# Question-1

**Check whether the following are quadratic equations:**

**(i) (x + 1)**

(ii) x

^{2}= 2(x â€“ 3)(ii) x

^{2}â€“ 2x = -2(3 â€“ x)**(iii) (x - 2)(x + 1) = (x - 1)(x + 3)**

**Solution:**

**(i) (x + 1)**

^{2}= 2(x â€“ 3)x^{2} + 2x + 1 = 2x â€“ 6

x^{2} + 7 = 0

This is of the form ax^{2} + bx + c = 0

âˆ´ It is a quadratic equation.

**(ii) x ^{2} â€“ 2x = -2(3 â€“ x) **

x^{2} â€“ 2x = -6 + 2x

x^{2} â€“ 2x -2x + 6 = 0

x^{2} â€“ 4x + 6. This is of the form ax^{2} + bx + c = 0

It is a quadratic equation.

**(iii) (x - 2)(x + 1) = (x - 1)(x + 3)**

x^{2} â€“ 2x + x - 2 = x^{2} + 3x â€“ x - 3

â€“ x â€“ 2x +1 = 0

- 3x + 1 = 0

It is not of the form ax^{2} + bx + c

It is not a quadratic equation.

# Question-2

**Check whether the following are quadratic equations:**

(i) (x â€“ 3) (2x + 1) = x (x + 5)

(i) (x â€“ 3) (2x + 1) = x (x + 5)

**(ii) (2x â€“ 1) (x â€“ 3) = (x + 5) (x â€“ 1) (iii) x**

^{2}+ 3x + 1 = (x â€“ 2)^{2}**(iv) (x+2)**

^{3}= 2x(x^{2}â€“ 1) (v) x^{3}â€“ 4x^{2}â€“ x + 1 = (x â€“ 2)^{3}

**Solution:**

**(i)(x â€“ 3) (2x + 1) = x (x + 5)**

2x^{2} + x â€“ 6x â€“ 3 = x^{2} + 5x

x^{2} - 3 = 0

This is of the form ax^{2} + bx + c = 0

Hence it is a quadratic equation.

**(ii)(2x â€“ 1) (x â€“ 3) = (x + 5) (x â€“ 1)**

2x^{2} â€“ 6x â€“ x + 3 = x^{2} â€“ x + 5x â€“ 5

2x^{2} â€“ 7x + 3 - x^{2} â€“4x + 5 = 0

x^{2} â€“ 11x + 8 = 0

This is of the form ax^{2} + bx + x = 0

Hence it is a quadratic equation.

**(iii) x ^{2} + 3x + 1 = (x â€“ 2)^{2}**

x^{2} + 3x + 1 = x^{2} â€“ 4x +4

7x â€“ 3 = 0

This is not of the form ax^{2} + bx + c = 0

âˆ´ It is not a quadratic equation.

**(iv) (x+2) ^{3} = 2x(x^{2} â€“ 1) **

x^{2} + 2^{3} + 3x^{2} Ã— 2 + 3x Ã— 4 = 2x^{3} â€“ 2x

x^{3} + 8 + 6x^{2 }+ 12x = 2x^{3} â€“ 2x

x^{3} â€“ 6x^{2} â€“ 14x â€“ 8 = 0

Its not of the form ax^{2} + bx + c = 0

It is not a quadratic equation.

**(v) x ^{3} â€“ 4x^{2} â€“ x + 1 = (x â€“ 2)^{3} **

x^{3} â€“ 4x^{2} â€“ x + 1 = x^{3} â€“ 8 - 3x^{2} Ã— 2 + 3x Ã— 2^{2}

x^{3} â€“ 4x^{2} â€“ x + 1 = x^{3} â€“ 8 â€“ 6x^{2} + 12x

2x^{2} â€“13x + 1 = 0

This of the form ax^{2} + bx + c = 0

Hence it is a quadratic equation.

# Question-3

**Represent the following situations in the form of quadratic equations:****(i)The area of a rectangular plot is 528 m**^{2}. 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.****Solution:**

(i)

**The area of a rectangular plot is 528 m**

^{2}. 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 l be length (in metres) and b be the breadth (in metres) of the rectangle.

Given

l = 1 + 2b â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

lb = 528 m

^{2}----- (2)

(Area of rectangle = length Ã— breadth)

Substituting (1) in (2)

(1+2b)b = 528

b +2b

^{2}= 528

2b

^{2}+ b â€“ 528 = 0

This is the quadratic equation where the breadth is in metres.

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

Let the 2 consecutive positive integer be x, x + 1

x(x + 1) = 306

x

^{2 }+ x - 306 = 0 is the quadratic equation where x is the smallest

positive integer

# Question-4

**Represent the following situations in the form of quadratic equations:****(i) Rohanâ€™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 Rohanâ€™s present age.****(ii) 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.****Solution:**

(i) Rohanâ€™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 Rohanâ€™s present age.

Let the Rohans present age be x

Rohans mothers present age = x + 26

After 3 years,Rohans age = x + 3

Rohans mothers age = x + 26 + 3 = x + 29

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

x

^{2}+ 3x + 29x + 87 = 360

x

^{2}+ 32x â€“ 273 = 0 where x is Rohans present age (in years)

(ii) 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 /hr

âˆ´ Time taken =

If speed = (x â€“ 8) km/h

Time taken for the same distance = hr

- = 3 (Given)

480x â€“ 480 (x â€“ 8) = 3(x â€“ 8)x

480x â€“ 480x + 480 x 8 = 3x

^{2}â€“ 24x

3x

^{2}â€“ 24x â€“ 3840 = 0

x

^{2}â€“ 8x - 1280 = 0 is the required quadratic equation where x is the speed of the train in km /hr.

# Question-5

**Find the roots of the following quadratic equations by factorization:****(i) x**

(iii) + 7x + 5 = 0 (iv) 2x^{2 }â€“ 3x â€“ 10 = 0 (ii) 2x^{2}+ x â€“ 6 = 0(iii) + 7x + 5 = 0 (iv) 2x

^{2}â€“ x + = 0**(v) 100x**^{2}â€“ 20x + 1 = 0**Solution:**

**(i)x**

^{2 }â€“ 3x â€“ 10 = 0x

^{2 }â€“ 5x + 2x - 10 = 0

x(x â€“ 5) + 2(x â€“ 5) = 0

(x â€“ 5)(x + 2) = 0

âˆ´ x = -2, x = 5

Thus â€“2, 5 are the roots of the equation.

**(ii) 2x**

^{2}+ x â€“ 6 = 02x

^{2}+ 4x â€“ 3x â€“ 6 = 0

2x(x + 2) â€“ 3(x + 2) = 0

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

â‡’ x = -2 (or) x =

**are the roots of the equation**

**(iii)**

**+ 7x + 5 = 0**

x

^{2 }+ 5x + 2x + 5 = 0

x() + (x + 5) = 0

(x + )(x + 5) = 0

âˆ´ x = -, x = -

âˆ´ The roots are -, -.

**(iv) 2x**

^{2}â€“ x + = 016x

^{2}â€“ 8x + 1 = 0

16x

^{2}â€“ 4x â€“ 4x + 1 = 0

4x(4x â€“ 1) â€“ (4x â€“ 1) = 0

(4x â€“ 1)

^{2}= 0

x = ,

âˆ´ The root of the equation are , .

**(v) 100x**

^{2}â€“ 20x + 1 = 0100x

^{2}â€“ 10x â€“ 10x + 1 = 0

10x(10x â€“ 1) â€“1 (10x â€“ 1) = 0

(10x â€“ 1)

^{2}= 0

x = ,

âˆ´ The roots of the equation are, .

# Question-6

**Solve the following situation mathematically:**

**(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.**

**Solution:**

(i) Let the number of marbles John had be x

Then the number of marbles Jivanti had = 45 â€“ x.

The number of marbles left with John, when he lost 5 marbles = x â€“ 5

The number of marbles left with Jivanti, when she lost 5 marbles = 45 â€“ x â€“ 5

= 40 â€“ x

Therefore, their product = (x â€“ 5)(40 â€“ x)

=40x â€“ x^{2} â€“ 200 + 5x

= -x^{2} + 45x â€“ 200

So, -x^{2} + 45x â€“ 200 = 124

i.e., -x^{2} + 45x â€“ 324 = 0

i.e., x^{2} - 45x + 324 = 0

Therefore, the number of marbles John had, satisfies the quadratic equation

x^{2} â€“ 45x + 324 = 0

Which is the required representation of the problem mathematically.

Now by solving we get, x^{2} â€“ 9x â€“ 36 x + 324 = 0

x(x â€“ 9) - 36(x â€“ 9) = 0

(x - 9)(x - 36) = 0

âˆ´ x = 9 (or) 36.

âˆ´ The number of marbles John had is 9 (or) 36.

(ii) Let the number of toys produced on that day be x.

Therefore, the cost of production (in rupees) of each toy that day = 55 â€“ x

So, the total cost of production (in rupees) that day = x(55 â€“ x)

Therefore, x(55 â€“ x) = 750

i.e., 55x â€“ x^{2} = 750

i.e., -x^{2} + 55x â€“ 750 = 0

i.e., x^{2} - 55x + 750 = 0

Therefore, the number of toys produced that day satisfies the quadratic equation

x^{2} â€“ 55x + 750 = 0

which is the required representation of the problem mathematically.

Now by solving we get, x^{2} â€“ 25x â€“ 30x + 750 = 0

x(x â€“ 25)- 30(x â€“ 25) = 0

(x - 25)(x - 30) = 0

x = 25, x = 30.

The number of toys produced in a day = 25 (or) 30.

# Question-7

**Find two numbers whose sum is 27 and product is 182.****Solution:**

Let the two numbers be x and y

x + y = 27 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

xy = 182 â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)

from (1) x = 27 â€“ y

Substituting in (2) we get,

(27 â€“ y)y = 182

27y â€“ y

^{2}= 182

y

^{2}â€“ 27y + 182 = 0

y

^{2}â€“ 13y - 14y + 182 = 0

y(y â€“ 13) â€“ 14(y â€“ 13) = 0

(y â€“ 14)(y - 13) = 0

y = 14 (or) y = 13

If y = 14, x = 27 â€“ 14 = 13

If y = 13, x = 27 â€“ 13 = 14

Hence the two numbers are 13 and 14.

# Question-8

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

**Solution:**

Let the 2 consecutive positive nos be x, x+1

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

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

2x^{2} + 2x â€“ 364 = 0

x ^{2} + x â€“ 182 = 0 (Dividing by 2)

â‡’ x^{2} + 14x â€“ 13x â€“ 182 = 0

x(x + 14) â€“ 13(x + 14) = 0

(x â€“ 13)(x + 14) = 0

â‡’ x = 13 (or) x = -14.

Since the positive integers cannot be -14

Hence the positive integers are 13 and 14.

# Question-9

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

**Solution:**

In the right angled triangle

Let (BC) the base be x cm and altitude (AB) be (x â€“ 7) cm

In the right angled triangle,

(AB)

^{2}+ (BC)

^{2}= (AC)

^{2}(Pythagoras theorem)

(x - 7)

^{2}+ x

^{2}= 13

^{2}

x

^{2}â€“ 14x + 49 + x

^{2}= 169

2x

^{2}â€“ 14x â€“ 120 = 0

x

^{2}â€“ 7x â€“ 60 = 0 (Dividing by 2 )

x

^{2}â€“ 12x +5x - 60 = 0

x(x â€“ 12) + 5(x â€“ 12) = 0

(x â€“ 12)(x + 5) = 0

=> x = -5 (or) x = 12

Since the side (base) cannot be negative, x â‰ -5

We assume x = 12

=> Base (BC) = 12 cm

Altitude = x â€“ 7 = 5 cm

Thus the two sides of the triangle are 5 cm and 12 cm.

# Question-10

**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**`**90, find the number of articles produced and the cost of each article.**

**Solution:**

Let the number of articles be x and the cost of each article be y

Given:

y = 2x + 3 â€¦â€¦â€¦â€¦â€¦â€¦..(1)

Total cost of production be xy = 90 â€¦â€¦â€¦â€¦â€¦.. (2)

Substuting y = 2x + 3 in (2), we get

x(2x + 3) = 90

2x

^{2}+ 3x â€“ 90 = 0

2x

^{2}+ 15x â€“ 12x â€“ 90 = 0

2x

^{2}+ 15x â€“ 12x â€“ 90 = 0

x(2x + 15) - 6(2x + 15) = 0

(2x + 15) (x â€“ 6) = 0

2x + 15 = 0 ; x â€“ 6 = 0

=>x = ; 6

Since the number of articles cannot be negative as well as a fraction x â‰

Hence the number of articles is 6 (i.e.,) x = 6

y = 2x + 3 from (1)

= 2(6) + 3 = 15

(i.e.,) Number of articles = 6

Cost of each article = `15

# Question-11

**Find the roots of the following quadratic equations, if they exist, by the method of completing the square:****(i) 2x**^{2}â€“ 7x + 3 = 0 (ii) 2x^{2}+ x â€“ 4 = 0**(iii) 4x**^{2}+4x + 3 = 0 (iv) 2x^{2}+ x + 4 = 0**Solution:**

**(i)2x**

^{2}â€“ 7x + 3 = 0Dividing by the coefficient of x

^{2}, we get

x

^{2}-

Comparing the quadratic equation ax

^{2}+ bx + c = 0

Here, a = 1, b = 7/2 and c = 3/2

Adding and subtracting the square of (b/2) = we get,

x

^{2}â€“ 2 Â´ Ã— x + = 0

= 0

= 0

= 0

Taking the square root we get,

When

x = =

When

(or) x =

**(ii) 2x**

^{2}+ x â€“ 4 = 0Dividing by 2 (coefficient of x

^{2}), we get

x

^{2}+ = 0

Adding and subtracting the square of(b/2) = we get,

x

^{2}+ 2(x) Ã— +

= 0

Taking the square root we get,

x =

âˆ´ x = ;

**(iii) 4x**

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

^{2}+ 2(2x) Ã— + ()

^{2}= 0

(2x +)

^{2}= 0

2x = -

x = -

âˆ´ Roots of equation are x = -

**(iv) 2x**

^{2}+ x + 4 = 0x

^{2}+ x/2 + 2 = 0

Adding and subtracting the square of (b/2) = we get,

x

^{2}+ 2(x) Ã—

= 0

The square of a number cannot be negative. Hence roots do not exist.

# Question-12

**Find the roots of the following quadratic equations by applying the quadratic formula.****(i) 2x**^{2}â€“ 7x + 3 = 0 (ii) 2x^{2}+ x â€“ 4 = 0**(iii) 4x**^{2}+ 4 = 0 (iv) 2x^{2}+ x + 4 = 0**Solution:**

**(i) 2x**

^{2}â€“ 7x + 3 = 0According to the quadratic formula: ax

^{2}+ bx + c = 0

a = 2

b = -7

c = 3 âˆ´ The roots are =

=

= =

=

= = 3 ;

âˆ´ The roots are 3, .

**(ii) 2x**

^{2}+ x â€“ 4 = 0According to the quadratic formula: ax

^{2}+ bx + c = 0

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

âˆ´ The roots are

=

=

âˆ´ The roots are ,

**(iii) 4x**

^{2}+ 4 = 0According to the quadratic formula: ax

^{2}+ bx + c = 0

a = 4, b = 4, c = 3

âˆ´ The roots are =

=

=

=

= ,

The roots are (,) =

**(iv)2x**

^{2}+ x + 4 = 0According to the quadratic formula: ax

^{2}+ bx + c = 0

a = 2, b = 1, c = 4

âˆ´ The roots are =

=

=

=

This is not possible, hence the roots do not exists.

# Question-13

**Find the roots of the following equations:**

**(i)**,**x â‰ 0****(ii)**,**x â‰ -4, 7****Solution:**

**(i)**,

**x â‰ 0**

x

^{2}â€“ 1 = 3x

x

^{2}â€“ 3x â€“ 1 = 0

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

The roots are x =

x =

=

(i.e.,) x = and

(ii)

(x+4)(x â€“ 7) = -30

x

^{2}â€“ 3x â€“ 28 = -30

x

^{2}â€“ 2x - x + 2 = 0

x(x â€“ 2) â€“1(x â€“ 2) = 0

(x â€“ 2) (x â€“ 1) = 0

x = 1 (or) 2 are the roots of the equation.

# Question-14

**The sum of the reciprocals of Rehmanâ€™s ages(in years) 3 years ago and 5 years from now is Find his present age.**

**Solution:**

Let Rehmanâ€™s present age be x years.

His age 3 years ago = x â€“ 3

5 years from now = x + 5

=

Taking LCM,

3(2x + 2) = (x â€“ 3) (x + 5)

6x + 6 = x^{2} + 2x -15

x^{2} â€“ 4x â€“ 21 = 0

x^{2} â€“ 7x + 3x - 21 = 0

x(x â€“ 7) + 3(x â€“ 7) = 0

â‡’ x = 7 or x = -3

Age cannot be negative thus x = 7.

âˆ´ His present age = 7 years

# Question-15

**In a class test, the sum of Shefaliâ€™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.****Solution:**

Let her marks in Maths = x and in English = y

Given x + y = 30 â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Given, (x + 2)(y â€“ 3) = 210 â€¦â€¦ (2)

From (1): x = 30 - y â€¦â€¦(3)

Substututing in (2) we get

(30 â€“ y + 2) (y â€“ 3) = 210

(32 â€“ y) (y â€“ 3) = 210

32y â€“ 96 â€“ y

^{2}+ 3y = 210

y

^{2}â€“ 35y + 306 = 0

y

^{2}â€“ 17y â€“ 18y + 306 = 0

y(y â€“ 17) â€“ 18(y â€“ 17) = 0

(y â€“ 18) (y â€“ 17) = 0

y = 18 (or) 17

â‡’ x = 13 (when y= 17 by substituting in (3))

x = 12 (when y= 18 by substituting in (3))

Her marks in Maths is 13 and Science is 17 (or) Maths is 12 and Science is 18.

# Question-16

**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.****Solution:**

Let x be the length of the shorter side

Longer side = (x + 30)

In the right angled Î” BCD,

(BD)

^{ 2}= (BC)

^{2}+ (CD)

^{2}(Pythogoras theorem)

(60 + x)

^{2}= (x + 30)

^{2}+ x

^{2}

(x + 60)

^{2}= x

^{2}+ x

^{2}+ 60x + 900

2x

^{2}+ 60x + 900 = x

^{2}+ 120x + 3600

x

^{2}- 60x â€“ 2700 = 0

x

^{2}- 90 x + 30x â€“ 2700 = 0

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

(x + 30)(x â€“ 90) = 0

x = -30; x = 90

Since the length of a side cannot be negative, x = 90

âˆ´ Shorter side = 90 m

âˆ´ Longer side = 90 + 30 = 120 m

# Question-17

**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****Solution:**

Let the two numbers be x, y

x

^{2}â€“ y

^{2}= 180 â€¦â€¦â€¦â€¦â€¦â€¦. (1)

y

^{2}= 8x â€¦â€¦â€¦â€¦â€¦â€¦. (2)

Substitute (2) in (1)

x

^{2}â€“ 8x = 180

x

^{2}â€“ 8x â€“ 180 = 0

x

^{2}+ 10x â€“ 18x â€“ 180 = 0

x(x + 10) â€“ 18(x + 10) = 0

(x + 10)(x â€“ 18) = 0

âˆ´ x = -10; x = 18

If x = -10 from (2)

y

^{2}= 8x= 8(-10) = -80

This is not possible as square of any cannot be negative.

If x = 18 from (2)

y

^{2}= 8x = 8(18) = 144

âˆ´ y = 12

Thus the numbers are 18, 12 (or) 18, -12.

# Question-18

**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**

**Solution:**

Let the speed of the train = x km/hr

Distance traveled = 360 km

Time taken = =

If the speed increases by 5km/hr the time required to reach the place reduces to one hour

Time taken =

= 1

360

360

1800 = x^{2} + 5x

x^{2} + 5x â€“ 1800 = 0

x^{2} + 45x â€“ 40x - 1800 = 0

x(x + 45) â€“ 40(x + 45) = 0

(x + 45)(x â€“ 40) = 0

âˆ´ x = 40 (or) x = -45

But speed of the train can only be positive, thus x = 40 km/hr

# Question-19

**Two water taps together can fill a tank in 9 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.****Solution:**

Let the time taken by smaller tap = x hrs

In 1 hr of the tank can filled by smaller tap.

Let the time taken by larger tap = (x â€“ 10) hrs

In 1 hr of the tank can be filled by larger tap.

& together they fill the tank in

=

=

75(2x - 10) = 8(x

^{2}â€“ 10x)

8x

^{2}â€“ 80x â€“ 150x + 750 = 0

8x

^{2}â€“ 230x + 750 = 0

4x

^{2}â€“ 115x + 375 = 0 (Dividing by 2)

4x

^{2}â€“ 100x â€“ 15x + 375 = 0

4x(x â€“ 25) â€“ 15(x â€“ 25) = 0

4x â€“ 15 = 0 (or) x â€“ 25 = 0

x = (or) x = 25

x = cannot satisfy the condition

Small tap will fill the tank in 25 hrs and larger tap in 15 hrs.

# Question-20

**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 11 km/h more than that of the passenger train, find the average speed of the two trains.****Solution:**

Let the average speed of the passenger train be x km/hr and the express train be (x + 11) km/hr

Time taken by passenger train =

Time taken by express train =

According to the question,

= 1

132

1452 = x

^{2}+ 11x

x

^{2}+11x â€“ 1452 = 0

x

^{2}+ 44x â€“ 33x â€“ 1452 = 0

x(x + 44) â€“ 33(x + 44) = 0

(x â€“ 33)(x + 44) = 0

x = 33, x = -44

Since the speed of train cannot be negative, x = 33 km/hr

Speed of the passenger train = 33 km/hr

Speed of the express train = 22km/hr

# Question-21

**Sum of the areas of two squares is 468 m**^{2}. If the difference of their perimeters is 24 m, find the sides of the two squares.**Solution:**

Let the side of the two squares be a, b.

Area of first square = a

^{2}

Area of second square = b

^{2}

âˆ´ a

^{2}+ b

^{2}= 468 â€¦â€¦â€¦â€¦â€¦.. (1)

Perimeter of first square = 4a

Perimeter of second square = 4b

4a - 4b = 24

â‡’ a â€“ b = 6

a = b + 6 â€¦â€¦â€¦â€¦â€¦â€¦.. (2)

substitute a in (1)

(b+6)

^{2}+ b

^{2}= 468

b

^{2}+ 12b + 36 + b

^{2}= 468

2b

^{2}+ 12b â€“ 432 = 0

Dividing by 2,

b

^{2}+ 6b â€“ 216 = 0

b

^{2}+ 18b â€“ 12b â€“ 216 = 0

b(b + 18) â€“12 (b +18) = 0

(b + 18)(b â€“ 12) = 0

âˆ´ b = -18, b = 12

Side cannot be negative thus b = 12 m

Substitute b = 12 in (2)

a = b + 6 = 12 + 6 = 18 m

Side of first square = 18 m

Side of second square = 12 m

# Question-22

**Find nature of roots of the following quadratic equation. If real roots exists find them.**

(i) 2x(i) 2x

^{2}â€“ 3x + 5 (ii) 3x^{2}- 4x + 4 (iii) 2x^{2}â€“ 6x + 3 = 0**Solution:**

**(i) 2x**

^{2}â€“ 3x + 5Here a = 2, b = -3, c = 5

Discriminant Î”= b

^{2}â€“ 4ac = 9 - (4 Ã— 2 Ã— 5)

= 9 â€“ 40

= -31

Equation 2x

^{2}â€“ 3x + 5 has imaginary roots.

**(ii) 3x**

^{2}- 4x + 4Here, a = 3, b = -4, c = 4

Î” = b

^{2}â€“ 4ac

= (16 Ã— 3) â€“ (4 Ã— 3 Ã— 4)

= (16 Ã— 3) â€“ (16 Ã— 3)

= 0

The roots are real and equal.

The roots are x =

=

=

=

=,

=,

= ,

**(iii) 2x**

^{2}â€“ 6x + 3 = 0a = 2 , b = -6, c = 3

Discriminant Î” = b

^{2}â€“ 4ac

= 36 â€“ (4 Ã— 2 Ã— 3)

= 36 â€“ 24

= 12

Î” > 0

âˆ´ Equation has real and distinct roots

x =

=

=

x =

x = (or) x =

# Question-23

**Find the value of k for each of the following quadratic equations, so that they have two equal roots.****(i) 2x**^{2}+ kx + 3 = 0**(ii) kx(x â€“ 2) + 6 = 0****Solution:**

**(i) 2x**

^{2}+ kx + 3 = 0As the equation has equal roots the value of Î” = 0

a = 2 , b = k, c = 3

Î” = b

^{2}â€“ 4ac

0 = k

^{2}â€“ (4 Ã— 2 Ã— 3)

0 = k

^{2}â€“ 24

k

^{2}= 24

k =

k = 2

**(ii) kx**

^{2}â€“ 2kx + 6 = 0As the equation has equal roots the value of D = 0

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

Î” = b

^{2}â€“ 4ac

0 = 4k

^{2}â€“ (4 Ã— k Ã— 6)

0 = 4k

^{2}â€“ 24k

0 = k

^{2}â€“ 6k

k(k â€“ 6) = 0

k = 0 (or) k = 6

Thus k = 6.

# Question-24

**Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m**^{2 }? If so, find its length and breadth.**Solution:**

Let length be x m and breadth be y m

Given, x = 2y .............(1)

ang xy = 800 m

^{2}..............(2)

substituting value of x from (1)

2y Ã— y = 800

2y

^{2}= 800

=> y

^{2}â€“ 400 = 0

y

^{2}= 400

y = 20 m â‡’ x = 40 m

Yes, it is possible to design a rectangular mango grove

**with length = 40 m and breadth = 20 m.**

# Question-25

**Is the following situtation possible, If so determine the present ages. The sum of ages of 2 friends is 20 years. Four years ago, the product of their ages in year was 48.****Solution:**

Let the present ages of friends be x, y.

x + y = 20 ................(1)

x = 20 â€“ y

The product of their ages 4 yrs ago,

(x â€“ 4)(y â€“ 4) = 48

xy â€“ 4y â€“ 4x +16 = 48

xy â€“ 4y â€“ 4x = 32

(20 â€“ y)y â€“ 4y â€“ 4(20 â€“ y) = 32

20y â€“ y

^{2}â€“ 4y â€“ 80 + 4y = 32

-y

^{2}+ 20y â€“ 112 = 0

Î” = b

^{2}â€“ 4ac

= 400 â€“ 4 Ã— 1 Ã— -112

= 400 â€“ 448

= -48

As Î” < 0 is not possible.

# Question-26

**Is it possible to design a rectangular park of perimeter 80 m and area 400 m**^{2}? If so find its length and breadth.**Solution:**

Area = Length Ã— Breadth

= l Ã— b = 400 m

^{2}...................(1)

Perimeter = 2(l + b) = 80 m

2(l + b) = 80

l + b = 40 ...................(2)

From (1), b =

substituting value of b in equation (2)

l + = 40

l

^{2}+ 400 = 40l

l

^{2}â€“ 40l + 400 = 0

Î” = b

^{2}â€“ 4ac

= (-40)

^{2}â€“ (4 Ã— 400 Ã— 1)

= 1600 â€“ 1600

= 0

Equation has equal roots.

As the equation has equal roots we can only design a square plot.