** ****Q.1 Find the roots of following equation:**

** x+1x=5**, with x ≠ 0.

**Sol.**

On simplifying the above equation we get:

** x ^{2 }– 5x + 1 = 0**

Here, a=1, b = – 5 and c = 1.

On putting the values of a, b and c in quadratic formulae we get:

** **

**Q.2 Find the roots of following equation:**

** ****Sol.**

On simplifying the above equation we get:

**7x ^{2} – 21x + 3 = 0**

Here, a = 7, b = -21 and c = 3

On putting the values of a, b and c in quadratic formulae we get:

** **

**Q.3 Find the nature of roots of the following quadratic equation:-**

** 7x ^{2 }+ x + 2 = 0**

** ****Sol.**

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

Since, discriminant (D) = b^{2 }– 4ac ………….. (1)

On putting the values of a, b and c in equation (1) we get: –

D = 1^{2} – 4(7×2)

Therefore D = -55

Since, **D < 0.** This equation will have** no real roots**.

**Q.4 Find the values of k in the following quadratic equation so that they have two equal real roots 3x ^{2 }+ kx + 5 = 0**

**Sol.**

Given, 3x^{2 }+ kx + 5 = 0

Here a = 3, b = k and c = 5

Now, If b^{2 }– 4ac = 0, then we will get two equal real roots.

Therefore,

k^{2} – 4×3×5 = 0

k^{2 }= 60

**Therefore for equal roots k=**

** **

**Q.5** **Find the values of k in the following quadratic equation so that they have two equal real roots. **

**kx(x – 9) + 3 = 0**

**Sol.**

On simplifying the given equation:

kx^{2} – 9kx +3 = 0

Here a = k, b = -9k and c = 3

Now, If b^{2 }– 4ac = 0, then we will get two equal real roots.

Therefore,

(-9k)^{2} – 4(k×3) = 0

81k^{2} – 12k = 0

(or)

k(81k – 12) = 0

Since value of k cannot be 0.

**Therefore for equal roots k=**

** **

**Q.6** **Find the nature of roots of the following quadratic equation, if real roots exist, Find them.**

**2x ^{2 }+ x – 4 = 0**

** ****Sol.**

Here a = 2, b = 1 and c = – 4

Since,

D = b^{2} – 4ac

Therefore,

D= 1^{2} – 4(- 4 × 2)

D= 1 + 32

Therefore, D= 33

**Since, D > o. The given quadratic equation will have real roots**.

On putting the values of a, b and c in the quadratic formulae we get:

** **

**Q.7 Deepak’s father ****is 27** **years older than him. 3 years from now the product of their ages will be 1224. Find the present age of Deepak****.**

**Sol.**

Let the present age of **Deepak** be **x years**

Then, present age of **Deepak’s father** will be = **(x + 27) years**

Now, their ages **after 3 years**: –

Deepak’s age = **(x + 3) years**

Deepak’s father age = x + 27 + 3 = **(x + 30) years**

According to given condition,

(x+3) (x+30) = 1224

x^{2} + 30x +3x +90 –1224 = 0

** x ^{2 }+ 33x – 1134 = 0** (Where x is the present age of Deepak in years)

**This is the required quadratic equation. **

Here, a = 1, b = 33 and c = -1134

On putting the values of a, b and c in the **quadratic formulae** we get:

**Therefore x=21 and x= -54**

Since, age cannot be negative. Therefore neglecting x = -54

Thus the **present age of Deepak = x = 21 years**

And the **present age of Deepak’s father will be = (x + 27) = 48 years**

** ****Q.8** **There is a rectangular park of area 860m ^{2}. The length (meters) of the park is three more than twice its breadth (meters). Find the length and breadth of this rectangular park.**

**Sol.**

Let, **x** be **length** (meters) and **y** be **breadth** (meter) of the rectangular plot

Now, according to given condition:

** x = 3 + 2y ……… (1)**

And

** x.y = 860m ^{2}…… (2)**

Now, substituting equation (1) in (2) we get

(3 + 2y) y = 860

2y^{2 }+ 3y = 860

**2y ^{2 }+ 3y – 860 = 0** where y = breadth (in meters)

**This is the required quadratic equation.**

Here, a = 2, b = 3 and c = -860

On putting the values of a, b and c in the **quadratic** **formulae** we get:

**Therefore y=20 and y= -21.5**

Since, length cannot be negative. Therefore neglecting y = -21.5

Thus **breadth of the rectangular park = y = 20m**

And **length of the rectangular park, x = (3 + 2y) = 43m**

** **

**Q.9** **A train is covering a distance of 540km from one city to another city at a uniform speed. If the speed of train had been reduced by 6km/h, then to cover the same distance the train would have taken 1hour more. Find the uniform speed of train.**

**Sol.**

Let the train travels at the uniform speed of **x km/hr**

Therefore,

Time taken to cover 540km =

Now,

When speed is reduced to **6km/h** then time taken to cover the same distance =

Now, according to given condition:

540x – 540 ( x – 6 ) = x ( x – 6 )

** **** x ^{2 }– 6x – 3240 = 0** Where x= speed of train in km/h

**This is the required quadratic equation.**

Here, a = 1, b = -6 and c = -3240

On putting the values of a, b and c in the **quadratic formulae** we get:

**Therefore x=60 and x= -54**

Since, speed cannot be negative. Therefore neglecting x = -54

**Thus, the uniform speed of train = x = 60km/h.**

**Q.10** **A chocolate factory produces a certain number of chocolates in a day. It was observed that on a particular day the cost of production of each chocolate (in rupees) was 5 more than thrice the number of chocolates produced on that day. If total cost of production on that day was 750 rupees. Find the number of chocolates produced and the cost of each chocolate.**

** ****Sol.**

Let **x** be number of **chocolates produced** and **y** be the **cost** of each chocolate.

Now, According to given conditions:

**y = 5 + 3x…………………. (1)**

And

**x.y = 750 ……………. (2)**

Substituting values of equation (1) in equation (2) we get:

x(3x + 5) = 750

**3x ^{2 }+ 5x – 750 = 0**

This is the required quadratic equation.

Here, a = 3, b = -5 and c = -750

On putting the values of a, b and c in the **quadratic** **formulae** we get:

**Therefore x=503 and x= -15**

Since, number of chocolates cannot be negative. Therefore neglecting x = -15

Hence number of **chocolates produced = x = 503 chocolates**

And **cost of each chocolate = y = (5 + 3x) = 55 Rs**

** **

**Q.11 Find two consecutive even positive integers, whose product is 168.**

** ****Sol.**

Let **x** be the **first** **number**, then **second** **number** will be **(x + 2).**

Now, according to the given condition:

x(x + 2) = 168

(or) **x ^{2} + 2x -168 = 0**

Here a =1, b = 2 and c = -168

Now, substituting the values of a, b and c in **quadratic** **formulae** we get:

**Therefore x=12 and x= -14**

Since x can’t be negative, therefore neglecting x = -14.

Therefore **first number= x = 12 and second number = (x + 2) = 14**

**Q.12** **Find two consecutive even positive integers, sum of whose squares is 100.**

**Sol.**

Let **x** be the **first** **number**, then **second** **number** will be **(x + 2)**

Now, According to the given condition:

x^{2 }+ (x+2)^{2} = 100

x^{2 }+ x^{2} + 4 + 4x = 100

2x^{2 }+ 4x – 96 = 0

** x ^{2 }+ 2x – 48 = 0**

Here, a =1, b = 2 and c = -48

Now, substituting the values of a, b and c in quadratic formulae we get:

**Therefore x=6 and x= -8**

Since x can’t be negative, therefore neglecting x = -8

Therefore **first number = x = 6 and second number = (x + 2) = 8**

**Q.13** **It takes 105mints more for a motor boat to go 35km upstream than to return downstream at the same spot. If speed of a motor boat in still water is 21 km/h. Find speed of the stream. **

**Sol.**

Since 60mints = 1 hour

Therefore 105mints =

Let the **speed of stream be x km/h**

Therefore, **speed** of **boat** in **upstream** = **(21-x) km/h**

And **speed** of **boat** in **downstream** = **(21+x) km/h**

Since

Therefore time taken by motorboat to go upstream =

And time taken by motorboat to go downstream =

Now, according to the given condition:

Now, from the above quadratic equation:- a = 1, b = 40 and c = -441

Substituting the values of a, b and c in **quadratic** **formulae** we get:

**Therefore x =9 and x = -49**

Since distance can’t be negative, therefore neglecting x = -49.

**Therefore, the speed of stream = 9 km/h.**

**Q.14** **If the difference in the parameter of two squares is 32m and sum of the areas of both squares is 544m ^{2}. Find the sides of both the squares.**

**Sol.**

Let **x** be the side of **1 ^{st} square** and

**y**be the side of

**2**.

^{nd}squareNow, according to the given conditions: –

4x – 4y = 32

Or, **x – y = 8 …..………………………….. (1)**

And

** x ^{2} + y^{2} = 544 …………………………… (2)**

Now, substituting equation (1) in equation (2) we get:

(8 – y)^{2} + y^{2} = 544

y^{2} + 64 – 16y + y^{2} = 544

2y^{2} -16y – 480 = 0

Or, **y ^{2} – 8y – 240 = 0**

Now, from the above quadratic equation:- a = 1, b = -8 and c = -240

Substituting the values of a, b and c in **quadratic** **formulae** we get:

**Therefore y=20 and y= -12**

Since distance can’t be negative, therefore neglecting y = -12.

Therefore, The **side of 2 ^{nd} square = y = 20m**

And the **side of 1 ^{st} square** = (8 + y) =

**28m**

**Q.15** **If the average speed of an express train is 15km/h more than that of a passenger train and for covering a distance of 150km between two stations A and B an express train takes 97hours less than a passenger train. Find the average speed of train A and train B.**

** ****Sol.**

Let **x km/h** be the average speed of a **passenger train**.

Therefore, Average speed of an **express train** will be **(x + 15) km/h.**

Now, time taken by an express train to cover 150km =

And, time taken by a passenger train to cover the same distance =

Now, according to the given condition:

**Therefore x2+15x−1750=0**

Now, from the above quadratic equation:- a = 1, b = 15 and c = -1750

Substituting the values of a, b and c in **quadratic** **formulae** we get:

**Therefore x=35 and x= -50**

Since speed can’t be negative, therefore neglecting x = -50.

**Therefore, speed of passenger train= x = 35km\h.**

** Andspeed of express train= (x + 15) = 50km\h.**

** **

**Q.16** **Two water taps with different diameters are used to fill a particular tank, both the taps together can fill a tank in 245hours.**

**However, if they are operated separately, tap with larger diameter can fill the tank 4 hours early as compared to the tap with smaller diameter. Find the time in which each tap can separately fill the tank.**

**Sol.**

Let, **x hours** be the time taken by the tap with smaller diameter to fill a tank separately.

Thus, in 1 hour it can fill ** 1x** part of the tank.

Let, the time taken by the tap with larger diameter to fill a tank separately is **(x-4) hours.**

Thus, in 1 hour it can fill ** 1x−4** part of the tank.

Both the taps together can fill tank in

Therefore in 1hour both the taps can fill

Now, according to the given condition:

Now, from the above quadratic equation:- a = 5, b = -68 and c = 96

Substituting the values of a, b and c in quadratic formulae we get:

x=

**Therefore x=12 and x= 1.6**

When x = 1.6, It doesn’t satisfy the given conditions, therefore neglecting x = 1.6

Therefore the **time taken by the tap with smaller diameter to fill the tank separately = x = 12 hours**

And the **time taken by the tap with larger diameter to fill the tank separately** = (x – 4) = **8 hours**

**Q.****17** **There is a requirement of a rectangular park of area 575m ^{2} and perimeter 100m. Check whether it’s possible or not? If possible then find length and breadth of that park.**

**Sol.**

Let **x** be **length** of the park and **y** be **breadth** of the park in meters.

Now, according to the given conditions:

2(x + y) = 100

(or)

x + y = 50 ……………………… (1)

And

x.y = 575 ……………………. (2)

Now, substituting equation (1) in equation (2) we get:

(50 – y)y = 575

**y ^{2} -50y +575=0**

Now, from the above quadratic equation: a = 1, b = -50 and c = 575

Since,

Therefore,

Therefore D=441

Since, **quadratic** **equation** will have **real** **roots** and hence the given situation is practically possible.

Now, from the quadratic formulae:

x=

x=

x=

D=

Therefore D =

Therefore D= 200

**Since, D≥0** Therefore

**The above quadratic will have real roots**and hence it is possible to design a park with given conditions.

Now,

y=

y=

y=

**Therefore y=25+52–√ and y=25−52–√**

Now,

If **breadth** of park =

Then, **Length** of park = (50 – y) =

And If **breadth** of park =

Then **Length** of park = (50 – y) =

**Q.18** **There is a circular park with diameter equal to 15m having two gates one for entry and other for exit fixed at diametrically opposite ends. A plant has to be planted on the boundary of a circular park in such a way that the difference of its distance from two opposite fixed gates P and Q from boundary is 3 metres.**

**Check whether the above give situation is possible or not? **

**And if it is possible to do so, then find the distance of plant from both the gates P and Q.**

**Sol.**

Let P be the location of the plant, A and C represents the location of gates of the park.

Now, According to the given condition:

PC – AP = 3

PC = 3 + AP

= 3 + x

Therefore, AC^{2 }= AP^{2 }+ PC^{2} (Pythagoras Theorem)

15^{2} = x^{2 }+ (x + 3)^{2}

225 = x^{2}+x^{2}+9+6x

2x^{2 }+ 6x – 216 = 0

(or) **x ^{2 }+ 3x – 108 = 0**

Now, from the above quadratic equation a = 1, b =3 and c = -108

Since,

Therefore

Therefore D=441

Since, **quadratic** **equation** will have **real** **roots** and hence the given situation is practically possible.

Now, from the quadratic formulae:

x=

x=

**Therefore x=9 and x= -12**

Since, the distance cannot be negative therefore neglecting x= -12

**Therefore, AP = x =9m and PC = (x+3) = 12m.**

**Q.19 Find the value of k, if one root of the quadratic equation k ^{2}x^{2 }– 21x + k + 9 = 0 is 1. **

**Sol.**

From the given quadratic equation: – a= k^{2}, b= -21 and c= (k + 9)

Let the other root of this quadratic equation be β

Therefore, sum of their roots = 1 + β =

Therefore

And product of their roots = 1 × β =

Therefore

Now, from equation (1) and equation (2) we get:

**Therefore k2+k−12=0**

Now, from the above quadratic equation: a = 1, b = 1 and c = -12

Substituting the values of a, b and c in **quadratic formulae** we get:

** k=3 and k= -4.**

Therefore, **the value of k is 3 or -4**

**Q.20 Determine the nature of roots of equation: 2x ^{2 }+ 3x + 9 = 0**

** ****Sol.**

From the given equation a = 2, b = 3 and c = 9.

Since, D = b^{2 }– 4ac

Therefore, D = 3^{3} – 4(9×2)= -63

**Since ****D < 0**, the given equation has no real roots.

**Q.21 If the ratio of roots of quadratic equation kx ^{2 }+ 18x + 15 = 0 is 1:5. Find the possible values of k.**

**Sol.**

From the given **quadratic** **equation**: a = k, b = 18 and c = 15

Since, roots of the given quadratic equation are in ratio **1:5**

Therefore **1 ^{st} root** will be

**α**and

**2**will be

^{nd}root**5α**.

Now, sum of roots = **α + 5α** =

Therefore,

Therefore

And product of roots = **α × 5α** =

Therefore

Now, on substituting equation (1) in equation (2) we get:

Therefore, **k = 2**