157-NCERT-10-4-Quadratic Equations - Ex-4.2

NCERT

10th Mathematics

Exercise 4.2

Topic: 4 Quadratic Equations

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EXERCISE 4.2

Q1. Find the roots of the following quadratic equations by factorisation:

(i) x² – 3x – 10 = 0     (ii) 2x² + x – 6 = 0

(iii) √2 x² + 7 x + 5√2 = 0     (iv) 2x² – x + 1/8 = 0

(v) 100x² – 20x + 1 = 0 

Explanation:

1) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

2) If α and β are the roots of the equation ax² + bx + c = 0, we have, 

aα² + bα + c = 0 and aβ² + bβ + c = 0. i.e. we say that, 

α and β satisfy the equation ax² + bx + c = 0.

3) We can solve this qardatic equation ax² + bx + c = 0 using factorisation method. 

Solution:

(i) x² – 3x – 10 = 0

1) Quadratic equation: x² – 3x – 10 = 0

here last term is 10 and its sign is minus, factorise 10 in such a way

that their difference will be - 3.

10 = (- 5) x (2) (- 5 + 2 = - 3).

x² – 5x + 2x – 10 = 0

x(x – 5) + 2(x – 5) = 0

(x – 5)(x + 2) = 0

2) So, (x – 5) = 0 or (x + 2) = 0.

3) So, x = 5 or x = - 2.

(ii) 2x² + x – 6 = 0

1) Quadratic equation: 2x² + x – 6 = 0

here last term is 6 and its sign is minus, factorise 2 x 6 in such a way

that their difference will be 1.

2 x 6 = (3) x (- 4) (4 - 3 = 1).

2x² – 3x + 4x – 6 = 0

x(2x – 3) + 2(2x – 3) = 0

(2x – 3)(x + 2) = 0

2) So, (2x – 3) = 0, or (x + 2) = 0

3) So, x = 3/2 or x = – 2.

(iii) √2x² + 7x + 5√2 = 0

1) Quadratic equation: √2x² + 7x + 5√2 = 0

here last term is 5√2 and its sign is plus, factorise √2 x 5√2 in such a way

that their sum will be 7.

√2 x 5√2 = 5 x 2 (5 + 2 = 7).

√2x² + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2(√2x + 5) = 0

(√2x + 5)(x + √2) = 0

2) So, (√2x + 5) = 0, or (x + √2) = 0

3) So, x = (- 5/√2) or x = - √2.

(iv) 2x² – x + 1/8 = 0

1) Quadratic equation: 2x² – x + 1/8 = 0

2) Multiply by 8, and we get: 16x² – 8x + 1 = 0

here last term is 1 and its sign is plus, factorise 16 x 1 in such a way

that their sum will be – 8.

16 x 1 = (– 4) x (–4) (– 4 – 4 = – 8).

16x² – 4x – 4x + 1 = 0

4x(4x – 1) – (4x – 1) = 0

(4x – 1)(4x – 1) = 0

3) So, (4x – 1) = 0, or (4x – 1) = 0

4) So, x = 1/4 or x = 1/4.

(v) 100x² – 20x + 1 = 0

1)  Quadratic equation: 100x² – 20x + 1 = 0

here last term is 1 and its sign is plus, factorise 100 x 1 in such a way

that their sum will be – 20.

100 x 1 = (– 10) x (– 10) (– 10 – 10 = – 20).

100x² – 10x – 10x + 1 = 0

10x(10x – 1) – (10x – 1) = 0

(10x – 1)(10x – 1) = 0

2) So, (10x – 1) = 0, or (10x – 1) = 0

3) So, x = 1/10 or x = 1/10.

Q2. 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.

(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.

Explanation:

1) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

2) Frame the quadratic equation from the given condition.

3) Solve this qardatic equation ax² + bx + c = 0 using factorisation method. 

Solution:

(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.

1) Let the number of marbles with John be x.

2) So, according to the problem, the number of marbles with Jivanti is (45 - x).

3) Both have lost 5 marbles, so now they have (x - 5) and (40 - x).

4) According to the problem,

(x - 5)(40 - x) = 124

x(40 - x) - 5(40 - x) = 124
40x - x² - 200 + 5x = 124
- x² + 40x + 5x - 200 - 124 = 0
- x² + 45x - 324 = 0
x² - 45x + 324 = 0

5)  Quadratic equation: x² – 45x + 324 = 0

here last term is 324 and its sign is plus, factorise 324 x 1 in such a way

that their sum will be – 45.

324 x 1 = (– 36) x (– 9) (– 36 – 9 = – 45).

x² – 36x – 9x + 324 = 0

x(x – 36) – 9(x – 36) = 0

(x – 36)(x – 9) = 0

6) So, (x – 36) = 0, or (x – 9) = 0

7) So, x = 36 or x = 9.

8) John has 36 marbles and Jivanti has 9 marbles.

(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.

1) Let the number of toys produced in a day be x.

2) So, each toy's production cost on a day will be Rs (55 - x).

3) On one particular day, the total cost was Rs 750.

4) According to the problem,

x(55 - x) = 750

55x - x² = 750
55x - x² - 750 = 0
- x² + 55x - 750 = 0

x² - 55x + 750 = 0

5)  Quadratic equation: x² - 55x + 750 = 0

here last term is 750 and its sign is plus, factorise 750 x 1 in such a way

that their sum will be – 55.

750 x 1 = (– 25) x (– 30) (– 25 – 30 = – 55).

x² – 25x – 30x + 750 = 0

x(x – 25) – 30(x – 25) = 0

(x – 25)(x – 30) = 0

6) So, (x – 25) = 0, or (x – 30) = 0

7) So, x = 25 or x = 30.

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

Q3. Find two numbers whose sum is 27 and whose product is 182.

1) Let the first number be x.

2) The second number will be (27 - x).

3) According to the problem,

x(27 - x) = 182

27x - x² = 182
27x - x² - 182 = 0
- x² + 27x - 182 = 0

x² - 27x + 182 = 0

4)  Quadratic equation: x² - 27x + 182 = 0

here last term is 182 and its sign is plus, factorise 182 x 1 in such a way

that their sum will be – 27.

182 x 1 = (– 14) x (– 13) (– 14 – 13 = – 27).

x² – 14x – 13x + 182 = 0

x(x – 14) – 13(x – 14) = 0

(x – 14)(x – 13) = 0

5) So, (x – 14) = 0, or (x – 13) = 0

6) So, x = 14 or x = 14.

7) Hence, the numbers are 13 and 14.

Q4. Find two consecutive positive integers, the sum of whose squares is 365.

1) Let the first positive integer be x.

2) The second number will be (x + 1).

3) According to the problem,

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x - 365 + 1 = 0

2x² + 2x - 364 = 0

x² + x - 182 = 0

4)  Quadratic equation: x² + x - 365 = 0

here last term is 182 and its sign is minus, factorise 182 x 1 in such a way

that their difference will be 1.

182 x 1 = (14) x (– 13) (14 – 13 = 1).

x² + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x + 14)(x – 13) = 0

6) So, (x + 14) = 0, or (x – 13) = 0

7) So, x = –14 or x = 13.

8) As our integer is positive, we have x = 13. 

9) So the integers are 13 and 14.

Q 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.

1) Here, a right triangle's altitude depends on its base, so let the base be x.

2) So, altitude will be (x - 7).

3) According to the problem,

x² + (x - 7)² = 13²

x² + x² - 14x + 49 = 13²

2x² - 14x + 49 = 169

2x² - 14x + 49 - 169 = 0

2x² - 14x - 120 = 0

x² - 7x - 60 = 0

4)  Quadratic equation: x² - 7x - 60 = 0

here last term is 60 and its sign is minus, factorise 60 x 1 in such a way

that their difference will be - 7.

60 x 1 = (12) x (5) (5 – 12 = - 7).

x² + 5x – 12x – 60 = 0

x(x + 5) – 12(x + 5) = 0

(x + 5)(x – 12) = 0

6) So, (x + 5) = 0, or (x – 12) = 0

7) So, x = – 5 or x = 12.

8) As the sides of a triangle can't be negative, ignore x = - 5, so we have x = 12. 

9) So the base of a triangle is 12 and the altitude is 7.

Q6. 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.

1) Let the number of pottery articles produced in a day be x.

2) So, the cost of production of each article will be Rs (2x + 3).

3) On one particular day, the total cost was Rs 90.

4) According to the problem,

x(2x + 3) = 90

2x² + 3x = 90

2x² + 3x - 90 = 0

2x² + 3x - 90 = 0

5) Quadratic equation: 2x² + 3x - 90 = 0

here last term is 90 and its sign is minus, factorise 90 x 2 in such a way

that their difference will be 3.

90 x 2 = (15) x (– 12) (15 – 12 = 3).

2x² + 15x – 12x + 90 = 0

x(2x + 15) – 6(2x + 15) = 0

(2x + 15)(x – 6) = 0

6) So, (2x + 15) = 0, or (x – 6) = 0

7) So, x = - 15/2 or x = 6.

8) As the number of articles produced can't be negative, so, x can't be - 15/2.

9) So the number of articles produced on that day is 6 and the cost of each article is

Rs 15.

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156-NCERT-10-4-Quadratic Equations - Ex-4.1

NCERT

10th Mathematics

Exercise 4.1

Topic: 4 Quadratic Equations

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EXERCISE 4.1

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

Explanation:

1) The quadratic polynomial is of the form ax² + bx + c, where a ≠ 0.

2) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

3) Equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a

quadratic equation.

4) The equation of the form ax² + bx + c = 0, where a ≠ 0 is known as the standard

form of a quadratic equation. The degree of this equation is 2.

Solution:

(i) (x + 1)² = 2(x – 3)

1) The given equation is:

(x + 1)² = 2(x – 3)

x² + 2x +1 = 2x – 6

x² + 2x +1 - 2x + 6 = 0

x² + 7 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation. 

(ii) x² – 2x = (–2) (3 – x)

1) The given equation is:

x² – 2x = (–2) (3 – x)

x² – 2x = – 6 + 2x

x² – 2x + 6 – 2x = 0

x² – 4x + 6 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation. 

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

1) The given equation is:

(x – 2)(x + 1) = (x – 1)(x + 3)

x(x + 1) – 2(x + 1) = x(x + 3) – (x + 3)

x² + x – 2x – 2 = x² + 3x – x – 3

x² – x – 2 = x² + 2x – 3
x² – x – 2 – x² – 2x + 3 = 0

– 3x + 1 = 0
3x – 1 = 0 -------------------equation 1

2) As the highest power of the variable x is 1, it is not the quadratic equation. 

(iv) (x – 3)(2x +1) = x(x + 5)

1) The given equation is:

(x – 3)(2x +1) = x(x + 5)

x(2x + 1) – 3(2x + 1) = x(x + 5)

2x² + x – 6x – 3 = x² + 5x

2x² + x – 6x – 3 – x² – 5x = 0

x² – 10x – 3 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

1) The given equation is:

(2x – 1)(x – 3) = (x + 5)(x – 1)

2x(x – 3) – (x – 3) = x(x – 1) + 5(x – 1)

2x² – 6x – x + 3 = x² – x + 5x – 5

2x² – 6x – x + 3 – x² + x – 5x + 5 = 0

2x² – x² – 6x – x + x – 5x + 3 + 5 = 0

x² – 11x + 8 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation.

(vi) x² + 3x + 1 = (x – 2)²

1) The given equation is:

x² + 3x + 1 = (x – 2)²

x² + 3x + 1 = x² – 4x + 4

x² + 3x + 1 – x² + 4x – 4 = 0

x² – x² + 3x + 4x + 1 – 4 = 0

7x – 3 = 0-------------------equation 1

2) As the highest power of the variable x is 1, it is not the quadratic equation.

(vii) (x + 2)³ = 2x (x² – 1)

1) We know that (a + b)³ = a³ + 3a² b + 3a b² + b³ 

(x + 2)³ = 2x (x² – 1)

x³ + 3x² (2) + 3x (2)² + 2³ = 2x³ – 2x

x³ + 6x² + 12x + 8 = 2x³ – 2x

x³ + 6x² + 12x + 8 – 2x³ + 2x = 0
x³ – 2x³ + 6x² + 12x + 2x + 8 = 0

– x³ + 6x² + 14x + 8 = 0 -------------------equation 1

2) As the highest power of the variable x is 3, it is not the quadratic equation.

(viii) x³ – 4x² – x + 1 = (x – 2)³

1) We know that (a - b)³ = a³ – 3a² b + 3a b² – b³ 

x³ – 4x² – x + 1 = (x – 2)³

x³ – 4x² – x + 1 = x³ – 3x² (2) + 3x (2)² – 2³

x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8

x³ – 4x² – x + 1 – x³ + 6x² – 12x + 8 = 0
x³ – x³ – 4x² + 6x² – x – 12x + 1 + 8 = 0

2x² – 13x + 9 = 0 -------------------equation 1

2) As the highest power of the variable x is 2, it is the quadratic equation.

Q2. 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.

Explanation:

1) Here, let x be any suitable variable.

2) According to the given equation, frame the quadratic equation.

3) Then solve this equation to get the required solutions. 

Solution:

(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.

1) Here, length depends on breadth, let the breadth be x m.

2) So, according to the problem, the length is more than twice its breadth,

breadth = (2x + 1) m

3) The area of the plot is 528 m², so we have,

x(2x + 1) = 528

2x² + x = 528

2x² + x – 528 = 0 ------------ equation 1

4) So, 2x² + x – 528 = 0 is the required quadratic equation.

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

1) Let the first integer be x.

2) So, the next integer will be (x + 1).

3) The product of the integers is 306, so we have,

x(x + 1) = 306

x² + x = 306

x² + x – 306 = 0 ------------ equation 1

4) So, x² + x – 306 = 0 is the required quadratic equation.

(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.

1) Here, the age of Rohan's mother is depending on him, so let Rohan's age be x.

2) So, according to the problem, the age of Rohan's mother will be (x + 26).

3) 3 years later, Rohan's age will be (x + 3), and his mother's age will be (x + 29). 

4) According to the problem,

(x + 3)(x + 29) = 360

x(x + 29) + 3(x + 29) = 360

x² + 29x + 3x + 87 = 360

x² + 32x + 87 = 360
x² + 32x + 87 – 360 = 0
x² + 32x – 273 = 0 ------------ equation 1

4) So, x² + 32x – 273 = 0 is the required quadratic equation.

(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.

1) Let the uniform speed of a train be x km/h.

2) To cover 480 km, the time taken will be (480/x) hrs.

3) If the speed is reduced by 8 km/h, then time will be increased by 3 hrs, so 

new speed = (x - 8) km/h, and new time = [(480/x) + 3] hrs.

4) According to the condition,

(x - 8)[(480/x) + 3] = 480

(x - 8)[(480 + 3x)/x] = 480

x(x - 8) = (480 + 3x)

x² - 8x = 480 + 3x

x² - 8x - 480 - 3x = 0
x² - 8x - 3x - 480 = 0

x² - 11x - 480 = 0 ------------ equation 1

5) So, x² - 11x - 480 = 0 is the required quadratic equation.

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