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**Ques 1: Find the roots of the following quadratic equations, if they exist, by the method of completing the square:**

**(i) 2x ^{2} âˆ’ 7x + 3 = 0(ii) 2x^{2} + x âˆ’ 4 = 0(iii) 4x^{2} + 4âˆš3 x+ 3 = 0(iv) 2x^{2} + x + 4 = 0**

**Sol: **(i)** **2x^{2 }âˆ’ 7x + 3 = 0

Dividing throughout by the co-efficient of x^{2}, we get

â‡’

â‡’

â‡’

â‡’

â‡’ **Case I:**

**Case II:**

Thus, required roots are

x = 3 and x = 1/2

(ii) 2x^{2} + xâˆ’ 4 = 0

We have:

2x^{2} + x âˆ’ 4 = 0

Dividing throughout by 2,

â‡’

â‡’

â‡’

â‡’

â‡’

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

Dividing throughout by 2, we have:

But the square of a number cannot be negative.

cannot be a real value.

So, no real roots exist.

â‡’ There is no real value of x satisfying the given equation.**Ques ****2: Find the roots of the following quadratic equations, using the quadratic formula:**

**(i) 2x ^{2} âˆ’ 7x + 3 = 0(ii) 2x^{2} + x âˆ’ 4 = 0(iii) 4x^{2} + 4âˆš3 x + 3 = 0(iv) 2x^{2} + x + 4 = 0**

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

a = 2

b = âˆ’ 7

c = 3

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 7^{)}2 âˆ’ 4 (2) (3)

= 49 âˆ’ 24 = 25 â‰¥ 0

Since b^{2} âˆ’ 4ac > 0

âˆ´ The given equation has real roots. The roots are given by

Thus, the roots of the given equation are

(ii) 2x^{2} + xâˆ’4 = 0

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

a = 2

b = 1 c = âˆ’ 4

âˆ´ b^{2} âˆ’ 4ac = (1)^{2} âˆ’ 4 (2) (âˆ’ 4) = 1 + 32

= 33 > 0

Since^{ }b^{2} âˆ’ 4ac > 0

âˆ´ The given equation has equal roots. The roots are given by

Thus, the required roots are:

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

a = 4

b = 43

c = 3

âˆ´

= [16 Ã— 3] âˆ’ 48

= 48 âˆ’ 48 = 0

Since b^{2} âˆ’ 4ac = 0

âˆ´ The given equation has real roots, which are given by:

âˆ´

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

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

a = 2

b = 1

c = 4

âˆ´ b^{2} âˆ’ 4ac = (1)^{2} âˆ’ 4 (2) (4)

= 1 âˆ’ 32 = âˆ’ 31 < 0

Since b^{2} âˆ’ 4ac is less than 0, therefore the given equation does not have real roots.**Ques ****3: Find the roots of the following equations:**

**Sol:**

Here, we have:

â‡’ x^{2} âˆ’ 1= 3 x

â‡’ x^{2} âˆ’ 3x âˆ’ 1 = 0 ...(1)

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

a = 1

b = âˆ’ 3

c = âˆ’ 1

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 3)^{2} âˆ’ 4 (1) (âˆ’ 1)

= 9 + 4 = 13 > 0

âˆ´

â‡’

Now, taking +ve sign,

Taking âˆ’ve sign,

Thus, the required roots of the given equation are:

We have:

â‡’ âˆ’ 11 Ã— 30 = 11 (x^{2} âˆ’ 3x âˆ’ 28)

â‡’ âˆ’ 30 = x^{2} âˆ’ 3x âˆ’ 28

â‡’ x^{2} âˆ’ 3x âˆ’ 28 + 30 = 0

â‡’ x^{2} âˆ’ 3x + 2 = 0

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

a =1

b = âˆ’ 3

c = 2

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 3)^{2} âˆ’ 4 (1) (2)

= 9 âˆ’ 8 = 1 > 0

âˆ´ The quadratic equation (1) has real roots, which are given by:

Taking +ve sign, we have:

Taking âˆ’ve sign, we have:

Thus, the roots of the given equation are:

x = 2 and x = 1.**Ques ****4: The sum of the reciprocals of Rehmanâ€™s ages, (in years) 3 years ago and 5 years from now is 1/3.**

Find his present age.**Sol:** Let the present age of Rehman = x

âˆ´ 3 years ago Rehmanâ€™s age = (x âˆ’ 3) years

5 years later Rehmanâ€™s age = (x + 5) years

Now, according to the condition,

â‡’ 3 [x + 5 + x âˆ’ 3] = (x âˆ’ 3) (x + 5)

â‡’ 3 [2x + 2] = x^{2} + 2x âˆ’ 15

â‡’ 6x + 6 = x^{2} + 2x âˆ’ 15

â‡’ x^{2} + 2x âˆ’ 6x âˆ’ 15 âˆ’ 6=0

â‡’ x^{2} âˆ’ 4x âˆ’ 21 = 0 ...(1)

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

a =1

b = âˆ’ 4

c = âˆ’ 21

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 4)^{2} âˆ’ 4 (1) (âˆ’ 21)

= 16 + 84

= 100

Since,

â‡’

Taking positive sign, we have:

Taking negative sign, we have:

Since age cannot be negative,

âˆ´ x â‰ âˆ’ 3 â‡’ x = 7

So, the present age of Rehman = 7 years.**Ques ****5: 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.**

**Sol:** Let, Shefaliâ€™s marks in Mathematics = x

âˆ´ Marks in English = (30 âˆ’ x) [âˆµ Sum of their marks in Eng. and Maths = 30]

Now, according to the condition,

(x + 2) Ã— [(30 âˆ’ x) âˆ’ 3] = 210

â‡’ (x + 2) Ã— (30 âˆ’ x âˆ’ 3) = 210

â‡’ (x + 2) (âˆ’x + 27) = 210

â‡’ âˆ’x^{2} + 25x + 54 = 210

â‡’ âˆ’x^{2} + 25x + 54 âˆ’ 210 = 0

â‡’ âˆ’x^{2} + 25x âˆ’ 156 = 0

â‡’ x^{2} âˆ’ 25x + 156 = 0 ...(1)

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

a = 1

b = âˆ’25

c = 156

âˆ´ b^{2} âˆ’ 4ac =(âˆ’25)2 âˆ’ 4(1) (156)

= 625 âˆ’ 624 = 1

Since,

â‡’

â‡’

Taking +ve sign, we have

Taking âˆ’ ve sign, we get

When x = 13, then 30 âˆ’ 13 = 17

When x = 12, then 30 âˆ’ 12 = 18

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

marks in Maths = 12, marks in English = 18**Ques ****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.**

**Sol:** Let the shorter side of rectangle be x m, then longer side be (x + 30) m and diagonal be (x + 60) m

In right-angled Î”ABC,

(x + 60)^{2} = (x + 30)^{2} + (x)^{2} [Pythagoras Theorem]

â‡’ x^{2} + 120x + 3600 = x^{2} + 60x + 900 + x^{2}

â‡’ x^{2} - 60x - 2700 = 0

â‡’ x^{2} - 90x + 30x - 2700 = 0

â‡’ x (x - 90) + 30 (x - 90)= 0

â‡’ (x + 30) (x - 90) = 0

â‡’ x + 30 = 0 or x - 90 = 0

â‡’ x = -30 (rejected) or x = 90

Hence, shorter side = 90 m

and longer side = 90 + 30 = 120 m**Ques ****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.****Sol: **Let the larger number be x.

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

â‡’ (smaller number)^{2} = 8x

â‡’ smaller number =

â‡’ x^{2} âˆ’ 8x = 180

â‡’ x^{2} âˆ’ 8x âˆ’ 180 = 0

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

a = 1

b = âˆ’ 8

c = âˆ’ 180

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 8)^{2} âˆ’ 4 (1) (âˆ’ 180)

= 64 + 720 = 784

Taking +ve sign, we get

Taking âˆ’ve sign, we get

But x = âˆ’ 10 is not admissible,

âˆ´ The smaller number = 18

Thus, the larger number = 12 or âˆ’ 12

Thus, the two numbers are:

18 and 12 or 18 and âˆ’12**Ques ****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.**

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

Since, time taken by the train

â‡’

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

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’

â‡’ â€“5 Ã— 360 = âˆ’ 1 (x^{2} + 5x)

â‡’ â€“5 Ã— 360 = âˆ’ x^{2} âˆ’ 5x âˆµ 360 Ã— 5 = 1800

â‡’ x^{2} + 5x âˆ’ 1800 = 0 ...(1)

Comparing (1) with

ax^{2} + bx + c = 0

â‡’ a = 1

b = 5

c = âˆ’1800

âˆ´ b^{2} âˆ’ 4ac = (5)^{2} âˆ’ 4 (1) (âˆ’ 1800)

= 25 + 7200

= 7225

â‡’

Taking +ve sign, we get:

Taking âˆ’ve sign, we get

Since, the speed of a vehicle cannot be negative,

âˆ´ x â‰ âˆ’ 45

Thus, x = 40 â‡’ speed of the train = 40 km/hr.**Ques ****9: Two water taps together can fill a tank in 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.**

**Sol:** Let the smaller tap fills the tank in x hours.

âˆ´ The larger tap fills the tank (x âˆ’ 10) hours.

â‡’ Time to fill the tank by smaller tap = 1/x hours.

Time to fill the tank by larger tap = 1/x-10 hours.

Since, the tank filled by the two taps together in 1 hour

Now, according to the condition,

â‡’

â‡’ 8x^{2 }âˆ’ 80x = 150x âˆ’ 750

â‡’ 8x^{2} âˆ’ 80x âˆ’ 150x= âˆ’ 750

â‡’ 8x^{2} âˆ’ 230x + 750 = 0 ..(1)

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

a =8

b = âˆ’230

c = 750

âˆ´ b^{2} âˆ’ 4ac =(âˆ’230)^{2} âˆ’ 4 (8) (750)

= 52900 âˆ’ 24000

= 28900

Since

â‡’

â‡’

Taking, the +ve sign, we get

Taking the âˆ’ve sign, we get

For

which is not possible, [âˆ´ Time cannot be negative]

âˆ´ x =25

â‡’ x âˆ’ 10 = 25 âˆ’ 10 = 15

Thus, time to fill the tank by the smaller tap alone = 25 hours and larger tap alone = 15 hours.**Ques ****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 11 km/h more than that of the passenger train, find the average speed of the two trains.**

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

âˆ´ Average speed of the express train = (x + 11) km/hr Total distance covered = 132 km

Also,

Time taken by the passenger train = 132/x hour

Time taken by the express train = 132/x + 11 hour

According to the condition, we get

â‡’ âˆ’ 11 (132) = âˆ’ 1 (x^{2} + 11x)

â‡’ âˆ’ 1452 = âˆ’ 1 (x^{2} + 11x)

â‡’ x^{2} + 11x âˆ’ 1452 = 0 ...(1)

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

a = 1

b = 11

c = âˆ’1452

âˆ´ b^{2} âˆ’ 4ac = (11)^{2} âˆ’ 4 (1) (âˆ’1452)

â‡’ b^{2} âˆ’ 4ac = 121 + 5808 = 5929

âˆ´ Taking +ve sign, we get

Taking âˆ’ve sign, we get

But average speed cannot be negative âˆ´

x â‰ âˆ’44

âˆ´ x = 33

â‡’ Average speed of the passenger train = 33 km/hr

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

= (33 + 11) km/hr

= 44 km/hr**Ques ****11: 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.**

**Sol: **Let the side of the smaller square be x m.

â‡’ Perimeter of the smaller square = 4x m.

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

â‡’ Side of the larger square = Perimeter / 4

âˆ´ Area of the smaller square = (side)2 = (x)^{2} m^{2}

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

According to the condition,

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

â‡’ x^{2} + x^{2} + 12x + 36 = 468

â‡’ 2x^{2} + 12x âˆ’ 432 = 0

â‡’ x^{2} + 6x âˆ’ 216 = 0 ...(1)

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

âˆ´ a = 1

b = 6

c = âˆ’ 216

âˆ´ b^{2} âˆ’ 4ac = (6)^{2} âˆ’ 4 (1) (âˆ’ 216)

= 36 + 864 = 900

Since,

âˆ´

â‡’

Taking +ve sign, we have:

Taking âˆ’ve sign, we get

But the length of a square cannot be negative

âˆ´ x = 12

â‡’ Length of the smaller square = 12 m

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

= 12 + 6

= 18 m

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