**1.** (i) (x + 1)^{2} = 2 (x – 3) {(a + b)^{2} = a^{2} + 2ab + b^{2}}

⇒ x^{2} + 1 + 2x = 2x – 6

⇒ x^{2} + 7 = 0

Here, degree of equation is 2. Therefore, it is a Quadratic Equation.

(ii) x^{2} – 2x = (-2) (3 – x)

⇒ x^{2} – 2x = -6 + 2x x^{2} – 2x – 2x + 6 = 0

⇒ x^{2} – 4x + 6 = 0

Here, degree of equation is 2. Therefore, it is a Quadratic Equation.

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

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

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

⇒ x – 2x – 2 – 3x + x + 3 = 0

⇒ -3x + 1 = 0

Here, degree of equation is 1.

Therefore, it is not a Quadratic Equation.

(iv) (x – 3) (2x + 1) = x (x + 5)

⇒ 2x^{2} + x – 6x – 3 = x^{2} + 5x

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

⇒ x^{2} – 10x – 3 = 0

Here, degree of equation is 2. Therefore, it is a quadratic equation.

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

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

⇒ 2x^{2}– 7x + 3 x^{2} + x – 5x + 5 = 0

⇒ x^{2} – 11x + 8 = 0

Here, degree of Equation is 2. Therefore, it is a Quadratic Equation.

(vi) x^{2} + 3x + 1 = (x-2)^{2} {(a – b)^{2} = a^{2} – 2ab + b^{2}}

⇒ x^{2} + 3x + 1 = x^{2} + 4 – 4x

⇒ x^{2} + 3x + 1 -x^{2} + 4x – 4 = 0

⇒ 7x – 3 = 0

Here, degree of equation is 1.

Therefore, it is not a Quadratic Equation.

(vii) (x + 2)^{3} = 2x (x^{2} – 1) {(a + b)^{3} = a^{3} + b^{3} + 3ab (a + b)}

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

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

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

⇒ x^{3} -6x^{2} – 14x – 8 = 0

Here, degree of Equation is 3.

Therefore, it is not a quadratic Equation.

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

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

⇒ -4x^{2} – x + 1 = -8 – 6x^{2} + 12x

⇒ 2x^{2}– 13x + 9 = 0

Here, degree of Equation is 2. Therefore, it is a Quadratic Equation.

**2.** (i) We are given that area of a rectangular plot is 528 m^{2}.

Let breadth of rectangular plot be x metres

Length is one more than twice its breadth.

Therefore length of rectangular plot is (2x + 1) metres

Area of rectangle = length x breadth

⇒ 528 = x (2x + 1) ⇒ 528 = 2x^{2} + x

⇒ 2x^{2} + x – 528 = 0

This is a Quadratic Equation.

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

It is given that x (x + 1) = 306

⇒ x^{2} + x = 306

⇒ x^{2} + x – 306 = 0

This is a Quadratic Equation.

(iii) Let present age of Rohan = x years

Let present age of Rohan’s mother = (x + 26) years

Age of Rohan after 3 years = (x + 3) years

Age of Rohan’s mother after 3 years = x + 26 + 3 = (x + 29) years

According to given condition:

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

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 32x 273 = 0

This is a Quadratic Equation.

(iv) Let speed of train be x km/h

Time taken by train to cover 480 km = 480x hours

If, speed had been 8km/h less then time taken would be (480x-8) hours

According to given condition, if speed had been 8km/h less then time taken is 3 hours less.

Therefore, 480x – 8 = 480x + 3

⇒ 480 (1x – 8 – 1x) = 3 ⇒ 480 (x – x + 8) (x) (x – 8) = 3

⇒ 480 × 8 = 3 (x) (x – 8) ⇒ 3840 = 3x^{2} – 24x

⇒ 3x^{2} – 24x – 3840 = 0

Dividing equation by 3, we get

⇒ x^{2}– 8x – 1280 = 0

This is a Quadratic Equation.

**3.** (i) x^{2} – 3x – 10 = 0

⇒ x^{2} – 5x + 2x – 10 = 0 ⇒ x (x – 5) + 2 (x – 5) = 0

⇒ (x – 5) (x + 2) = 0 ⇒ x = 5, -2

(ii) 2x^{2}+ x – 6 = 0

⇒ 2x^{2} + 4x – 3x – 6 = 0 ⇒ 2x (x + 2) – 3 (x + 2) = 0

⇒ 16x^{2} – 4x – 4x + 1 = 0 ⇒ 4x (4x – 1) – 1 (4x – 1) = 0

⇒ (4x – 1) (4x – 1) = 0 x = 1/4, 1/4

(v) 100x^{2} – 20x + 1 = 0

⇒ 100x^{2} – 10x 1.0x + 1 = 0 ⇒ 10x (10x – 1) – 1 (10x – 1) = 0

**4.** (i) x^{2} – 45x + 324 = 0

⇒ x^{2} – 36x 9x + 324 = 0 ⇒ x (x – 36) – 9 (x – 36) = 0

⇒ (x – 9) (x – 36) = 0 ⇒ x = 9, 36

(ii) x^{2} – 55x + 750 = 0

⇒ x^{2} – 25x – 30x + 750 = 0 ⇒ x (x – 25)- 30 (x – 25) = 0

⇒ (x – 30) (x – 25) = 0 ⇒ x = 30, 25

**5.** Let first number be x and let second number be (27 – x)

According to given condition, the product of two numbers is 182.

Therefore,

x (27 – x) = 182

⇒ 27x – x^{2}= 182 ⇒ x^{2} – 27x + 182 = 0

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

⇒ (x – 14) (x – 13) = 0 ⇒ x = 14, 13

Therefore, the first number is equal to 14 or 13

And, second number is = 27 – x = 27 – 14 = 13 or Second number = 27 – 13 = 14 Therefore two numbers are 13 and 14.

**6.** Let first number be x and let second number be (x + 1)

According to given condition,

⇒ x^{2} + (x + 1)^{2} = 365 ⇒ {(a + b)^{2} = a^{2} + b^{2}+ 2ab}

⇒ x^{2} + x^{2} + 1 + 2x = 365 ⇒ 2x^{2} + 2x – 364 = 0

Dividing equation by 2

⇒ x^{2}+ x – 182 = 0 ⇒ x^{2} + 14x – 13x – 182 = 0

⇒ x (x + 14) – 13 (x + 14) = 0 ⇒ (x + 14) (x – 13) = 0

⇒ x = 13, -14

Therefore first number = 13 (We discard -14 because it is negative number) Second number = x + 1 = 13 + 1 = 14

Therefore two consecutive positive integers are 13 and 14 whose sum of squares is equal to 365.

**7.** Let base of triangle be x cm and let altitude of triangle be (x – 7) cm It is given that hypotenuse of triangle is 13 cm

According to Pythagoras Theorem,

13^{2} = x^{2} + (x – 7)^{2} (a + b)^{2} = a^{2} + b^{2} + 2ab

⇒169 = x^{2} + x^{2}+ 49 – 14x ⇒ 169 = 2x^{2} – 14x + 49

⇒2x^{2} – 14x – 120 = 0

Dividing equation by 2

⇒ x^{2} – 7x – 60 = 0 ⇒ x^{2} – 12x + 5x – 60 = 0

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

⇒ x = -5, 12

We discard x = -5 because length of side of triangle cannot be negative. Therefore, base of triangle = 12 cm

Altitude of triangle = (x – 7) = 12 – 7 = 5 cm

**8.** Let cost of production of each article be Rs x

We are given total cost of production on that particular day = Rs 90 Therefore, total number of articles produced that day = 90/x According to the given conditions,

⇒ x^{2} = 180 + 3x ⇒ x^{2} – 3x – 180 = 0

⇒ x^{2} – 15x + 12x – 180 = 0 ⇒ x (x – 15) + 12 (x – 15) = 0

⇒ (x – 15) (x + 12) = 0 ⇒ x = 15, -12

Cost cannot be in negative, therefore, we discard x = -12

Therefore, x = Rs 15 which is the cost of production of each article.

Number of articles produced on that particular day

**9. **(i)** **2x^{2} – 7x + 3 = 0

First we divide equation by 2 to make coefficient of x^{2} equal to 1,

We divide middle term of the equation by 2x, we get

We add and subtract square of from the equation

Taking Square root on both sides,

(ii) 2x^{2} + x – 4 = 0

Dividing equation by 2,

Following procedure of completing square,

Taking square root on both sides,

Dividing equation by 4,

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

Dividing equation by 2,

Following the procedure of completing square,

Taking square root on both sides

Right hand side does not exist because square root of negative number does not exist.

Therefore, there is no solution for quadratic equation 2x^{2}+x+4 = 0

**10.** (i) 2x^{2} – 7x + 3 = 0

Comparing quadratic equation 2x^{2} – 7x + 3 = 0 with general form axe + bx + c = 0, we get a = 2, b = -7 and c = 3

(ii) 2x^{2} + x – 4 = 0

Comparing quadratic equation 2x^{2} + x – 4 = 0 with the general form ax^{2} + bx + c = 0, we get a = 2, b = 1 and c = -4

(iii) 4x^{2} + 4√3x +3 = 0

Comparing quadratic equation 4x^{2} + 4√3x + 3 = 0 with the general form ax^{2} + bx + c = 0, we get a = 4, b = 4√3 and c = 3

A quadratic equation has two roots. Here, both the roots are equal.

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

Comparing quadratic equation 2x^{2} + x + 4 = 0 with the general form ax^{2} + bx + c =0, we get a = 2, b = 1 and c = 4

But, square root of negative number is not defined.

Therefore, Quadratic Equation 2x^{2} + x + 4 = 0 has no solution.

Comparing equation x^{2} – 3x – 1 = 0 with general form ax^{2} + bx + c = 0, We get a = 1, b = -3 and c = -1

Comparing equation x^{2} – 3x + 2 = 0 with general form ax^{2} + bx + c = 0, We get a=1,b=-3 and c=2

**12.** Let present age of Rehman = x years

Age of Rehman 3 years ago = (x – 3) years.

Age of Rehman after 5 years = (x + 5) years

According to the given condition:

Comparing quadratic equation x^{2} – 4x – 21 = 0 with general form ax^{2} + bx + c = 0, We get a = 1, b = -4 and c = -21

We discard x=-3. Since age cannot be in negative.

Therefore, present age of Rehman is 7 years.

**13.** Let Shefali’s marks in Mathematics = x

Let Shefali’s marks in English = 30 – x

If, she had got 2 marks more in Mathematics, her marks would be = x + 2

If, she had got 3 marks less in English, her marks in English would be = 30 – x – 3 = 27 – x

According to given condition:

(x + 2) (27 – x) = 210

27x – x^{2} + 54 – 2x = 210 ⇒ x^{2} – 25x + 156 = 0

Comparing quadratic equation x^{2} – 25x + 156 = 0 with general form ax^{2} + bx + c = 0, We get a = b = -25 and c = 156

Therefore, Shefali’s marks in Mathematics = 13 or 12 Shefali’s marks in English = 30 – x = 30 – 13 = 17

Or Shefali’s marks in English = 30 – x = 30 – 12 = 18

Therefore her marks in Mathematics and English are (13, 17) or (12, 18).

**14.** Let shorter side of rectangle = x metres

Let diagonal of rectangle = (x + 60) metres

Let longer side of rectangle = (x + 30) metres

According to pythagoras theorem,

(x+ 60)^{2} _{=}(x+ 30)^{2} + x2 ⇒ x^{2} + 3600 + 120x = x^{2} + 900 + 60x + x^{2}

⇒ x^{2} – 60x – 2700 = 0

Comparing equation x^{2} – 60x – 2700 = 0 with standard form ax2 + bx + c = 0,

We get a = b = -60 and c = -2700

We ignore -30. Since length cannot be in negative.

Therefore x = 90 which means length of shorter side = 90 metres

And length of longer side = x + 30 = 90 + 30 = 120 metres

Therefore, length of sides are 90 and 120 in metres.

**15.** Let smaller number = x and let larger number =y

According to condition:

y^{2} – x^{2} = 180 … (1)

Also, we are given that square of smaller number is 8 times the larger number.

⇒ x^{2} = 8y … (2)

Putting equation (2) in (1), we get

y^{2} – 8y = 180 ⇒ y^{2} – 8y – 180 = 0

Comparing equation y^{2} – 8y – 180 = 0 with general form aye + by + c = 0, We get a = 1, b = -8 and c = -180

⇒ y = 18, -10

Using equation (2) to find smaller number:

x^{2} = 8y

⇒ x^{2} = 8y = 8 × 18 = 144 ⇒ x = ±12

And, x^{2} = 8y = 8 × -10 = -80 {No real solution for x}

Therefore two numbers are (12, 18) or (-12, 18)

**16.** Let the speed of the train = x km/hr

If, speed had been 5 km/hr more, train would have taken 1 hour less. So, according to this condition

Comparing equation x^{2} + 5x – 1800 = 0 with general equation axe + bx + We get a = 1, b = 5 and c = -1800

Since speed of train cannot be in negative. Therefore, we discard x = -45 Therefore, speed of train = 40 km/hr

**17.** Let time taken by tap of smaller diameter to fill the tank = x hours

Let time taken by tap of larger diameter to fill the tank = (x – 10) hours

It means that tap of smaller diameter fills part of tank in 1 hour. …(1)

And, tap of larger diameter fills part of tank in 1 hour. …(2)

When two taps are used together, they fill tank in 758 hours.

In 1 hour, they fill part of tank …..(3)

From (1), (2) and (3),

⇒ 75 (2x – 10) = 8 (x^{2} – 10x) ⇒ 150x – 750= 8x^{2 }– 80x

⇒ 8x – 80x – 150x + 750 = 04 ⇒ x^{2} – 115x + 375= 0

Comparing equation 4x^{2} – 115x + 375 = 0 with general equation axe + bx + c = 0, We get a = 4, b = -115 and c = 375

Time taken by larger tap = x – 10 = 3.75 – 10 = -6.25 hours

Time cannot be in negative. Therefore, we ignore this value.

Time taken by larger tap = x – 10 = 25 – 10 = 15 hours

Therefore, time taken by larger tap is 15 hours and time taken by smaller tap is 25 hours.

**18.** Let average speed of passenger train = x km/h

Let average speed of express train = (x + 11) km/h

Time taken by passenger train to cover 132 km = hours

Time taken by express train to cover 132 km hours

According to the given condition,

Comparing equation x^{2} + 11x – 1452 = 0 with general quadratic equation ax^{2} + bx + c = 0, we get a = 1, b = 11 and c = -1452

As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h

And, speed of express train = x + 11 = 33 + 11 = 44 km/h 11.

**19.** Let perimeter of first square = x metres

Let perimeter of second square = (x + 24) metres

Length of side of first square = ^{ }metres {Perimeter of square = 4 x length of side}

Length of side of second square = metres

Area of first square = side × side

According to given condition:

Comparing equation x^{2} + 24x – 3456 = 0 with standard form ax^{2} + bx + c = 0, We get a = 1, b =24 and c = -3456

Perimeter of square cannot be in negative. Therefore, we discard x=-72. Therefore, perimeter of first square = 48 metres

And, Perimeter of second square = x + 24 = 48 + 24 = 72 metres

**20. **(i) 2x^{2} – 3x + 5 = 0

Comparing this equation with general equation ax^{2} + bx + c = 0,

We get a = 2, b = -3 and c = 5

Discriminant = b^{2} – 4ac = (-3)^{2} – 4 (2) (5) = 9 – 40 = -31

Discriminant is less than 0 which means equation has no real roots.

(ii) 3x^{2 }– 4√3x +4 = 0

Comparing this equation with general equation ax^{2} + bx + c = 0,

We get a = 3, b = -4√3 and c = 4

Discriminant = b^{2} – 4ac = (-4√3)^{2} – 4 (3) (4) = 48 – 48 = 0

Discriminant is equal to zero which means equations has equal real roots.

Because, equation has two equal roots, it means

(iii) 2x^{2} – 6x + 3 = 0

Comparing equation with general equation ax^{2} + bx + c = 0, We get a = 2, b = -6, and c = 3

Discriminant = b^{2} – 4ac = (-6)^{2} – 4 (2) (3) = 36 – 24 = 12 Value of discriminant is greater than zero.

Therefore, equation has distinct and real roots.

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

We know that quadratic equation has two equal roots only when the value

of discriminant is equal to zero.

Comparing equation 2x^{2} + kx + 3 = 0 with general quadratic equation ax^{2} + bx + c= 0, we get a = 2, b = k and c = 3

Discriminant = b^{2} – 4ac = k^{2} – 4 (2) (3) = k^{2} – 24 Putting discriminant equal to zero

k^{2} – 24 = 0 ⇒ k^{2} = 24

⇒ k = ±√24 = ±2√6 = k = 2√6, – 2√6

(ii) kx (x – 2) + 6 = 0

⇒ kx^{2} – 2kx + 6 = 0

Comparing quadratic equation kx^{2} – 2kx + 6 = 0 with general form ax^{2} + bx + c = 0, we get a = k, b = -2k and c = 6

Discriminant = b^{2} – 4ac = (-2k)^{2} – 4 (k) (6) = 4k^{2} – 24k

We know that two roots of quadratic equation are equal only if discriminant is equal to zero.

Putting discriminant equal to zero

4k^{2} – 24k = 0

⇒ 4k (k – 6) = 0 ⇒ k = 0, 6

The basic definition of quadratic equation says that quadratic equation is the equation of the form ax^{2} + bx + c = 0, where a ≠ 0.

Therefore, in equation kx^{2} – 2kx + 6 = 0, we cannot have k = 0.

Therefore, we discard k = 0.

Hence the answer is k = 6.

**22.** Let breadth of rectangular mango grove = x metres

Let length of rectangular mango grove = 2x metres

Area of rectangle = length x breadth = x x 2x = 2x^{2} m^{2}

According to given condition:

2x^{2} = 800

⇒ 2x^{2} – 800 = 0 ⇒ x^{2} – 400 = 0

Comparing equation x^{2} – 400 = 0 with general form of quadratic equation ax^{2} + bx + c =0, we get a = 1, b = 0 and c = -400

Discriminant = b^{2} – 4ac = (0)^{2} – 4 (1) (-400) = 1600

Discriminant is greater than 0 means that equation has two disctinct real roots. Therefore, it is possible to design a rectangular grove.

We discard negative value of x because breadth of rectangle cannot be in negative. Therefore, x = breadth of rectangle = 20 metres

Length of rectangle = 2x = 2 x 20 = 40 metres

**23.** Let age of first friend = x years and let age of second friend = (20 – x) years Four years ago, age of first friend = (x – 4) years

Four years ago, age of second friend = (20 – x) – 4 = (16 – x) years

According to given condition,

(x – 4) (16 – x) = 48 ⇒ 16x – x^{2} – 64 + 4x = 48

⇒ 20x – x^{2 }– 112 = 0 ⇒ x^{2} – 20x + 112 = 0

Comparing equation, x^{2} – 20x + 112 = 0 with general quadratic equation ax^{2} + bx + c = 0, we get a = 1, b = -20 and c = 112

Discriminant = b^{2} – 4ac = (-20)^{2} – 4 (1) (112) = 400 – 448 = -48 < 0

Discriminant is less than zero which means we have no real roots for this equation.

Therefore, the give situation is not possible.

**24.** Let length of park = x metres

We are given area of rectangular park = 400 m^{2}

Therefore, breadth of park = {Area of rectangle = length x breadth}

Perimeter of rectangular park = 2 (length + breath) = metres

We are given perimeter of rectangle = 80 metres

According to condition:

Comparing equation, x^{2} – 40x + 400 = 0 with general quadratic equation axe + bx + c =

0, we get a = 1, b = -40 and c = 400

Discriminant = b^{2} – 4ac = (-40)^{2} – 4 (1) (400) = 1600 – 1600 = 0

Discriminant is equal to 0.

Therefore, two roots of equation are real and equal which means that it is possible to design a rectangular park of perimeter 80 metres and area 400 m^{2}.

Here, both the roots are equal to 20.

Therefore, length of rectangular park = 20 metres