45𝑥𝑥 − 40𝑥𝑥 − 1800 = 0

⇒ 𝑥𝑥(𝑥𝑥 + 45)− 40(𝑥𝑥 + 45) = 0

⇒ (𝑥𝑥 + 45)(𝑥𝑥 − 40) = 0

⇒ 𝑥𝑥 = 40 or 𝑥𝑥 = −45

But, speed cannot be negative.

Thus, the speed of the train is 40 km/h

9. Two water taps together can fill a tank in 9 38 hours. The tap of larger diameter takes

10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution:

Let the time taken by the smaller diameter tap to fill the tank be 𝑥𝑥 hr.

Time taken by the larger diameter tap = (𝑥𝑥 − 10)hr

Volume of tank filled by smaller tap in 1hour = 1𝑥𝑥

Volume of tank filled by larger tap in 1 hour = 1𝑥𝑥−10

As per the question, the tank can be filled in 9 38

= 758

hours by both the taps together.

Hence,

1𝑥𝑥 +

1𝑥𝑥 − 10 =

875

𝑥𝑥 − 10 + 𝑥𝑥𝑥𝑥(𝑥𝑥 − 10) =

875

⇒2𝑥𝑥 − 10𝑥𝑥(𝑥𝑥 − 10) =

875

⇒ 75(2𝑥𝑥 − 10) = 8𝑥𝑥2 − 80𝑥𝑥





⇒ 150𝑥𝑥 − 750 = 8𝑥𝑥2 − 80𝑥𝑥

⇒ 8𝑥𝑥2 − 230𝑥𝑥 + 750 = 0

⇒ 8𝑥𝑥2 − 200𝑥𝑥 − 30𝑥𝑥 + 750 = 0

⇒ 8𝑥𝑥(𝑥𝑥 − 25) − 30(𝑥𝑥 − 25) = 0

⇒ (𝑥𝑥 − 25)(8𝑥𝑥 − 30) = 0

𝑖𝑖. 𝑒𝑒. ,𝑥𝑥 = 25 or 𝑥𝑥 = 308

= 154

Taking 𝑥𝑥 = 154

Time taken by smaller tap

= 𝑥𝑥 = 154

hrs

Time taken by larger tap = 𝑥𝑥 − 10

= 154− 10 = 15−40

4= − 25

4

Since time is negative,

𝑥𝑥 = 154

is not the solution

Thus, the time taken separately by the smaller diameter tap and the larger diameter tap will be 25 hours and (25− 10) = 15 hours respectively.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km h⁄ more than that of the passenger train, find the average speed of the two trains.

Solution:

Let the average speed of passenger train be 𝑥𝑥 km/h.

Average speed of express train = (𝑥𝑥 + 11) km/h (Given in question)

According to the question, the time taken by the express train to cover 132 km is

1 hour less than a passenger train.

Therefore, time taken by passenger train – time taken by express train = 1 hour

∴ 132𝑥𝑥− 132

𝑥𝑥+11= 1 �Total time = Distance

Average Speed�

⇒ 132 �𝑥𝑥 + 11 − 𝑥𝑥𝑥𝑥(𝑥𝑥 + 11) � = 1





⇒132 × 11𝑥𝑥(𝑥𝑥 + 11) = 1

⇒ 132 × 11 = 𝑥𝑥(𝑥𝑥 + 11)

⇒ 𝑥𝑥2 + 11𝑥𝑥 − 1452 = 0

⇒ 𝑥𝑥2 + 44𝑥𝑥 − 33𝑥𝑥 − 1452 = 0

⇒ 𝑥𝑥(𝑥𝑥 + 44)− 339𝑥𝑥 + 44) = 0

⇒ (𝑥𝑥 + 44)(𝑥𝑥 − 33) = 0

⇒ 𝑥𝑥 = −44, 33

Average speed of passenger train cannot be negative. Hence, the speed of the passenger train is 33km/h and thus, the speed of the express train will be 33 +11 = 44 km/h.

11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:

Let the sides of the two squares be 𝑥𝑥 m and 𝑦𝑦 m.

So, their perimeter will be 4𝑥𝑥 and 4𝑦𝑦 respectively and their areas will be 𝑥𝑥2and 𝑦𝑦2 respectively.

According to the question, 4𝑥𝑥 − 4𝑦𝑦 = 24

⇒ 𝑥𝑥 − 𝑦𝑦 = 6

⇒ 𝑥𝑥 = 𝑦𝑦 + 6

Also, 𝑥𝑥2 + 𝑦𝑦2 = 468

⇒ (6 + 𝑦𝑦)2 + 𝑦𝑦2 = 468

⇒ 36 + 𝑦𝑦2 + 12𝑦𝑦 + 𝑦𝑦2 = 468

⇒ 2𝑦𝑦2 + 12𝑦𝑦 − 432 = 0

⇒ 𝑦𝑦2 + 6𝑦𝑦 − 216 = 0

⇒ 𝑦𝑦2 + 18𝑦𝑦 − 12𝑦𝑦 − 216 = 0

⇒ 𝑦𝑦(𝑦𝑦 + 18)− 12(𝑦𝑦 + 18) = 0

⇒ (𝑦𝑦 + 18)(𝑦𝑦 − 12) = 0

⇒ 𝑦𝑦 = −18 or 12.

But, side of a square cannot be negative.

Hence, 𝑦𝑦 = 12 & 𝑥𝑥 = 12 + 6 = 18





Therefore, the sides of the squares are 12 m and 18 m

EXERCISE 4.4

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) 2𝑥𝑥2 − 3𝑥𝑥 + 5 = 0

(ii) 3𝑥𝑥2 − 4√3𝑥𝑥 + 4 = 0

(iii) 2𝑥𝑥2 − 6𝑥𝑥 + 3 = 0

Solution:

We know that for a quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, discriminant is

𝑏𝑏2 − 4𝑎𝑎𝑐𝑐.

(A) If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 > 0 implies two distinct real roots

(B) If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0 implies two equal real roots

(C) If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 < 0 implies imaginary roots

(i) 2𝑥𝑥2 − 3𝑥𝑥 + 5 = 0

Comparing this equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we obtain 𝑎𝑎 = 2, 𝑏𝑏 =−3, 𝑐𝑐 = 5

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = (−3)2 − 4(2)(5) = 9− 40 = −31

As 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 < 0,

Hence, no real root is possible for the given equation.

(ii) 3𝑥𝑥2 − 4√3𝑥𝑥 + 4 = 0

Comparing this equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we obtain

𝑎𝑎 = 3,𝑏𝑏 = −4√3, 𝑐𝑐 = 4

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = �−4√3�2− 4(3)(4) = 48− 48 = 0

As 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0,

So, real roots exist for the given equation and they are equal to each other and the roots will be − 𝑏𝑏

2𝑎𝑎 and − 𝑏𝑏

2𝑎𝑎.

−𝑏𝑏

2𝑎𝑎 =−�−4√3�

2 × 3 =4√3

6 =2√3

3 =2√3





Hence, the roots are 2√3

and 2√3

.

(iii) 2𝑥𝑥2 − 6𝑥𝑥 + 3 = 0

Comparing this equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we obtain 𝑎𝑎 = 2, 𝑏𝑏 =−6, 𝑐𝑐 = 3

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = (−6)2 − 4(2)(3) = 36− 24 = 12

As 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 > 0,

So, two distinct real roots exist for this equation as follows:


2𝑎𝑎

=−(−6) ± �(−6)2 − 4(2)(3)

2(2)

=6 ± √12

4 =6 ± 2√3

4

=3 ± √3

2

Hence, the roots are 3+√32

or 3−√32

.

2. Find the values of 𝑘𝑘 for each of the following quadratic equations, so that they have two equal roots.

(i) 2𝑥𝑥2 + 𝑘𝑘𝑥𝑥 + 3 = 0

(ii) 𝑘𝑘𝑥𝑥(𝑥𝑥 − 2) + 6 = 0

Solution:

We know that if an equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 has two equal roots, its discriminant (𝑏𝑏2 − 4𝑎𝑎𝑐𝑐) will be 0.

(i) 2𝑥𝑥2 + 𝑘𝑘𝑥𝑥 + 3 = 0

Comparing this equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we get 𝑎𝑎 = 2, 𝑏𝑏 = 𝑘𝑘, 𝑐𝑐 =3

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = (𝑘𝑘)2 − 4(2)(3) = 𝑘𝑘2 − 24

For equal roots,

Discriminant = 0

𝑘𝑘2 − 24 = 0

⇒ 𝑘𝑘2 = 24





⇒ 𝑘𝑘 = ±√24 = ±2√6

(ii) 𝑘𝑘𝑥𝑥(𝑥𝑥 − 2) + 6 = 0 or 𝑘𝑘𝑥𝑥2 − 2𝑘𝑘𝑥𝑥 + 6 = 0

Comparing the equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we get 𝑎𝑎 = 𝑘𝑘, 𝑏𝑏 =−2𝑘𝑘, 𝑐𝑐 = 6

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = (−2𝑘𝑘)2 − 4(𝑘𝑘)(6) = 4𝑘𝑘2 − 24𝑘𝑘

For equal roots,

𝐷𝐷 = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0

4𝑘𝑘2 − 24𝑘𝑘 = 0

4𝑘𝑘(𝑘𝑘 − 6) = 0

Either 4𝑘𝑘 = 0 or 𝑘𝑘 − 6 = 0

⇒ 𝑘𝑘 = 0 or 𝑘𝑘 = 6

But, if 𝑘𝑘 = 0, then the equation will not have the terms ‘𝑥𝑥2’ and ‘𝑥𝑥’.

Hence, if this quadratic equation has two equal roots, then 𝑘𝑘 should be 6 only.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solution:

Let the breadth of mango grove be 𝑥𝑥.

Length of mango grove will be 2𝑥𝑥.

Area of mango grove = (2𝑥𝑥)(𝑥𝑥) = 2𝑥𝑥2

Hence, 2𝑥𝑥2 = 800

⇒ 𝑥𝑥2 =800

2

⇒ 𝑥𝑥2 = 400

Cancelling square on both the sides, we get 𝑥𝑥 = ±20





But length cannot be negative.

So, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m

4. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution:

Let the age of one friend be 𝑥𝑥 years.

Age of the other friend will be = (20− 𝑥𝑥) years. (Given in question)

Four years ago, age of 1st friend = (𝑥𝑥 − 4) years

and, age of 2nd friend = ((20− 𝑥𝑥)− 4) = (16 − 𝑥𝑥) years

As per the question,

(𝑥𝑥 − 4)(16− 𝑥𝑥) = 48

16𝑥𝑥 − 64 − 𝑥𝑥2 + 4𝑥𝑥 = 48

−𝑥𝑥2 + 20𝑥𝑥 − 112 = 0

𝑥𝑥2 − 20𝑥𝑥 + 112 = 0

Comparing this equation with 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, we obtain

𝑎𝑎 = 1,𝑏𝑏 = −20, 𝑐𝑐 = 112

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = (−20)2 − 4(1)(112)

= 400− 448 = −48

As 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 < 0,

Thus, no real roots are possible for this equation and hence, this situation is not possible.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.

Solution:

Let the length and breadth of the rectangular park be 𝑙𝑙 and b.

Perimeter = 2(𝑙𝑙 + 𝑏𝑏) = 80

𝑙𝑙 + 𝑏𝑏 = 40

or, 𝑏𝑏 = 40− 𝑙𝑙





Area = 𝑙𝑙 × 𝑏𝑏 = 𝑙𝑙(40− 𝑙𝑙) = 40𝑙𝑙 − 𝑙𝑙2

40𝑙𝑙 − 𝑙𝑙2 = 400

𝑙𝑙2 − 40𝑙𝑙 + 400 = 0

Comparing this equation with 𝑎𝑎𝑙𝑙2 + 𝑏𝑏𝑙𝑙 + 𝑐𝑐 = 0, we obtain

𝑎𝑎 = 1,𝑏𝑏 = −40, 𝑐𝑐 = 400

Discriminant = 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = (−40)2 − 4(1)(400)

= 1600− 1600 = 0

As 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 = 0,

Thus, this equation has equal real roots, and hence, this situation is possible.

Root of this equation,

𝑙𝑙 = −𝑏𝑏

2𝑎𝑎

𝑙𝑙 = −(−40)2(1) =

402 = 20

So, length of park, 𝑙𝑙 = 20 m

and breadth of park, 𝑏𝑏 = 40 − 𝑙𝑙 = 40− 20 = 20 m

♦ ♦ ♦



Documents

CBSE NCERT Solutions for Class 10 Mathematics Chapter 4€¦ · Thus, the given equation is a quadratic equation as 𝑎𝑎≠0. 2. Represent the following situations in the form