MAK LABS — Class 10 Maths Chapter 4 Quadratic Equations — NCERT Exercise 4.1, 4.2 & 4.3 Solutions

1Check whether the following are quadratic equations:

An equation is quadratic if it can be written as ax² + bx + c = 0, where a ≠ 0.

(i) (x + 1)² = 2(x – 3)

x² + 2x + 1 = 2x – 6 ⇒ x² + 7 = 0

It is of the form ax² + bx + c = 0. Yes, it is a quadratic equation.

(ii) x² – 2x = (–2)(3 – x)

x² – 2x = –6 + 2x ⇒ x² – 4x + 6 = 0

It is of the form ax² + bx + c = 0. Yes, it is a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

x² – x – 2 = x² + 2x – 3 ⇒ –3x + 1 = 0

The highest power of x is 1. No, it is not a quadratic equation.

(iv) (x – 3)(2x + 1) = x(x + 5)

2x² – 5x – 3 = x² + 5x ⇒ x² – 10x – 3 = 0

It is of the form ax² + bx + c = 0. Yes, it is a quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

2x² – 7x + 3 = x² + 4x – 5 ⇒ x² – 11x + 8 = 0

It is of the form ax² + bx + c = 0. Yes, it is a quadratic equation.

(vi) x² + 3x + 1 = (x – 2)²

x² + 3x + 1 = x² – 4x + 4 ⇒ 7x – 3 = 0

The highest power of x is 1. No, it is not a quadratic equation.

(vii) (x + 2)³ = 2x(x² – 1)

x³ + 6x² + 12x + 8 = 2x³ – 2x ⇒ x³ – 6x² – 14x – 8 = 0

The highest power of x is 3. No, it is not a quadratic equation.

(viii) x³ – 4x² – x + 1 = (x – 2)³

x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8 ⇒ 2x² – 13x + 9 = 0

It is of the form ax² + bx + c = 0. Yes, it is a quadratic equation.

2Represent 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.

Let the breadth be x m. Then the length = (2x + 1) m.

Area = x(2x + 1) = 528 ⇒ 2x² + x – 528 = 0

Required equation 2x² + x – 528 = 0

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

Let the integers be x and x + 1.

x(x + 1) = 306 ⇒ x² + x – 306 = 0

Required equation x² + x – 306 = 0

(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.

Let Rohan’s present age be x years. Then his mother’s age = (x + 26) years.

After 3 years, their ages will be (x + 3) and (x + 29) years.

(x + 3)(x + 29) = 360 ⇒ x² + 32x + 87 = 360 ⇒ x² + 32x – 273 = 0

Required equation x² + 32x – 273 = 0

(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.

Let the speed of the train be x km/h.

Time taken = 480x hours ; at reduced speed, time = 480x – 8 hours

480x – 8 – 480x = 3 ⇒ 480x – 480(x – 8) = 3x(x – 8)

3840 = 3x² – 24x ⇒ x² – 8x – 1280 = 0

Required equation x² – 8x – 1280 = 0

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

(i) x² – 3x – 10 = 0

x² – 5x + 2x – 10 = 0 ⇒ x(x – 5) + 2(x – 5) = 0 ⇒ (x – 5)(x + 2) = 0

Roots 5 and –2

(ii) 2x² + x – 6 = 0

2x² + 4x – 3x – 6 = 0 ⇒ 2x(x + 2) – 3(x + 2) = 0 ⇒ (x + 2)(2x – 3) = 0

Roots 32 and –2

(iii) √2x² + 7x + 5√2 = 0

√2x² + 5x + 2x + 5√2 = 0 ⇒ x(√2x + 5) + √2(√2x + 5) = 0 ⇒ (√2x + 5)(x + √2) = 0

Roots –5√2 and –√2

(iv) 2x² – x + 18 = 0

Multiplying by 8: 16x² – 8x + 1 = 0.

16x² – 4x – 4x + 1 = 0 ⇒ 4x(4x – 1) – 1(4x – 1) = 0 ⇒ (4x – 1)(4x – 1) = 0

Roots 14 and 14

(v) 100x² – 20x + 1 = 0

100x² – 10x – 10x + 1 = 0 ⇒ 10x(10x – 1) – 1(10x – 1) = 0 ⇒ (10x – 1)(10x – 1) = 0

Roots 110 and 110

2Solve 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. Find out how many marbles they had to start with.

From Example 1, the equation is x² – 45x + 324 = 0, where x is the number of marbles John had.

x² – 36x – 9x + 324 = 0 ⇒ (x – 36)(x – 9) = 0 ⇒ x = 36 or x = 9

If John had 36 marbles, Jivanti had 45 – 36 = 9. If John had 9, Jivanti had 36.

John had 36 and Jivanti had 9 marbles (or John 9 and Jivanti 36)

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

From Example 1, the equation is x² – 55x + 750 = 0, where x is the number of toys.

x² – 25x – 30x + 750 = 0 ⇒ (x – 25)(x – 30) = 0 ⇒ x = 25 or x = 30

Number of toys produced 25 or 30

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

Let one number be x. Then the other number = 27 – x.

x(27 – x) = 182 ⇒ x² – 27x + 182 = 0

x² – 13x – 14x + 182 = 0 ⇒ (x – 13)(x – 14) = 0 ⇒ x = 13 or 14

The numbers are 13 and 14

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

Let the integers be x and x + 1.

x² + (x + 1)² = 365 ⇒ 2x² + 2x – 364 = 0 ⇒ x² + x – 182 = 0

x² + 14x – 13x – 182 = 0 ⇒ (x + 14)(x – 13) = 0 ⇒ x = 13 or x = –14

The integers are positive, so x = 13.

The integers are 13 and 14

5The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let the base be x cm. Then the altitude = (x – 7) cm.

By Pythagoras theorem:

x² + (x – 7)² = 13² ⇒ 2x² – 14x + 49 = 169 ⇒ x² – 7x – 60 = 0

(x – 12)(x + 5) = 0 ⇒ x = 12 or x = –5

Length cannot be negative, so x = 12. Altitude = 12 – 7 = 5.

Base = 12 cm and altitude = 5 cm

6A 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.

Let the number of articles be x. Then the cost of each article = ₹ (2x + 3).

x(2x + 3) = 90 ⇒ 2x² + 3x – 90 = 0

2x² + 15x – 12x – 90 = 0 ⇒ (2x + 15)(x – 6) = 0 ⇒ x = 6 or x = –152

The number of articles cannot be negative, so x = 6. Cost of each article = 2(6) + 3 = ₹ 15.

Number of articles = 6 and cost of each article = ₹ 15

1Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x² – 3x + 5 = 0 (ii) 3x² – 4√3x + 4 = 0 (iii) 2x² – 6x + 3 = 0

For ax² + bx + c = 0, the discriminant is b² – 4ac.

(i) 2x² – 3x + 5 = 0

a = 2, b = –3, c = 5 ; b² – 4ac = 9 – 40 = –31 < 0

No real roots

(ii) 3x² – 4√3x + 4 = 0

a = 3, b = –4√3, c = 4 ; b² – 4ac = 48 – 48 = 0

So, the equation has two equal real roots.

x = –b2a = 4√36 = 2√3

Two equal real roots 2√3 and 2√3

(iii) 2x² – 6x + 3 = 0

a = 2, b = –6, c = 3 ; b² – 4ac = 36 – 24 = 12 > 0

So, the equation has two distinct real roots.

x = –b ± √b² – 4ac2a = 6 ± √124 = 6 ± 2√34 = 3 ± √32

Two distinct real roots 3 + √32 and 3 – √32

2Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x² + kx + 3 = 0 (ii) kx(x – 2) + 6 = 0

For two equal roots, b² – 4ac = 0.

(i) 2x² + kx + 3 = 0

a = 2, b = k, c = 3 ; k² – 4(2)(3) = 0 ⇒ k² = 24 ⇒ k = ±2√6

k = 2√6 or –2√6

(ii) kx(x – 2) + 6 = 0

kx² – 2kx + 6 = 0 ; a = k, b = –2k, c = 6

4k² – 24k = 0 ⇒ 4k(k – 6) = 0 ⇒ k = 0 or k = 6

If k = 0, the equation is not quadratic. So, k = 6.

k = 6

3Is 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.

Let the breadth be x m. Then the length = 2x m.

2x × x = 800 ⇒ 2x² – 800 = 0 ⇒ x² = 400

Discriminant = 0² – 4(2)(–800) = 6400 > 0, so real roots exist.

x = ±20 ; breadth cannot be negative, so x = 20

Yes, it is possible. Length = 40 m and breadth = 20 m

4Is 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.

Let the age of one friend be x years. Then the other is (20 – x) years.

Four years ago, their ages were (x – 4) and (16 – x) years.

(x – 4)(16 – x) = 48 ⇒ –x² + 20x – 64 = 48 ⇒ x² – 20x + 112 = 0

Discriminant = (–20)² – 4(1)(112) = 400 – 448 = –48 < 0

There are no real roots.

The situation is not possible.

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

Perimeter = 80 m, so length + breadth = 40 m.

Let the length be x m. Then the breadth = (40 – x) m.

x(40 – x) = 400 ⇒ x² – 40x + 400 = 0

Discriminant = (–40)² – 4(1)(400) = 1600 – 1600 = 0

So, there are two equal real roots.

x = 402 = 20 ; breadth = 40 – 20 = 20

Yes, it is possible. Length = 20 m and breadth = 20 m

The park will be a square of side 20 m.

4. Quadratic Equations

Exercise 4.1

Exercise 4.2

Exercise 4.3