NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Exercise 4.1

Q.1) Check whether the following are quadratic equations:
(i) (𝑥 + 1)2 = 2(𝑥 − 3)                           (ii) 𝑥2 − 2𝑥 = (−2)(3 − 𝑥)
(iii) (𝑥 − 2)(𝑥 + 1) = (𝑥 − 1)(𝑥 + 3)      (iv) (𝑥 − 3)(2𝑥 + 1) = 𝑥(𝑥 + 5)
(v) (2𝑥 − 1)(𝑥 − 3) = (𝑥 + 5)(𝑥 − 1)     (vi) 𝑥2 + 3𝑥 + 1 = (𝑥 − 2)2
(vii) (𝑥 + 2)3 = 2𝑥(𝑥2 − 1)                  (viii) 𝑥3 − 4𝑥2 − 𝑥 + 1 = (𝑥 − 2)3
Sol.1) (i) (𝑥 + 1)2 = 2(𝑥 − 3)
LHS: (𝑥 + 1)2 = 𝑥2 + 2𝑥 + 1
(Using (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏2)
RHS: 2(𝑥 – 3) = 2𝑥 – 6
Now; 𝑥2 + 2𝑥 + 1 = 2𝑥 – 6
Or, 𝑥2 + 2𝑥 + 1 – 2𝑥 + 6 = 0
Or, 𝑥2 + 7 = 0
Since the equation is in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is a quadratic equation.

(ii) 𝑥2 − 2𝑥 = (−2)(3 – 𝑥)
𝑥2– 2𝑥 = (−2)(3 – 𝑥)
Or, 𝑥2– 2𝑥 = −6 – 2𝑥
Or, 𝑥2 – 2𝑥 + 2𝑥 + 6 = 0
Or, 𝑥2 + 6 = 0
Since the equation is in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is a quadratic equation.

(iii) (𝑥 − 2)(𝑥 + 1) = (𝑥 − 1)(𝑥 + 3)
LHS: (𝑥 – 2)(𝑥 + 1)
= 𝑥2 + 𝑥 – 2𝑥 – 2
= 𝑥2 – 𝑥 – 2
RHS: (𝑥 – 1)(𝑥 + 3)
= 𝑥2 + 3𝑥 – 𝑥 + 3
= 𝑥2 + 2𝑥 + 3
Now; 𝑥2– 𝑥 – 2 = 𝑥2 + 2𝑥 + 3
Or, 𝑥2 – 𝑥 – 2 – 𝑥2 – 2𝑥 – 3 = 0
Or, − 3𝑥 – 5 = 0
Since the equation is not in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is not a quadratic equation.

(iv) (𝑥 − 3)(2𝑥 + 1) = 𝑥(𝑥 + 5)
LHS: (𝑥 – 3)(2𝑥 + 1)
= 2𝑥2 + 𝑥 – 6𝑥 – 6
= 2𝑥2 – 5𝑥 – 6
RHS: 𝑥(𝑥 + 5)

= 𝑥2 + 5𝑥
Now; 2𝑥2 – 5𝑥 – 6 = 𝑥2 + 5𝑥
Or, 2𝑥2– 5𝑥 – 6 – 𝑥2– 5𝑥 = 0
Or, 𝑥2– 10𝑥 – 6 = 0
Since the equation is in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is a quadratic equation.

(v) (2𝑥 − 1)(𝑥 − 3) = (𝑥 + 5)(𝑥 − 1)
LHS: (2𝑥 – 1)(𝑥 – 3)
= 2𝑥2 – 6𝑥 – 𝑥 + 3
= 2𝑥2 – 7𝑥 + 3
RHS: (𝑥 + 5)(𝑥 – 1)
= 𝑥2 – 𝑥 + 5𝑥 – 5
= 𝑥2 + 4𝑥 – 5
Now; 2𝑥2 – 7𝑥 + 3 = 𝑥2 + 4𝑥 – 5
Or, 2𝑥2 – 7𝑥 + 3 – 𝑥2 − 4𝑥 + 5 = 0
Or, 𝑥2– 11𝑥 + 8 = 0
Since the equation is in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is a quadratic equation.

(vi) 𝑥2 + 3𝑥 + 1 = (𝑥 − 2)2
𝑥2 + 3𝑥 + 1 = (𝑥 – 2)2
Or, 𝑥2 + 3𝑥 + 1 = 𝑥2– 4𝑥 + 4
Or, 𝑥2 + 3𝑥 + 1 – 𝑥2 + 4𝑥 – 4 = 0
Or, 7𝑥 – 3 = 0
Since the equation is not in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is not a quadratic equation.

(vii) (𝑥 + 2)3 = 2𝑥(𝑥2 − 1)
LHS: (𝑥 + 2)3
Using (𝑎 + 𝑏)3 = 𝑎3 + 3𝑎2𝑏 + 3𝑎𝑏2 + 𝑏3, we get;
(𝑥 + 2)3 = 𝑥3 + 6𝑥2 + 12𝑥 + 8
RHS: 2𝑥(𝑥2– 1)
= 2𝑥3 – 2𝑥
Now; 𝑥3 + 6𝑥2 + 12𝑥 + 8 = 2𝑥3 – 2𝑥
Or, 𝑥3 + 6𝑥2 + 12𝑥 + 8 – 2𝑥3 + 2𝑥 = 0
Or, − 𝑥3 + 6𝑥2 + 14𝑥 + 8 = 0
Since the equation is not in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0;
hence it is not a quadratic equation.

(viii) 𝑥3 − 4𝑥2 − 𝑥 + 1 = (𝑥 − 2)3
LHS: 𝑥3 – 4𝑥2 – 𝑥 + 1
RHS: (𝑥 – 2)3
= 𝑥3– 8 – 6𝑥2 + 12𝑥
Now; 𝑥3 – 4𝑥2 – 𝑥 + 1 = 𝑥3 – 6𝑥2 + 12𝑥 – 8
Or, 𝑥3– 4𝑥2 – 𝑥 + 1 – 𝑥3 + 6𝑥2 – 12𝑥 + 8 = 0
Or, 2𝑥2– 13𝑥 + 9 = 0
Since the equation is in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0; 
hence it is a quadratic equation.

Q.2) Represent the following situation in the form of quadratic equation:
(i) The area of a rectangular plot is 528 𝑚2. The length of the plot (in meters) 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 the Rohan’s age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Sol.2) i) Let us assume breadth = 𝑥
Therefore; length = 2𝑥 + 1
Since area = 𝑙𝑒𝑛𝑔𝑡ℎ × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ
Hence; 𝑥(2𝑥 + 1) = 528
⇒ 2𝑥2 + 𝑥 = 528
⇒ 2𝑥2 + 𝑥 – 528 = 0

ii) Let us assume the first integer = 𝑥
Hence; second integer = 𝑥 + 1
As per question; 𝑥(𝑥 + 1) = 306
⇒ 𝑥2 + 𝑥 = 306
⇒ 𝑥2 + 𝑥 – 306 = 0

iii) Let us assume, Rohan’s present age = 𝑥
So, his mother’s present age = 𝑥 + 26
Three years from now, Rohan’s age = 𝑥 + 3
Three years from now, mother’s age = 𝑥 + 29
As per question; (𝑥 + 3)(𝑥 + 29) = 360
⇒ 𝑥2 + 29𝑥 + 3𝑥 + 87 = 360
⇒ 𝑥2 + 32𝑥 + 87 = 360
⇒ 𝑥2 + 32𝑥 + 87 – 360 = 0
⇒ 𝑥2 + 32𝑥 – 273 = 0

iv) Let us assume, speed of train = 𝑥 𝑘𝑚/ℎ
Therefore; reduced speed = 𝑥 – 8 𝑘𝑚/ℎ
We know, time = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑠𝑝𝑒𝑒𝑑
Hence;
⇒ 𝑡 = 480/𝑥                       ….. (i)
In case of reduced speed,
⇒ 𝑡 + 3 = 480/𝑥−8
⇒ 𝑡 = (480/𝑥−8) − 3          …… (ii)
From equations (1) and (2);
⇒ 480/𝑥 = (480/𝑥−8) − 3
⇒ 480/𝑥 = 480−3(𝑥−8)/𝑥−8
⇒ 480/𝑥 = 480−3𝑥+24/𝑥−8
⇒ 480(𝑥 − 8) = 𝑥(504 − 3𝑥)
⇒ 480𝑥 − 3840 = 504𝑥 − 3𝑥2
⇒ 480𝑥 − 3840 − 504𝑥 + 3𝑥2 = 0
⇒ 3𝑥2 − 24𝑥 − 3840 = 0
⇒ 𝑥2 − 8𝑥 − 1280 = 0

Exercise 4.2

Q.1) Find the roots of the following quadratic equations by factorization:
Sol.1) (i) 𝑥2– 3𝑥 – 10 = 0
𝑥2 – 3𝑥 – 10 = 0
⇒ 𝑥2– 5𝑥 + 2𝑥 – 10 = 0
⇒ 𝑥(𝑥 – 5) + 2(𝑥 – 5) = 0
⇒ (𝑥 + 2)(𝑥 – 5) = 0
Now; case 1: (𝑥 + 2) = 0
⇒ 𝑥 = − 2
Case 2: (𝑥 – 5) = 0
⇒ 𝑥 = 5
Hence, roots are: - 2 and 5

(ii) 2𝑥2 + 𝑥 – 6
2𝑥2 + 𝑥 – 6 = 0
⇒ 2𝑥2 + 4𝑥 – 3𝑥 – 6 = 0
⇒ 2𝑥(𝑥 + 2) – 3(𝑥 + 2) = 0
⇒ (2𝑥 – 3)(𝑥 + 2) = 0
Case 1: (2𝑥 – 3) = 0
⇒ 2𝑥 = 3
⇒ 𝑥 = 3/2
Case 2: (𝑥 + 2) = 0
⇒ 𝑥 = − 2
Hence, roots are – 2 and 3/2

(iii) √2𝑥2 + 7𝑥 + 5√2 = 0
√2𝑥2 + 7𝑥 + 5√2 = 0
⇒ √2𝑥2 + 2𝑥 + 5𝑥 + 5√2 = 0
⇒ √2𝑥(𝑥 + √2) + 5(𝑥 + √2) = 0
⇒ (√2𝑥 + 5)(𝑥 + √2) = 0
Case 1: (√2𝑥 + 5) = 0
⇒ √2𝑥 = 5

⇒ 100𝑥2 − 20𝑥 + 1 = 0
⇒ 100𝑥2 − 20𝑥 + 1 = 0
⇒ 10𝑥(10𝑥 − 1) − 1(10𝑥 − 1) = 0
⇒ (10𝑥 − 1)(10𝑥 − 1) = 0
Hence, 𝑥 = 1/10

Q.2) i) Solve the problems given in Example 1.

(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.
Sol.2) i) Given, John and Jivanti together have number of marbles = 45
After losing of 5 marbles by each of them, number of marble = 45 – 5 – 5 = 45 – 10 = 35
Let us assume, John has 𝑥 marbles
Hence; marbles with Jivanti = 35 – 𝑥
As per question; product of marbles after loss = 124
herefore; 𝑥(35 – 𝑥) = 124
⇒ 35𝑥 – 𝑥2 = 124
⇒ − 𝑥2 + 35𝑥 – 124 = 0
⇒ 𝑥2 – 35𝑥 + 124 = 0
⇒ 𝑥2 – 4𝑥 – 31𝑥 + 124 = 0
⇒ 𝑥(𝑥 – 4) – 31 (𝑥 – 4) = 0
⇒ (𝑥 – 31)(𝑥 – 4) = 0
Hence, 𝑥 = 31 and 𝑥 = 4
One person has 31 marbles and another has 4 marbles
At the beginning; one person had 36 marbles and another had 9 marbles.

Q.2) 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 Rs. 750. We would like to find out the number of toys produced on that day.
Sol.2) ii) Let us assume, number of toys = 𝑥
Then, cost of production of each toy = 𝑥 – 55
Hence, total cost of production = 𝑥(55 – 𝑥) = 750
⇒ 55𝑥 – 𝑥2 = 750
⇒ 𝑥2 − 55𝑥 + 750 = 0
⇒ 𝑥2 − 30𝑥 − 25𝑥 + 750 = 0
⇒ 𝑥(𝑥 − 30) – 25(𝑥 – 30) = 0
⇒ (𝑥 – 25)(𝑥 – 30) = 0
Hence, 𝑥 = 25 and 𝑥 = 30
Thus, number of toys is 25 or 30

Q.3) Find two numbers whose sum is 27 and product is 182.
Sol.3) Let us assume, one of the numbers = 𝑥
Hence, second number = 27 – 𝑥
As per question; 𝑥(27 – 𝑥) = 182
⇒ 27𝑥 – 𝑥2 = 182
⇒ 27𝑥 – 𝑥2 – 182 = 0
⇒ 𝑥2 – 27𝑥 + 182 = 0
⇒ 𝑥2 – 14𝑥 – 13𝑥 + 182 = 0
⇒ 𝑥(𝑥 – 14) – 13(𝑥 – 14) = 0
⇒ (𝑥 – 13)(𝑥 – 14) = 0
Hence, 𝑥 = 13 and 𝑥 = 14
Hence, the numbers are 13 and 14

Q.4) Find two consecutive positive integers, sum of whose squares is 365.
Sol.4) Let us assume, first integer = 𝑥
Then, second integer = 𝑥 + 1
As per question; 𝑥2 + (𝑥 + 1)2 = 365
⇒ 𝑥2 + 𝑥2 + 2𝑥 + 1 = 365
⇒ 2𝑥2 + 2𝑥 + 1 – 365 = 0
⇒ 2𝑥2 + 2𝑥 – 364 = 0
⇒ 𝑥2 + 𝑥 – 182 = 0
⇒ 𝑥2 + 14𝑥 – 13𝑥 – 182 = 0
⇒ 𝑥(𝑥 + 14) – 13(𝑥 + 14) = 0
⇒ (𝑥 – 13)(𝑥 + 14) = 0
Hence, 𝑥 = 13 and 𝑥 = − 14
Since integers are positive, hence they are 13 and 14

Q.5) The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Sol.5) Let us assume, base = 𝑥
Therefore; altitude = 𝑥 – 7
As per question; using Pythagoras Theorem:
132 = 𝑥2 + (𝑥 – 7)2
⇒ 169 = 𝑥2 + 𝑥2– 14𝑥 + 49
⇒ 2𝑥2 – 14𝑥 + 49 – 169 = 0
⇒ 2𝑥2 – 14𝑥 – 120 = 0
⇒ 𝑥2– 7𝑥 – 60 = 0
⇒ 𝑥2 – 12𝑥 + 5𝑥 – 60 = 0
⇒ 𝑥(𝑥 – 12) + 5(𝑥 – 12) = 0
⇒ (𝑥 + 5)(𝑥 – 12) = 0
Hence, 𝑥 = − 5 and 𝑥 = 12
Ruling out the negative value; we have 𝑥 = 12
So, altitude = 12 – 5 = 7
Thus, two sides are 12 𝑐𝑚 and 5 𝑐𝑚

Q.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.
Sol.6) Let us assume, number of pottery in a day = 𝑥
So, cost of production of each article = 2𝑥 + 3
As per question; 𝑥(2𝑥 + 3) = 90
⇒ 2𝑥2 + 3𝑥 = 90
⇒ 2𝑥2 + 3𝑥 – 90 = 0
⇒ 2𝑥2 – 12𝑥 + 15𝑥 – 90 = 0
⇒ 2𝑥(𝑥 – 6) + 15(𝑥 – 6) = 0
⇒ (2𝑥 + 15)(𝑥 – 6) = 0
Hence, 𝑥 = − 15/2 and 𝑥 = 6
Ruling out the negative value; 𝑥 = 6
Cost of article = 𝑅𝑠. 15

Exercise 4.3

Q.1) Find the roots of the following quadratic equations, if they exist, by the method of completing square.
Sol.1) (i) 2𝑥2 – 7𝑥 + 3 = 0
Checking the existence of roots:
We know;
𝐷 = 𝑏2 − 4𝑎𝑐
= (−7)2 − 4 × 2 × 3
= 49 − 24 = 25
Since 𝐷 > 0; hence two different roots are possible for this equation.
Now; 2𝑥2 – 7𝑥 + 3 can be written as follows:

ii) 2𝑥2 + 𝑥 − 4 = 0
Checking the existence of roots:
We know;
𝐷 = 𝑏2 − 4𝑎𝑐
= 12 − 4 × 2 × (−4)
= 1 + 32 = 33
Since 𝐷 > 0; hence roots are possible for this equation.
By dividing the equation by 2; we get following equation:
⇒ 𝑥2 + 𝑥/2 − 2 = 0
⇒ 𝑥2 + 2 (1/4) 𝑥 − 2 = 0
⇒ 𝑥2 + 2 (1/4) 𝑥 = 2
⇒ 𝑥2 + 2 (1/4) 𝑥 + (1/4)2 = 2 + (1/4)2
Assuming 𝑥 = 𝑎 and ¼ = 𝑏; the above equation can be written in the form of

(𝑎 + 𝑏)2 

(iii) 4𝑥2 + 4√3𝑥 + 3 = 0

checking the existence of roots
we know,
𝐷 = 𝑏2 − 4𝑎𝑐
= (43)2 − 4 × 4 × 3
= 48 − 48 = 0
Since D = 0; hence roots are possible for this equation.
After dividing by 4; the equation can be written as follows:

(iv) 2𝑥2 + 𝑥 + 4

Checking the existence of roots:
We know;
𝐷 = 𝑏2 − 4𝑎𝑐
= 12 − 4 × 2 × 4
= −31
Since 𝐷 < 0; hence roots are not possible for this equation.

Q.2) Find the roots of the quadratic equations given in Q 1 above by applying the quadratic formula.
Sol.2) (i) 2𝑥2 – 7𝑥 + 3
We have; 𝑎 = 2, 𝑏 = − 7 and 𝑐 = 3
D can be calculated as follows:
𝐷 = 𝑏2 − 4𝑎𝑐
= (−7)2 − 4 × 2 × 3
= 49 − 24 = 25

(ii) 4𝑥2 + 4√3𝑥 + 3 = 0
We have; 𝑎 = 4, 𝑏 = 4√3 and 𝑐 = 3
D can be calculated as follows:
𝐷 = 𝑏2 − 4𝑎𝑐
= (4√3)2 − 4 × 4 × 3
= 48 − 48 = 0
Now; root can be calculated as follows:
𝑅𝑜𝑜𝑡 = − 𝑏/2𝑎 = − 4√3/2×4
= − √3/2

Q.3) Find the roots of the following equations:
Sol.3) i) 𝑥 − 1/𝑥
= 3; 𝑥 ≠ 0. 𝑥2−1/𝑥 = 3
𝑥2 − 3𝑥 − 1 = 0
We have, 𝑎 = 1, 𝑏 = −3 & 𝑐 = −1
Root can be calculated as follows:

⇒ 𝑥2 − 3𝑥 − 28 = −30

⇒ 𝑥2 − 3𝑥 − 28 + 30 = 0
⇒ 𝑥2 − 3𝑥 + 2 = 0
⇒ 𝑥2 − 2𝑥 − 𝑥 + 2 = 0
⇒ 𝑥(𝑥 − 2) − 1(𝑥 − 2) = 0
⇒ (𝑥 − 1)(𝑥 − 2) = 0
Hence, roots are 1 & 2

Q.4) The sum of reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age
Sol.4) Let us assume, Rehman’s present age = 𝑥
Therefore, 3 years ago, Rehman’s age = 𝑥 – 3
And, 5 years from now, Rehman’s age = 𝑥 + 5
As per question;

⇒ 6𝑥 + 6 = 𝑥2 + 2𝑥 − 15
⇒ 𝑥2 + 2𝑥 − 15 − 6𝑥 − 6 = 0
⇒ 𝑥2 − 4𝑥 − 21 = 0
⇒ 𝑥2 − 7𝑥 + 3𝑥 − 21 = 0
⇒ 𝑥(𝑥 − 7) + 3(𝑥 − 7) = 0
⇒ (𝑥 + 3)(𝑥 − 7) = 0
Ruling out the negative value; Rehman’s Age = 7 𝑦𝑒𝑎𝑟

Q.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.
Sol.5) Let us assume, marks in Mathematics = 𝑥
Therefore, marks in English = 30 – 𝑥
If she scores 2 marks more in Mathematics; then marks in mathematics = 𝑥 + 2
And if she scores 3 marks less in English, the marks in English = 30 – 𝑥 – 3 = 27 – 𝑥
As per question;
⇒ (𝑥 + 2)(27 – 𝑥) = 210
⇒ 27𝑥 – 𝑥2 + 54 – 2𝑥 = 210
⇒ 25𝑥 – 𝑥2 + 54 – 210 = 0
⇒ 25𝑥 – 𝑥2 – 156 = 0
⇒ 𝑥2 – 25𝑥 + 156 = 0
⇒ 𝑥2 – 12𝑥 – 13𝑥 + 156 = 0
⇒ 𝑥(𝑥 – 12) – 13(𝑥 – 12) = 0
⇒ (𝑥 – 12)(𝑥 – 13) = 0
Hence, 𝑥 = 12 and 𝑥 = 13
Case 1: If 𝑥 = 13, then marks in English = 30 – 13 = 17
Case 2: If 𝑥 = 12, then marks in English = 30 – 12 = 18
In both the cases; after adding 2 marks to mathematics and deducting 3 marks from
English; the product of resultants is 210

Q.6) The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Sol.6) Let us assume, the shorter side = 𝑥
Then; longer side = 𝑥 + 30 and diagonal = 𝑥 + 60
Using Pythagoras Theorem, we get following equation:
⇒ (𝑥 + 60)2 = 𝑥2 + (𝑥 + 30)2
⇒ 𝑥2 + 120𝑥 + 3600
⇒ 𝑥2 + 𝑥2 + 60𝑥 + 900
⇒ − 𝑥2 + 60𝑥 + 2700 = 0
⇒ 𝑥2 − 60𝑥 − 2700 = 0
⇒ 𝑥2 − 90𝑥 + 30𝑥 − 2700 = 0
⇒ 𝑥(𝑥 − 90) + 30(𝑥 − 90) = 0
⇒ (𝑥 + 30)(𝑥 − 90) = 0
⇒ 𝑥 = −30 and 𝑥 = 90
Discarding the negative value, we have 𝑥 = 90 𝑚, longer side = 120 𝑚 and diagonal = 150 𝑚

Q.7) The difference of squares of two numbers is 180. The square of the smaller number is 8 times the large number. Find the two numbers.
Sol.7) Let us assume, larger number = 𝑥
Hence, square of smaller number = 8𝑥
As per question;
⇒ 𝑥2– 8𝑥 = 180
⇒ 𝑥2 – 8𝑥 – 180 = 0
⇒ 𝑥2 – 18𝑥 + 10𝑥 – 180 = 0
⇒ 𝑥(𝑥 – 18) + 10(𝑥 – 18) = 0
⇒ (𝑥 + 10)(𝑥 – 18) = 0
Hence, 𝑥 = − 10 and 𝑥 = 18
Discarding the negative value; 𝑥 = 18
Smaller number
= √8 × 18 = √144 = 12
Hence, the numbers are; 12 and 18

Q.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.
Sol.8) Let us assume, speed of train = 𝑥
We know; time = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒/𝑠𝑝𝑒𝑒𝑑
In case of normal speed;
⇒ 𝑡 = 360/𝑥
In case of increased speed

⇒ 360𝑥 + 1800 = 𝑥2 + 365𝑥
⇒ 𝑥2 + 5𝑥 − 1800 = 0
⇒ 𝑥2 + 45𝑥 − 40𝑥 − 1800 = 0
⇒ 𝑥(𝑥 + 45) − 40(𝑥 + 45) = 0
⇒ (𝑥 − 40)(𝑥 + 45) = 0
⇒ 𝑥 = 40 and 𝑥 = −45
Discarding the negative value; we have speed of train = 40 𝑘𝑚/ℎ

Q.9) Two water taps together can fill a tank 9 and 3/8 hours. The tap of larger diameter takes 10 hour less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Sol.9) Let us assume, smaller tap takes x hours to fill the tank
Then, time taken by larger tap = 𝑥 – 10
In 1 hour, the smaller tap will fill 1/𝑥 of tank
In 1 hour, the larger tap will fill 1/𝑥−10 of tank.
As per question

⇒ 150𝑥 − 750 = 8𝑥2 − 80𝑥
⇒ 8𝑥2 − 80𝑥 − 150𝑥 − 750 = 0
⇒ 8𝑥2 − 230𝑥 + 750 = 0
⇒ 4𝑥2 ​​​​​​​ − 115𝑥 + 375 = 0
⇒ 4𝑥2 ​​​​​​​ − 100𝑥 − 15𝑥 + 375 = 0
⇒ 4𝑥(𝑥 − 25) − 15(𝑥 − 25) = 0
⇒ (4𝑥 − 15)(𝑥 − 25) = 0
⇒ 𝑥 = 15/4 and 𝑥 = 25
Since 15/4
is less than the difference in their individual timings hence time taken by smaller tap = 25 ℎ𝑜𝑢𝑟𝑠 and that by larger tap = 15 ℎ𝑜𝑢𝑟𝑠

Q.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.
Sol.10) Let us assume, speed of passenger train = 𝑥, then speed of express train = 𝑥 + 11

⇒ 132𝑥 + 1452 = 𝑥2 + 143𝑥
⇒ 𝑥2 + 143𝑥 − 132𝑥 − 1452 = 0
⇒ 𝑥2 + 11𝑥 − 1452 = 0
⇒ 𝑥2 + 44𝑥 − 33𝑥 − 1452 = 0
⇒ 𝑥(𝑥 + 44) − 33(𝑥 + 44) = 0
⇒ (𝑥 − 33)(𝑥 + 44) = 0
Hence, 𝑥 = 33 and 𝑥 = − 44
Discarding the negative value, speed of passenger train = 33 km/h and speed of express train = 44 km/h
Average speed can be calculated as follows:
33+44/2 = 38.5 𝑘𝑚/ℎ

Q.11) Sum of the areas of two squares is 468 square meter. If the difference of the perimeters is 24 𝑚, find the sides of the two squares.
Sol.11) We know perimeter = 4 × 𝑠𝑖𝑑𝑒
If 𝑥 and 𝑦 are the sides of two squares, then;
⇒ 4𝑥 – 4𝑦 = 24
⇒ 𝑥 – 𝑦 = 6
⇒ 𝑦 = 𝑥 – 6
Now sum of areas can be given by following equation:
⇒ 𝑥2 + (𝑥 − 6)2 = 468
⇒ 𝑥2 + 𝑥2 − 12𝑥 + 36 = 468
⇒ 2𝑥2 − 12𝑥 + 36 − 468 = 0
⇒ 2𝑥2 − 12𝑥 − 432 = 0
⇒ 𝑥2 − 6𝑥 − 216 = 0
⇒ 𝑥2 − 18𝑥 + 12𝑥 − 216 = 0
⇒ 𝑥(𝑥 − 18) + 12(𝑥 − 18) = 0
⇒ (𝑥 + 12)(𝑥 − 18) = 0
Hence; 𝑥 = −12 and 𝑥 = 18
Hence, sides of squares are; 12 𝑚 and 18 𝑚

Exercise 4.4

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

Sol.1) i) 2𝑥2 − 3𝑥 + 5
We have a = 2, b = - 3 and c = 5
𝐷 = 𝑏2 − 4𝑎𝑐
= (−3)2 − 4 × 2 × 5
= 9 − 40 = −31
Here; 𝐷 < 0; hence no real root is possible.
(ii): 3𝑥2 − 4√3 𝑥 + 4 = 0
Solution: We have; 𝑎 = 3, 𝑏 = − 4√3 and 𝑐 = 4
𝐷 = 𝑏2 − 4𝑎𝑐
= (−4√3)2 − 4 × 3 × 4
= 48 − 48 = 0
Here; 𝐷 = 0; hence root are equal and real.
Root can be calculated as follows:

Q.2) Find the value of 𝑘 for each of the following quadratic equations, so that they have two equal roots.
Sol.2) (i) 2𝑥2 + 𝑘𝑥 + 3 = 0
We have; 𝑎 = 2, 𝑏 = 𝑘 and 𝑐 = 3
For equal roots; D should be zero.
Hence; 𝑏2 − 4𝑎𝑐 = 0
⇒ 𝑘2 − 4 × 2 × 3 = 0
⇒ 𝑘2 − 24 = 0
⇒ 𝑘2 = 24
𝑘 = 2√6

(ii) 𝑘𝑥(𝑥 – 2) + 6 = 0
⇒ 𝑘𝑥(𝑥 – 2) + 6 = 0
⇒ 𝑘𝑥2– 2𝑘𝑥 + 6 = 0
Here; 𝑎 = 𝑘, 𝑏 = − 2𝑘 and 𝑐 = 6
For equal roots, D should be zero
𝑏2 − 4𝑎𝑐 = 0
⇒ (−2𝑘)2 − 4 × 𝑘 × 6 = 0
⇒ 4𝑘2 − 24𝑘 = 0
⇒ 4𝑘2 = 24𝑘
⇒ 𝑘2 = 6𝑘
𝑘 = 6

Q.3) Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 square meter? If so, find its length and breadth.
Sol.3) Let us assume, breadth = 𝑥, then length = 2𝑥
As per question;
⇒ 2𝑥2 = 800
⇒ 𝑥2 = 400
⇒ 𝑥 = 20
Hence, length = 40 𝑚 and breadth = 20 𝑚

Q.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.
Sol.4) Let us assume, age of one friend = 𝑥, then age of another friend = 20 – 𝑥
Four years ago, age of first friend = 𝑥 – 4
Four years ago, age of second friend = 16 – 𝑥
As per question;
⇒ (𝑥 – 4)(16 – 𝑥) = 48
⇒ 16𝑥 – 𝑥2 – 64 + 4𝑥 = 48
⇒ 20𝑥 – 𝑥2 – 64 – 48 = 0
⇒ 20𝑥 – 𝑥2 – 112 = 0
⇒ 𝑥2– 20𝑥 + 112 = 0
Let us check the existence of root;
𝐷 = 𝑏2 − 4𝑎𝑐
= (−20)2 − 4 × 112
= 400 − 448 = −48
Here; 𝐷 < 0, hence no real root is possible. The given situation is not possible.

Q.5) Is it possible to design a rectangular park of perimeter 80m and area 400 square meter?
If so, find its length and breadth.
Sol.5) Perimeter = 2(𝑙𝑒𝑛𝑔𝑡ℎ + 𝑏𝑟𝑒𝑎𝑑𝑡ℎ)
2(length + breadth) = 80 𝑚
length + breadth = 40 𝑚
If length is assumed to be 𝑥, then breadth = 40 – 𝑥
As per question;
⇒ 𝑥(40 – 𝑥) = 400
⇒ 40𝑥 – 𝑥2 = 400
⇒ 40𝑥 – 𝑥2 – 400 = 0
⇒ 𝑥2– 40𝑥 + 400 = 0
Let us check the existence of roots:
𝐷 = 𝑏2 − 4𝑎𝑐
= (−40)2 − 4 × 400
= 1600 − 1600 = 0
Here; D = 0, hence roots are possible.
Now, root can be calculated as follows:
𝑅𝑜𝑜𝑡 = − 𝑏/2𝑎 = 40/2 = 20𝑚
This is a square with side 20 𝑚

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