NCERT Solution for Quadratic Equations

(i) (x + 1)2 = 2(x – 3)

(ii) x^{2} – 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) x^{2} + 3x +1 = (x – 2)^{2}

(vii) (x + 2)^{3} = 2x(x^{2} – 1)

(viii) x^{3} – 4x^{2} – × + 1 = (x – 2)^{3}

Sol. (i) (x + 1)^{2} = 2(x – 3)

We have:

(x + 1)^{2} = 2 (x – 3) x^{2} + 2x + 1 = 2x – 6

⇒ x^{2} + 2x + 1 – 2x + 6 = 0

⇒ x^{2} + 70

Since x^{2} + 7 is a quadratic polynomial

∴ (x + 1)^{2} = 2(x – 3) is a quadratic equation.

(ii) x^{2}– 2x = (–2) (3 – x)

We have:

x^{2} – 2x = (– 2) (3 – x)

⇒ x^{2} – 2x = –6 + 2x

⇒ x^{2} – 2x – 2x + 6 = 0

⇒ x^{2} – 4x + 6 = 0

Since x^{2} – 4x + 6 is a quadratic polynomial

∴ x^{2} – 2x = (–2) (3 – x) is a quadratic equation.

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

We have:

(x – 2) (x + 1) = (x – 1) (x + 3)

⇒ x^{2} – x – 2 = x^{2} + 2x – 3

⇒ x^{2} – x – 2 – x^{2} – 2x + 3 = 0

⇒ –3x + 1 = 0

Since –3x + 1 is a linear polynomial

∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.

(iv) (x – 3) (2x + 1) = x(x + 5)

We have:

(x – 3) (2x + 1) = x(x + 5)

⇒ 2x^{2} + x – 6x – 3 = x^{2} + 5x

⇒ 2x^{2} – 5x – 3 – x^{2} – 5x – 0

⇒ x^{2} + 10x – 3 = 0

Since x^{2} + 10x – 3 is a quadratic polynomial

∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.

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

We have:

(2x – 1) (x – 3) = (x + 5) (x – 1)

⇒ 2x^{2} – 6x – x + 3 = x^{2} – x + 5x – 5

⇒ 2x^{2} – x^{2} – 6x – x + x – 5x + 3 + 5 = 0

⇒ x^{2} – 11x + 8 = 0

Since x^{2} – 11x + 8 is a quadratic polynomial

∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a quadratic equation.

(vi) x^{2} + 3x + 1 = (x – 2)^{2}

We have:

x^{2} + 3x + 1 = (x – 2)^{2}

⇒ x^{2} + 3x + 1 = x^{2} – 4x + 4

⇒ x^{2} + 3x + 1 – x^{2} + 4x – 4 =0

⇒ 7x – 3 = 0

Since 7x – 3 is a linear polynomial.

∴ x^{2} + 3x + 1 = (x – 2)^{2} is not a quadratic equation.

(vii) (x + 2)^{3} = 2x(x^{2} – 1)

We have:

(x + 2)^{3} = 2x(x^{2} – 1)

x^{3} + 3x^{2}(2) + 3x(2)^{2} + (2)^{3} = 2x^{3} – 2x

⇒ x^{3} + 6x^{2} + 12x + 8 = 2x^{3} – 2x

⇒ x^{3} + 6x^{2} + 12x + 8 – 2x^{3} + 2x = 0

⇒ –x^{3} + 6x^{2} + 14x + 8 = 0

Since –x^{3} + 6x^{2} + 14x + 8 is a polynomial of degree 3

∴ (x + 2)^{3} = 2x(x^{2} – 1) is not a quadratic equation.

(viii) x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}

We have:

x^{3} – 4x^{2} – x + 1 = (x – 2)^{3}

⇒ x^{3} – 4x^{2} – x + 1 = x^{3} + 3x^{2}(– 2) + 3x(– 2)^{2} + (– 2)^{3}

⇒ x^{3} – 4x^{2} – x + 1 = x^{3} – 6x^{2} + 12x – 8

⇒ x^{3} – 4x^{2} – x – 1 – x^{3} + 6x^{2} – 12x + 8 = 0

2x^{2} – 13x + 9 = 0

Since 2x^{2} – 13x + 9 is a quadratic polynomial

∴ x^{3} – 4x^{2} – x + 1 = (x – 2)^{3} is a quadratic equation.

(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.

Sol. (i) Let the breadth = x metres

Length = 2 (Breadth) + 1

Length = (2x + 1) metres

Since Length × Breadth = Area

∴ (2x + 1) × x = 528

2x^{2} + x = 528

2x^{2} + x – 528 = 0

Thus, the required quadratic equation is

2x^{2} + x – 528 = 0

(ii) Let the two consecutive numbers be x and (x + 1).

∵ Product of the numbers = 306

∴ x (x + 1) = 306

⇒ x^{2} + x = 306

⇒ x^{2} + x – 306 = 0

Thus, the required equdratic equation is

x^{2} + x – 306 = 0

(iii) Let the present age = x

∴ Mother’s age = (x + 26) years

After 3 years

His age = (x + 3) years

Mother’s age = [(x + 26) + 3] years

= (x + 29) years

According to the condition,

⇒ (x + 3) × (x + 29) = 360

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 29x + 3x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

Thus, the required quadratic equation is

x^{2} + 32x – 273 = 0

(iv) Let the speed of the tram = u km/hr

Distance covered = 480 km

Time taken = Distance + Speed

= (480 ÷ u) hours

Speed = (u – 8) km/ hour

According to the condition,

⇒ 480u – 480(u – 8) = 3u(u – 8)

⇒ 480u – 480u + 3840 = 3u^{2} – 24u

⇒ 3840 – 3u^{2} + 24u = 0

⇒ 1280 – u^{2} + 8u = 0

⇒ –1280 + u^{2} – 8u = 0

⇒ u^{2} – 8u – 1280 = 0

Thus, the required quadratic equation is

u^{2} – 8u – 1280 = 0

(i) x^{2} – 3x – 10 = 0

(ii) 2x^{2} + x – 6 = 0

(iii)

(iv) 2x^{2} – x + 8 = 0

(v) 100x^{2} – 20x + 1 = 0

Sol. (i) x^{2} – 3x – 10 = 0

We have:

x^{2} – 3x – 10 = 0

⇒ x^{2} – 5x + 2x – 10 = 0

⇒ x (x – 5) + 2(x – 5) = 0

⇒ (x – 5) (x + 2) = 0

Either x – 5 = 0 ⇒ x = 5

or x + 2 = 0 ⇒ x = –2

Thus, the required roots are x = 5 and x = –2.

(ii) 2x^{2} + x – 6 = 0

We have:

2x^{2} + x – 6 = 0

2x^{2} + 4x – 3x – 6 = 0

2x(x + 2) – 3 (x + 2) = 0

(x + 2) (2x – 3) = 0

Either x + 2 = 0 ⇒ x = –2

or 2x – 3 = 0 ⇒ x = –2

Thus, the required roots are x = –2 and

(v) 100x^{2} – 20x + 1 = 0

We have:

100x^{2} – 20x + 1 = 0

100x^{2} – 10x – 10x + 1 = 0

10x(10x – 1) – 1 (10x – 1) = 0

(10x – 1) (10x – 1) = 0

(10x – 1) = 0 and (10x – 1) = 0

Sol. (i) We have:

x^{2} – 45x + 324 = 0

x^{2} – 9x – 36x + 324 = 0

x(x – 9) – 36(x – 9) = 0

(x – 9) (x – 36) = 0

Either x – 9 = 0 ⇒ x = 9

or x – 36 = 0 ⇒ x = 36

Thus, x = 9 and × = 36

(ii) We have:

x^{2} – 55x + 750 = 0

x^{2} – 30x – 25x + 750 = 0 ∵(– 30) × (– 25) = 750 and

x (x – 30) – 25 (x – 30) = 0 (– 30) + (– 25) = – 55

(x – 30) (x – 25) = 0

Either x – 30 = 0 ⇒ x = 30

or x – 25 = 0 ⇒ x = 25

Thus, x = 30 and x = 25.

Sol. Here, sum of the numbers is 27.

Let one of the numbers be x.

∴ Other number = 27 – x

According to the condition,

Product of the numbers = 182

⇒ x (27 – x) = 182

⇒ 27x – x^{2} = 182

⇒ –x^{2} + 27x – 182 = 0

⇒ x^{2} – 27x + 182 = 0

⇒ x^{2} – 13x – 14x + 182 = 0 –27 = (–13) + (– 14) and

⇒ x (x – 13) – 14 (x – 13) = 0 (– 13) × (– 14) = 182

⇒ (x – 13) (x – 14) = 0

Either x – 13 = 0 ⇒ x = 13

or x – 14 = 0 ⇒ x = 14

Thus, the required numbers are 13 and 14.

Sol. Let the two consecutive positive integers be x and (x + 1).

Since the sum of the squares of the numbers = 365

x^{2} + (x + 1)^{2} = 365

x^{2} + [x^{2} + 2x + 1] = 365

x^{2} + x^{2} + 2x + 1 = 365

2x^{2} + a + 1 – 365 = 0

2x^{2} + 2x – 364 =0

x^{2} + x – 182 = 0

x^{2} + 14x – 13x – 182 = 0 ∵ +14 –13 = 1 and 14 × (–13) = ‐ 182

x(x + 14) –13 (x + 14) = 0

(x +14) (x – 13) = 0

Either x + 14 = 0 ⇒ x = – 14

or x – 13 = 0 = x = 13

Since x has to be a positive integer

∴ x = 13

⇒ x + 1 = 13 + 1 =1 4

Thus, the required consecutive positive integers are 13 and 14.

Sol. Let the base of the given right triangle be ‘x’ cm.

∴ Its height = (x – 7) cm

Squaring both sides, we get

169 = x^{2} + (x – 7)^{2}

⇒ 169 = x^{2} + x^{2} – 14x + 49

⇒ 14x + 49 – 169 = 0

⇒ 2x^{2} – 14x – 120 = 0

⇒ x^{2} – 7x – 60 = 0

⇒ x^{2} – 12x + 5x – 60 = 0

⇒ x (x – 12) + 5 (x – 12) 0

⇒ (x–12) (x + 5) = 0

Either x –12 = 0 ⇒ x=12

or x + 5 = 0 ⇒ x = –5

But the side of triangle can never be negative,

⇒ x = 12

∴ Length of the base = 12 cm

⇒ Length of the height = (12 – 7) cm = 5 cm

Thus, the required base = 12 cm and height = 5 cm,

Sol. Let the number of articles produced in a day = x

∴ Cost of production of each article = Z (2x + 3)

According to the condition,

Total cost = 90

⇒ x × (2x + 3) = 90

⇒ 2x^{2} + 3x = 90

⇒ 2x^{2} + 3x – 90 = 0

⇒ 2x^{2} – 12x + 15x – 90 = 0

⇒ 2x(x – 6) + 15(x – 6) = 0

⇒ (x – 6) (2x+15) = 0

⇒ x – 6 = 0 ⇒ x = 6

But the number of articles cannot be negative.

⇒ x = 6

∴ Cost of each article = Rs (2 × 6 + 3) = Rs. 15

Thus, the required number of articles produced is 6 and the cost of each article is Rs. 15.

Exercise 4.3

(i) 2x^{2} – 7x + 3 = 0

(ii) 2x^{2} + x – 4 = 0

(iii) 4x^{2} + 4+ 3 = 0

(iv) 2x^{2} + x + 4 = 0

Sol. (i) 2x^{2} – 7x + 3 = 0

(ii) 2x^{2} + x – 4 = 0

We have:

2x^{2} + x – 4 = 0

Dividing throughout by 2,

(i) 2x^{2} – 7x + 3 = 0

(ii) 2x^{2} + x – 4 = 0

(iii)

(iv) 2x^{2} + x + 4 = 0

Sol. (i) 2x^{2} – 7x + 3 = 0

Comparing the given equation with ax^{2} + bx + c = 0, we have

a = 2

b = –7

c = 3

∴ b^{2} – 4ac = (–7)^{2} – 4 (2)(3)

= 49 – 24 = 25 0

Since b^{2} – 4ac > 0

∴ The given equation has real roots. The roots are given by

Taking +ve sign,

Taking –ve sign,

Thus, the roots of the given equation are

(ii) 2x^{2} + x – 4 = 0

Comparing the given equation with ax^{2} + bx + c = 0 we have:

a = 2

b = 1

c = –4

∴ b^{2} – 4ac = (1)^{2} – 4(2)(–4)

= 1 + 32

= 33 > 0

Since b^{2} – 4ac > 0

∴ The given equation has equal roots. The roots are given by

Since b^{2} – 4ac = 0

∴ The given equation has real roots, which are given by:

(iv) 2x^{2} + x + 4 = 0

Comparing the given equation with ax^{2} + bx + c = 0 we have:

a = 2

b = 1

c = 4

∴ b^{2} – 4ac = (1)^{2} – 4(2)(4)

= 1 – 32

= –31 > 0

Since b^{2} – 4ac is less than 0, therefore the given equation does not have real roots.

Find his present age.

Sol. Let the present age of Rehman = x

∴ 3 years ago Rehman’s age = (x – 3) years

5 years later Rehman’s age = (x + 5) years

Now, according to the condition,

⇒ 3 [x + 5 + x – 3] = (x – 3) (x + 5)

⇒ 3[2x + 2] = x^{2} + 2x – 15

⇒ 6x + 6 = x^{2} + 2x – 15

⇒ x^{2} + 2x – 6x – 15 – 6 = 0

⇒ x^{2} – 4x – 21 = 0

Now, comparing (1) with ax^{2} + bx + c = 0, we have:

a = 1

b = – 4

c = – 21

b^{2} – 4ac = ( – 4)^{2} – 4 (1) ( – 21)

= 16 + 84

= 100

Since,

Sol. Let, Shefali’s marks in Mathematics = x

∴ Marks in English = (30 – x) [∵Sum of their marks in Eng. and Maths = 30]

Now, according to the condition,

(x + 2) × [(30 – x) – 3] = 210

⇒ (x + 2) × (30 – × – 3) = 210

⇒(x + 2) ( – x + 27) 210

⇒ – x^{2} + 25x + 54 = 210

⇒ – x^{2} + 25x + 54 – 210 = 0

⇒ – x^{2} + 25x – 156 = 0

⇒ x^{2} – 25x + 156 = 0 ...(1)

Now, comparing (1) with ax^{2} + bx + c = 0

a = 1

b = – 25

c = 156

b^{2} – 4ac = ( – 25)^{2} – 4(1) (156)

= 625 – 624 = 1

Since,

When x = 13, then 30 – 13 = 17

When x = 12, then 30 – 12 = 18

Thus, marks in Maths = 13, marks in English = 17

marks in Maths = 12, marks in English = 18

Sol. Let the shorter side (i.e., breadth) = x metres.

∴ The longer side (length) = (x + 30) metres.

In a rectangle,

⇒ (x + 60)^{2} = 2x^{2} + 60x + 900

⇒ x^{2} + 120x + 3600 = 2x^{2} + 60x + 900

⇒ 2x^{2} – x^{2} + 60x – 120x + 900 – 3600 = 0

⇒ x^{2} – 60x – 2700 = 0

Comparing (1) with ax^{2} + bx + c = 0

a = 1

b= –60

c = –2700

∴ b^{2} – 4ac = (–60)^{2} – 4(1) (–2700)

⇒ b^{2} – 4ac = 3600 + 10800

⇒ b^{2} – 4ac = 14400

Since,

Since breadth cannot be negative,

∴ x ≠ – 30 ⇒ x = 90

∴ x + 30 = 90 + 30=120

Thus, the shorter side = 90 m

The longer side = 120 m.

Sol. Let the larger number be x. Since,

(smaller number)^{2} = 8 (larger number)

⇒ (smaller number)^{2} = 8x

Sol. Let the uniform speed of the train be x km/hr.

When speed is 5 km/hr more then time is 1 hour less.

Sol. Let the smaller tap fills the tank in × hours.

∴ The larger tap fills the tank (x – 10) hours.

Sol. Let the average speed of the passenger train be x km/ hr

∴ Average speed of the express train = (x + 11) km/hr

Total distance covered = 132 km

According to the condition, we get

But average speed cannot be negative

∴ x – 44

∴ x = 33

⇒ Average speed of the passenger train = 33 kin/hr

∴ Average speed of the express train = (x + 11) km/hr

= (33 + 11) km/hr

= 44 km/hr

Sol. Let the side of the smaller square be × m.

⇒ Perimeter of the smaller square = 4x m.

∴ Perimeter of the larger square = (4x + 24) m

∴ Area of the smaller square = (side)^{2} = (x)^{2} m2

Area of the larger square = (x + 6)^{2} m2

According to the condition,

x^{2} + (x + 6)^{2} = 468

⇒ x^{2} + x^{2} + 12x + 36 = 468

⇒ 2x^{2} + 12x – 432 = 0

⇒ x^{2} + 6x – 216 = 0

Comparing (1) with ax^{2} + bx + c = 0

∴ a = 1

b = 6

c = – 216

∴ b^{2} – 4ac = (6)^{2} – 4 (1) (–216)

= 36 + 864 = 900

Since,

But the length of a square cannot be negative

∴ x = 12

⇒ Length of the smaller square = 12 m

Thus, the length of the larger square = x + 6

= 12 + 6

= 18m

(i) 2x^{2} – 3x + 5 = 0 (ii)

(iii) 2x^{2} – 6x + 3 – 0

Sol. (i) 2x^{2} – 3x + 5 = 0

Comparing the given quadratic equation with ax^{2} + bx + c = 0, we have:

a = 2

b= – 3

c = 5

∴ The discriminant = b^{2} – 4ac

= ( – 3)^{2} – 4(2)(5)

= – 9 – 40

= – 31 < 0

Since b^{2} – 4ac is negative.

∴ The given quadratic equation has no real roots.

(i)

Comparing the given quadratic equation with ax^{2} + bx + c= 0, we get

Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:

(iii) 2x^{2} – 6x + 3 = 0

Comparing it with the general quadratic equation, we have:

a = 2

b = –6

c = 3

∴ The given quadratic equation has two real and distinct roots, which are given by

(i) 2x^{2} + kx + 3 = 0

(ii) kx(x – 2) + 6 = 0

Sol. (i) 2x^{2} + kx + 3 = 0

Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get

a = 2

b = k

c = 3

b^{2} – 4ac = ( – k)^{2} – 4 (2) (3)

= k2 – 24

∵ For a quadratic equation to have equal roots,

b^{2} – 4ac = 0

(ii) kx(x – 2) + 6 = 0

Comparing kx (x – 2) + 6 = 0 i.e., kx^{2} – 2kx + 6 = 0 with ax^{2} + bx + c = 0, we get

a = k

b = – a

c = 6

∴ b^{2} – 4ac = ( – 2k)^{2} – 4 (k) (6)

= 4k2 – 24k

Since, the roots are real and equal,

∴ b^{2} – 4ac = 0

⇒ 4k2 – 24k = 0

⇒ 4k(K – 6) = 0

⇒ 4k = 0 or k – 6=0

⇒ k = 0 or k = 6

But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.

Sol. Let the breadth be x metres.

∴ Length = 2x metres

Now, Area = Length × Breadth

= 2x × x metre2

= 2x^{2} sq. metre.

According to the given condition,

2x^{2} = 800

Therefore, x = 20 and x = – 20

But x = – 20 is possible (∵ breadth cannot be negative).

∴ x = 20

⇒ 2x = 2 × 20 = 40

Thus, length = 40 m and breadth = 20 m

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

Sol. Let the age of one friend = x years

∴ The age of the other friend = (20 – x – 4) years [∵ Sum of their ages is 20 years]

Four years ago

Age of one friend = (x – 4) years

Age of the other friend = (20 – x – 4) years

= (16 – x) years

According to the condition,

(x – 4) × (16 – x) = 48

⇒ 16x – 64 – x^{2} – 4x = 48

⇒ – x^{2} – 20x – 64 – 48 = 0

⇒ – x^{2} – 20x – 112 = 0

⇒ x^{2} + 20x + 112 – 0 (1)

Here, a = 1, b = 20 and c = 112

∴ b^{2} – 4ac = (20)^{2} – 4 (1) (112)

= 400 – 448

= – 48 < 0

Since b^{2} – 4ac is Is than O.

∴ The quadratic equation (1) has no real roots.

Thus, the given equation is not possible.

Sol. Let the breadth of the rectangle be x metres.

Since, the parimetee of the rectangle = 80 m

∴ 2 [Length + Breadth] = 80

2 [Length + x] 80

⇒ Length = (40 – x) metres

∴ Area of the rectangle = Length × breadth

= (40 – x) × x sq. m

= 40x – x^{2}

Now, according to the given condition,

Area of the rectangle = 400 m2

∴ 40x – x^{2} = 400

⇒ – x^{2} + 40x – 400 = 0

⇒ x^{2} – 40x + 400 = 0

Comparing (1) with axe + bx + c = 0, we get

a = 1

b = –40

c = 400

∴ b^{2} – 4ac = ( – 40)^{2} – 4 (1) (400)

= 1600 – 1600 – 0

Thus, the equation (1) has two equal and real roots.

∴ Breadth, x = 20 m

∴ Length = (40 – x) = (40 – 20) m = 20 m.