Class 10 - Maths - Quadratic Equations

Exercise 4.1

Question 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

Answer:

(i) (x + 1)2 = 2(x – 3)

=> x2 + 2x + 1 = 2x – 6

=> x2 + 2x + 1 - 2x + 6 = 0

=> x2 + 7 = 0

=> x2 + 0x + 7 = 0

This is an equation of the form ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

(ii) x2 – 2x = (–2) (3 – x)

=> x2 – 2x = -6 + 2x

=> x2 -2x – 2x + 6 = 0

=> x2 – 4x + 6 = 0

This is an equation of the form ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

=> x2 – 2x + x – 2 = x2 – x + 3x – 3

=> x2 – x – 2 = x2 + 2x – 3

=> x2 – x – 2 - x2 - 2x + 3 = 0

=> -3x + 1 = 0

=> 3x – 1 = 0

This is not an equation of the form ax2 + bx + c = 0

Hence, the given equation is not a quadratic equation.

(iv) (x – 3)(2x +1) = x(x + 5)

=> 2x2 – 6x + x – 3 = x2 + 5x

=> 2x2 – 5x – 3 = x2 + 5x

=> 2x2 – 5x – 3 - x2 - 5x = 0

=> x2 – 10x – 3 = 0

This is an equation of the form ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

=> 2x2 – x – 6x + 3 = x2 + 5x – x – 5

=> 2x2 – 7x + 3 = x2 + 4x – 5

=> 2x2 – 7x + 3 - x2 - 4x + 5 = 0

=> x2 – 11x + 8 = 0

This is an equation of the form ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

(vi) x2 + 3x + 1 = (x – 2)2

=> x2 + 3x + 1 = x2 + 4 – 4x

=> x2 + 3x + 1 - x2 - 4 + 4x = 0

=> 7x - 3 = 0

This is not an equation of the form ax2 + bx + c = 0

Hence, the given equation is not a quadratic equation.

(vii) (x + 2)3 = 2x (x2 – 1)

=> x3 + 6x2 + 12x + 8 = 2x3 – 2x

=> x3 + 6x2 + 12x + 8 - 2x3 + 2x = 0

=> -x3 + 6x2 + 14x + 8 = 0

=> x3 - 6x2 - 14x - 8 = 0

This is not an equation of the form ax2 + bx + c = 0

Hence, the given equation is not a quadratic equation.

(viii) x3 – 4x2 – x + 1 = (x – 2)3

=> x3 – 4x2 – x + 1 = x3 - 6x2 + 12x - 8

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

=> 2x2 – 13x + 9 = 0

This is an equation of the form ax2 + bx + c = 0

Hence, the given equation is a quadratic equation.

Question 2:

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.

Answer:

(i) Let the breadth = x metres

Length = 2(Breadth) + 1

Length = (2x + 1) metres

Since Length * Breadth = Area

=> (2x + 1) * x = 528

=> 2x2 + x = 528

=> 2x2 + x – 528 = 0

Thus, the required quadratic equation is:    2x2 + x – 528 = 0

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

Given, product of the numbers = 306

=> x(x + 1) = 306

=> x2 + x = 306

=> x2 + x – 306 = 0

Thus, the required quadratic equation is:    x2 + x – 306 = 0

(iii) Let the present age of Rohan = x

So, his mother’s age = (x + 26) years

After 3 years,

Rohan’s age = (x + 3) years

His mother’s age = [(x + 26) + 3] years = (x + 29) years

According to the condition,

The product of their ages (in years) 3 years from now will be 360

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

=> x2 + 29x + 3x + 87 = 360

=> x2 + 29x + 3x + 87 – 360 = 0

=> x2 + 32x – 273 = 0

Thus, the required quadratic equation is:   x2 + 32x – 273 = 0

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

Distance covered = 480 km

Time taken = Distance/Speed

= (480/u) hours

In second case,

Speed = (u – 8) km/ hour

So, time taken = Distance/speed = 480/(u – 8) hours

According to the condition,

480/(u - 8) – 480/u = 3

=> 480u – 480(u – 8) = 3u(u – 8)

=> 480u – 480u + 3840 = 3u2 – 24u

=> 3840 – 3u2 + 24u = 0

=> 1280 – u2 + 8u = 0

=> –1280 + u2 – 8u = 0

=> u2 – 8u – 1280 = 0

Thus, the required quadratic equation is: u2 – 8u – 1280 = 0

Exercise 4.2

Question 1:

Find the roots of the following quadratic equations by factorization:

(i) x2 – 3x – 10 = 0                        (ii) 2x2 + x – 6 = 0                             (iii) √2x2 + 7x + 5√2 = 0

(iv) 2x2 – x +1/8 = 0                     (v) 100x2 – 20x + 1 = 0

Answer:

(i) x2 – 3x – 10 = 0

=> x2 – 5x + 2x – 10 = 0

=> x(x - 5) + 2(x - 5) = 0

=> (x - 5)(x  +2) = 0

=> x = -2, 5

Hence, the required roots are: -2 and 5

(ii) 2x2 + x – 6 = 0

=> 2x2 + 4x – 3x – 6 = 0

=> 2x(x + 2) – 3(x + 2) = 0

=> (x + 2)(2x - 3) = 0

=> x = -2, 3/2

Hence, the required roots are: -2 and 3/2

(iii) √2x2 + 7x + 5√2 = 0

=> √2x2 + 2x + 5x + 5√2 = 0

=> √2x2 + √2 * √2x + 5x + 5√2 = 0

=> √2x(x + √2) + 5(x + √2) = 0

=> (x + √2)(√2x + 5) = 0

=> x = -√2, -5/√2

Hence, the required roots are: -√2 and -5/√2

(iv) 2x2 – x + 1/8 = 0

=> (16x2 – 8x + 1)/8 = 0

=> 16x2 – 8x + 1 = 0

=> 16x2 – 4x – 4x + 1 = 0

=> 4x(4x - 1) – 1(4x - 1) = 0

=> (4x - 1)(4x - 1) = 0

=> x = 1/4, 1/4

Hence, the required roots are: 1/4 and 1/4

(v) 100x2 – 20x + 1 = 0

=> 100x2 – 10x – 10x + 1 = 0

=> 10x(10x - 1) – 1(10x - 1) = 0

=> (10x - 1)(10x - 1) = 0

=> x = 1/0, 1/10

Hence, the required roots are: 1/10 and 1/10

Question 2:

Solve the problems given in Example 1.

Answer:

(i) We have:

=> x2 – 45x + 324 = 0

=> x2 – 9x – 36x + 324 = 0

=> x(x – 9) – 36(x – 9) = 0

=> (x – 9) (x – 36) = 0

=> x = 9, 36

Thus, x = 9 and × = 36

(ii) We have:

=> x2 – 55x + 750 = 0

=> x2 – 30x – 25x + 750 = 0

=> x (x – 30) – 25 (x – 30) = 0

=> (x – 30) (x – 25) = 0

=> x = 30, 25

Thus, x = 30 and x = 25.

Question 3:

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

Answer:

Here, sum of the numbers is 27.

Let one of the numbers be x.

So, other number = 27 – x

According to the condition,

Product of the numbers = 182

=> x(27 – x) = 182

=> 27x – x2 = 182

=> - x2 + 27x – 182 = 0

=> x2 – 27x + 182 = 0

=> x2 – 13x – 14x + 182 = 0

=> x (x – 13) – 14 (x – 13) = 0

=> (x – 13) (x – 14) = 0

=> x = 13, 14

Thus, the required numbers are 13 and 14.

Question 4:

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

Answer:

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

Since the sum of the squares of the numbers = 365

=> 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 +14) (x – 13) = 0

=> x = -14, 13

Since x has to be a positive integer

So, x = 13

Now, x + 1 = 13 + 1 =1 4                   T

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

Question 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.

Answer:

Let the base of the given right triangle be x cm.

So, its height = (x – 7) cm

Now, Hypotenuse = √{(Base)2 + (Height)2}

=> 13 = √{x2 + (x - 7)2}

Squaring both sides, we get

=> 169 = x2 + (x – 7)2

=> 169 = x2 + x2 – 14x + 49

=> 2x2 – 14x + 49 - 160

=> 2x2 – 14x – 120 = 0

=> x2 – 7x – 60 = 0

=> x2 – 12x + 5x – 60 = 0

=> x(x – 12) + 5 (x – 12) = 0

=> (x–12) (x + 5) = 0

=> x = 12, -5

But the side of triangle can never be negative,

So, x = 12

So, 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.

Question 6:

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the c

ost 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.

Answer:

Let the number of articles produced in a day = x

Cost of production of each article = 2x + 3

According to the condition,

Given, total cost = 90

=> x(2x + 3) = 90

=> 2x2 + 3x = 90            => 2x2 + 3x – 90 = 0

=> 2x2 – 12x + 15x – 90 = 0

=> 2x(x – 6) + 15(x – 6) = 0

=> (x – 6) (2x + 15) = 0           => x = 6, -15/2

But the number of articles cannot be negative.

So, x = 6

Hence, the 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

Question 1:

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x2 – 7x + 3 = 0       (ii) 2x2 + x – 4 = 0        (iii) 4x2 + 4√3x + 3 = 0          (iv) 2x2 + x + 4 = 0

Answer:

(i) 2x2 – 7x + 3 = 0

Divide by 2 on both sides, we get

2x2/2 – 7x/2 + 3/2 = 0/2

=> x2 – 7x/2 + 3/2 = 0

=> x2 – 7x/2 + 3/2 + (7/4)2 – (7/4)2 = 0

=> x2 – 7x/2 + (7/4)2 + 3/2 – (7/4)2 = 0

=> (x – 7/4)2 + 3/2 – 49/16 = 0

=> (x – 7/4)2 + (3 * 8 – 49 * 1)/16 = 0

=> (x – 7/4)2 + (24 – 49)/16 = 0

=> (x – 7/4)2 - 25/16 = 0

=> (x – 7/4)2 = 25/16

=> x – 7/4 = ±√(25/16)

=> x – 7/4 = ±5/4

Case 1: When 5/4 is positive

x – 7/4 = 5/4

=> x = 5/4 + 7/4

=> x = (5 + 7)/4

=> x = 12/4

=> x = 3

Case 2: When 5/4 is negative

x – 7/4 = -5/4

=> x = -5/4 + 7/4

=> x = (-5 + 7)/4

=> x = 2/4

=> x = 1/2

Hence, the required roots are: 3 and 1/2

(ii) 2x2 + x – 4 = 0

Divide by 2 on both sides, we get

2x2/2 + x/2 – 4/2 = 0/2

=> x2 + x/2 – 2 = 0

=> x2 + x/2 – 2 + (1/4)2 – (1/4)2 = 0

=> x2 + x/2 + (1/4)2 – 2 - (1/4)2 = 0

=> (x + 1/4)2 – 2 – 1/16 = 0

=> (x + 1/4)2 – 33/16 = 0

=> (x + 1/4)2 = 33/16

=> x + 1/4 = ±√(33/16)

=> x + 1/4 = ±√33/4

Case 1: When √33/4 is positive

=> x + 1/4 = √33/4

=> x = √33/4 – 1/4

=> x = (√33 - 1)/4

Case 2: When √33/4 is negative

=> x + 1/4 = -√33/4

=> x = -√33/4 – 1/4

=> x = -(√33 + 1)/4

Hence, the required roots are: (√33 - 1)/4 and -(√33 + 1)/4

(iii) 4x2 + 4√3x + 3 = 0

Divide by 4 on both sides, we get

4x2/4 + 4√3x/4 + 3/4 = 0/4

=> x2 + √3x + 3/4 = 0

=> x2 + √3x + 3/4 + (√3/2)2 - (√3/2)2 = 0

=> (x + √3/2)2 + 3/4 - 3/4 = 0

=> (x + √3/2)2 = 0

=> (x + √3/2) (x + √3/2) = 0

=> x = -√3/2, -√3/2

Hence, the required roots are: √3/2 and -√3/2

(iv) 2x2 + x + 4 = 0

Divide by 2 on both sides, we get

2x2/2 + x/2 + 4/2 = 0/2

=> x2 + x/2 + 2 = 0

=> x2 + x/2 + 2 + (1/4)2 - (1/4)2 = 0

=> (x + 1/4)2 + 2 – 1/16 = 0

=> (x + 1/4)2 + 31/16 = 0

=> (x + 1/4)2 = -31/16

Since square of a number cannot be negative,

So, (x + 1/4)2 cannot be a real value.

Hence, there is no real value of x satisfying the given equation.

Question 2:

Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Answer:

(i) 2x2 – 7x + 3 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 2, b= – 7, c = 3

Now discriminant = b2 – 4ac

= (-7) 2 – 4 * 2 * 3

= 49 – 24

= 25 > 0

Since b2 – 4ac is positive.

Hence, the given quadratic equation has real roots. The roots are given as:

x = {-b ± √(b2 – 4 * a * c)}/2a

=> x = [-(-7) ± √{(-7)2 – 4 * 2 * 3)}]/(2 * 2)

=> x = {7 ± √(49 – 24)}/4

=> x = (7 ± √25)/4

=> x = (7 ± 5)/4

Taking positive sign, we get

=> x = (7 + 5)/4

=> x = 12/4

=> x = 3

Taking negative sign, we get

=> x = (7 - 5)/4

=> x = 2/4

=> x = 1/2

Thus, the roots are: 3 and 1/2

(ii) 2x2 + x – 4 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 2, b= 1, c = -4

Now discriminant = b2 – 4ac

= 1 2 – 4 * 2 * (-4)

= 1 + 32

= 33 > 0

Since b2 – 4ac is positive.

Hence, the given quadratic equation has real roots. The roots are given as:

x = {-b ± √(b2 – 4 * a * c)}/2a

=> x = [-1 ± √{12 – 4 * 2 * (-4)}]/(2 * 2)

=> x = {-1 ± √(1 + 32)}/4

=> x = (-1 ± √33)/4

Taking positive sign, we get

=> x = (-1 + √33)/4

Taking negative sign, we get

=> x = (-1 - √33)/4

Thus, the roots are: (-1 + √33)/4 and (-1 - √33)/4

(iii) 4x2 + 4√3x + 3 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 4, b= 4√3, c = 3

Now discriminant = b2 – 4ac

= (4√3) 2 – 4 * 4 * 3

= 48 – 48

= 0

Since b2 – 4ac is zero.

Hence, the given quadratic equation has equal roots. The roots are given as:

x = {-b ± √(b2 – 4 * a * c)}/2a

=> x = [-4√3 ± √{(4√3)2 – 4 * 4 * 3)}]/(2 * 4)

=> x = {-4√3 ± √(48 – 48)}/8

=> x = (-4√3 ± 0)/8

=> x = -4√3/8

=> x = -√3/2

Thus, the roots are: -√3/2 and -√3/2

(iv) 2x2 + x + 4 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 2, b= 1, c = 4

Now discriminant = b2 – 4ac

= 1 2 – 4 * 2 * 4

= 1 - 32

= -31 < 0

Since b2 – 4ac is negative.

Hence, the given quadratic equation has no real roots.

Question 3:

Find the roots of the following equations:

(i) x – 1/x = 3, x ≠ 0                       (ii) 1/(x + 4) – 1/(x - 7) = 11/30, x ≠ -4, 7

Answer:

(i) Given, x – 1/x = 3

=> (x2 - 1)/x = 3

=> x2 - 1 = 3x

=> x2 – 3x – 1 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 1, b= -3, c = -1

Now discriminant = b2 – 4ac

= (-3) 2 – 4 * 1 * (-1)

= 9 + 4

= 13 > 0

Since b2 – 4ac is positive.

Hence, the given quadratic equation has real roots. The roots are given as:

x = {-b ± √(b2 – 4 * a * c)}/2a

=> x = [-(-3) ± √{(-3)2 – 4 * 1 * (-1)}]/(2 * 1)

=> x = {3 ± √(9 + 4)}/2

=> x = (3 ± √13)/2

Taking positive sign, we get

=> x = (3 + √13)/2

Taking negative sign, we get

=> x = (3 - √13)/2

Thus, the roots are: (3 + √13)/2 and (3 - √13)/2

(ii) Given, 1/(x + 4) – 1/(x - 7) = 11/30

=> {(x - 7) – (x + 4)}/{(x + 4)(x - 3)} = 11/30

=> (x - 7 – x - 4)/(x2 – 3x - 28) = 11/30

=> -11/(x2 – 3x - 28) = 11/30

=> -1/(x2 – 3x - 28) = 1/30

=> x2 – 3x – 28 = -30

=> x2 – 3x – 28 + 30 = 0

=> x2 – 3x + 2 = 0

=> x2 – x – 2x + 2 = 0

=> x(x - 1) – 2(x - 1) = 0

=> (x - 1)(x - 2) = 0

=> x = 1, 2

Thus, the roots of the equation are: 1, 2

Question 4:

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Answer:

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,

1/(x - 3) + 1/(x + 5) = 1/3

=> {(x + 5) + (x - 3)}/ {(x + 5)(x - 3)} = 1/3

=> 3[x + 5 + x – 3] = (x – 3) (x + 5)

=> 3[2x + 2] = x2 + 2x – 15

=> 6x + 6 = x2 + 2x – 15

=> x2 + 2x – 6x – 15 – 6 = 0

⇒ x2 – 4x – 21 = 0

=> x2 – 7x + 3x – 21 = 0

=> x(x – 7) + 3(x - 7) = 0

=> (x - 7)(x + 3) = 0

=> x = 7, -3

Since age cannot be negative,

So, x = 7

Hence, the present age of Rehman = 7 years

Question 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.

Answer:

Let, Shefali’s marks in Mathematics = x

So, the marks in English = (30 – x)                 [Since sum of their marks in Eng. and Maths = 30]

Now, according to the condition,

(x + 2) * [(30 – x) – 3] = 210

=> (x + 2) * (30 – x – 3) = 210

=> (x + 2) * (-x + 27) = 210

=> -x2 + 27x – 2x + 54 = 210

=> -x2 + 25x + 54 = 210

=> -x2 + 25x + 54 – 210 = 0

=> -x2 + 25x – 156 = 0

=> x2 - 25x + 156 = 0

=> x2 - 13x – 12x + 156 = 0

=> x(x – 13) – 12(x - 13) = 0

=> (x - 13)(x - 12) = 0

=> x = 12, 13

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

Question 6:

The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side,

find the sides of the field.

Answer:

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

So, the longer side (length) = (x + 30) metres.

In a rectangle,

Diagonal = √{(breadth)2 + (length)2}

=> x + 60 = √{x2 + (x + 30)2}

Squaring on both sides, we get

(x + 60)2 = x2 + (x + 30)2

=> (x + 60)2 = x2 + x2 + 60x + 900

=> (x + 60)2 = 2x2 + 60x + 900

=> x2 + 120x + 3600 = 2x2 + 60x + 900

=> 2x2 – x2 + 60x – 120x + 900 – 3600 = 0

=> x2 – 60x – 2700 = 0

=> x2 – 90x + 30x – 2700 = 0

=> x(x - 90) + 30(x - 90) = 0

=> (x - 90)(x + 30) = 0

=> x = 90, -30

Since breadth cannot be negative

So, x = 90

Now, x + 30 = 90 + 30=120

Thus, the shorter side = 90 m

The longer side = 120 m.

Question 7:

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Answer:

Let the larger number be x.

Since (smaller number) 2 = 8 (larger number)

=> (smaller number)2 = 8x

=> smaller number = √(8x)

According to question,

x2 - √(8x)2 = 180

=> x2 - 8x = 180

=> x2 - 8x – 180 = 0

=> x2 - 18x + 10x – 180 = 0

=> x(x - 18) + 10(x - 18) = 0

=> (x - 18)(x + 10) = 0

=> x = 18, -10

Since x ≠ -10 as x is the larger of two numbers.

So, x = 18

Hence, the larger number = x = 18

Smaller number = √(8 * 18) = √144 = 12

Question 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.

Answer:

Let the speed of the train is x km/hr

Given, the train travelling at a uniform speed for 360 km would have taken 48 minutes less to

travel the same distance

if its speed work 5 km per hour more.

=> 360/x - 360/(x + 5) = 48/60           {since 48 minutes = 48/60 hours}

=> 360{1/x - 1/(x + 5)} = 4/5

=> 360[{(x + 5) - x}/{x(x + 5)}] = 4/5

=> 360[{x + 5 - x}/{x(x + 5)}] = 4/5

=> 360[5/{x2 + 5x}] = 4/5

=> 1800/{x2 + 5x}] = 4/5

=> 1800*5 = 4(x2 + 5x)

=> (1800*5)/4 = x2 + 5x

=> x2 + 5x = 450 * 5

=> x2 + 5x = 2250

=> x2 + 5x - 2250 = 0

=> x2 - 45x + 50x - 2250 = 0

=> x(x - 45) + 50(x - 45) = 0

=> (x - 45)*(x + 50) = 0

=> x = 45, -50

Since speed cannot be negative

So, x = 45

Hence, the speed of the train is 45 km/hr

Question 9:

Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately.

Find the time in which each tap can separately fill the tank.

Answer:

Let the larger tap fills the tank in x hours.

So, the smaller tap fills the tank (x + 10) hours.

Time taken to fill tank by larger tap in one hour = 1/x hours

Time taken to fill tank by Smaller tap in one hour = 1/(x + 10) hours

Now, the tank filled by the two taps together in one hour = 1/x + 1/(x + 10)

= (x + 10 + x)/(x2 + 10x)

= (2x + 10)/(x2 + 10x)

Now, according to question,

(2x + 10)/(x2 + 10x) = 1/9 3/8

=> (2x + 10)/(x2 + 10x) = 1/(75/8)

=> (2x + 10)/(x2 + 10x) = 8/75

=> 75(2x + 10) = 8(x2 + 10x)

=> 150x + 750 = 8x2 + 80x

=> 8x2 + 80x - 150x  - 750 = 0

=> 8x2 – 70x – 750 = 0

=> 4x2 – 35x – 375 = 0              [divide by 4]

=> 4x2 – 60x + 25x – 375 = 0

=> 4x(x - 15) + 25(x - 15) = 0

=> (x - 15)(4x + 25) = 0

=> x = 15, -25/4

Since x is the time taken to fill the tank cannot be negative

So, x = 15

Hence, the larger tap fills the tank in 15 hours.

and the smaller tap fills the tank in (x + 10) = (15 + 10) = 25 hours.

Question 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 11km/h

more than that of the passenger train, find the average speed of the two trains.

Answer:

Let the speed of the passenger train = x km/h

Therefore, the speed of express train = (x + 11) km/h

Distance travelled = 132 km

Time taken by passenger train = 132/x hours

Time taken by express train = 132/(x + 11) hours

According to question,

132/x = 132/(x + 11) + 1

=> 132/x - 132/(x + 11) = 1

=> {132(x + 11) – 132x}/{x(x + 11)} = 1

=> 132x + 1452 – 132x = x(x + 11)

=> 1452 = x(x + 11)

=> x2 + 11x – 1452 = 0

=> x2 + 44x – 33x - 1452 = 0

=> x(x + 44) – 33(x + 44) = 0

=> (x - 33)(x + 44) = 0

=> x = 33, -44

Since, the speed of the train cannot be negative

So, x = 33

Hence, the speed of the passenger train = 33 km/h

Therefore, the speed of express train = (33 + 11) = 44 km/h

Question 11:

Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Answer:

Let the side of larger square = x m

And side of the smaller square = y m

According to question,

x2 + y2 = 468   …………1

Difference between perimeters is

4x – 4y = 24

=> 4(x - y) = 24

=> x – y = 24/4

=> x  - y = 6

=> x = y + 6  ………2

Put value of x in equation 1, we get

(y + 6)2 + y2 = 468

=> y2 + 36 + 12y + y2 = 468

=> 2y2 + 12y + 36 – 486 = 0

=> 2y2 + 12y - 432 = 0

=> y2 + 6y - 216 = 0                 [Divide by 2]

=> y2 + 18y – 12y - 216 = 0

=> y(y + 18) – 12(y + 18) = 0

=> (y + 18)(y - 12) = 0

=> y = 12, -18

Since length of square cannot be negative

So, y = 12

So, the side of the smaller square = y = 12 m

and the side of larger square = x = y + 6 = 12 + 6 = 18 m

Exercise 4.4

Question 1:

Find 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

Answer:

(i) 2x2 – 3x + 5 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 2, b= – 3, c = 5

Now discriminant = b2 – 4ac

= (-3) 2 – 4 * 2 * 5

= 9 – 40

= -31 < 0

Since b2 – 4ac is negative.

Hence, the given quadratic equation has no real roots.

(ii) 3x2 – 4√3x + 4 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 3, b= -4√3, c = 4

Now discriminant = b2 – 4ac

= (-4√3) 2 – 4* 3 * 4

= 48 – 48

= 0

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

Here, the roots are: -b/2a and –b/2a

i.e -(-4√3)/(2 * 3) and -(-4√3)/(2 * 3)

=> 4√3/6 and 4√3/6

=> 2√3/3 and 2√3/3

=> 2√3/(√3 * √3) and 2√3/(√3 * √3)

=> 2/√3 and 2/√3

(iii) 2x2 – 6x + 3 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we have:

a = 2, b= -6, c = 3

Now discriminant = b2 – 4ac

= (-6) 2 – 4* 2 * 3

= 36 – 24

= 12 > 0

Thus, the given quadratic equation has two real and distinct roots which are given as:

x = {-b ± √(b2 – 4 * a * c)}/2a

=> x = [-(-6) ± √{(-6)2 – 4 * 2 * 3)}]/(2 * 2)

=> x = {6 ± √(36 – 24)}/4

=> x = (6 ± √12)/4

=> x = (6 ± 2√3)/4

=> x = 2(3 ± √3)/4

=> x = (3 ± √3)/2

Thus, the roots are: (3 + √3)/2 and (3 - √3)/2

Question 2:

Find 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

Answer:

(i) 2x2 + kx + 3 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = 2, b = k, c = 3

For a quadratic equation to have equal roots,

Discriminant = 0

=> b2 – 4ac = 0

=> k2 - 4 * 2 * 3 = 0

=> k2 – 24 = 0

=> k2 = 24

=> k = ±√24

=> k = ±2√6

(ii) kx(x - 2) + 6 = 0

=> kx2 – 2kx + 6 = 0

Comparing the given quadratic equation with ax2 + bx + c = 0, we get

a = k, b = -2k, c = 6

For a quadratic equation to have equal roots,

Discriminant = 0

=> b2 – 4ac = 0

=> (-2k)2 - 4 * k * 6 = 0

=> 4k2 – 24k = 0

=> 4k(k – 6) = 0

=> k = 0, 6

But k cannot be 0 otherwise, the given equation is no more quadratic.

Thus, the required value of k = 6

Question 3:

Is 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.

Answer:

Let the breadth be x metres.

So, Length = 2x metres

Now, Area = Length × Breadth

= 2x * x

= 2x2 metre2

According to the given condition,

2x2 = 800

=> x2 = 800/2

=> x2 = 400

=> x = ±√400

=> x = 20 and x = -20

But x = -20 is possible                                (Since breadth cannot be negative).

So, x = 20

Now 2x = 2 * 20 = 40

Thus, length = 40 m and breadth = 20 m

Question 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.

Answer:

Let the age of one friend = x years

The age of the other friend = (20 – x) years             [Since 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 – x2 – 4x = 48

=> -x2 – 20x – 64 – 48 = 0

=> -x2 – 20x – 112 = 0

=> x2 + 20x + 112 = 0   …………….1

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

Now, b2 – 4ac = (20) 2 – 4 * 1 * 112

= 400 – 448

= – 48 < 0

Since b2 – 4ac is less than 0.

So, the quadratic equation 1 has no real roots.

Thus, the given equation is not possible.

Question 5:

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

Answer:

Let the breadth of the rectangle be x metres.

Given, the perimeter of the rectangle = 80 m

=> 2(Length + Breadth) = 80

=> 2(Length + x) = 80

=> Length + x = 80/2

=> Length + x = 40

=> Length = (40 – x) metres

Again, area of the rectangle = Length * breadth

= (40 – x) * x

= 40x – x2

Now, according to the given condition,

Area of the rectangle = 400

=> 40x – x2 = 400

=> -x2 + 40x – 400 = 0

=> x2 – 40x + 400 = 0    …………….1

Comparing equation (1) with ax2 + bx + c = 0, we get

a = 1, b = –40, c = 400

Now, b2 – 4ac = (-40)2 – 4 * 1 * 400

= 1600 – 1600 – 0

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

So, x = -b/2a and –b/2a

=> x = -(-40)/(2 * 1) and -(-40)/(2 * 1)

=> x = 40/2 and 40/2

=> x = 20 and 20

So, Breadth = x = 20 m

and length = (40 – x) = (40 – 20) m = 20 m.

.