# Ncert Solutions For Class 10 Maths Ex 4.4

## Ncert Solutions For Class 10 Maths Chapter 4 Ex 4.4

Q.1 Find the roots of following equation:

x+1x=5$x+\frac{1}{x}=5$, with x ≠ 0.

Sol.

On simplifying the above equation we get:

x2 – 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:

x=(5)+(5)24(1×1)2×1$x=\frac{-(-5)+\sqrt{(-5)^{2}-4(1\times 1)}}{2\times 1}$   and   x=(5)(5)24(1×1)2×1$x=\frac{-(-5)-\sqrt{(-5)^{2}-4(1\times 1)}}{2\times 1}$

x=5+2542$x=\frac{5+\sqrt{25-4}}{2}$   and   x=52542$x=\frac{5-\sqrt{25-4}}{2}$

x=5+212$x=\frac{5+\sqrt{21}}{2}$   and   x=5212$x= \frac{5-\sqrt{21}}{2}$

Q.2 Find the roots of following equation:

1x1x3=7$\frac{1}{x}-\frac{1}{x-3}=7$, where x0,3$x\neq 0,3$

Sol.

On simplifying the above equation we get:

7x2 – 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:

x=(21)+(21)24(7×3)2×7$x=\frac{-(-21)+\sqrt{(-21)^{2}-4(7\times 3)}}{2\times 7}$   and   x=(21)(21)24(7×3)2×7$x=\frac{-(-21)-\sqrt{(-21)^{2}-4(7\times 3)}}{2\times 7}$

x=21+4418414$x=\frac{21+\sqrt{441-84}}{14}$   and   x=214418414$x=\frac{21-\sqrt{441-84}}{14}$

x=21+35714$x=\frac{21+\sqrt{357}}{14}$   and   x=2135714$x= \frac{21-\sqrt{357}}{14}$

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

7x2 + x + 2 = 0

Sol.

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

Since, discriminant (D) = b2 – 4ac ………….. (1)

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

D = 12 – 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 3x2 + kx + 5 = 0

Sol.

Given,                 3x2 + kx + 5 = 0

Here          a = 3, b = k and c = 5

Now, If b2 – 4ac = 0, then we will get two equal real roots.

Therefore,

k2 – 4×3×5 = 0

k2 = 60

Therefore for equal roots k= 215$2\sqrt{15}$

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:

kx2 – 9kx +3 = 0

Here        a = k, b = -9k and c = 3

Now, If b2 – 4ac = 0, then we will get two equal real roots.

Therefore,

(-9k)2 – 4(k×3) = 0

81k2 – 12k = 0

(or)

k(81k – 12) = 0

Since value of k cannot be 0.

Therefore for equal roots  k= 1281$\frac{12}{81}$

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

2x2 + x – 4 = 0

Sol.

Here        a = 2, b = 1 and c = – 4

Since,

D = b2 – 4ac

Therefore,

D= 12 – 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:

x=(1)+(1)24(2×4)2×2$x=\frac{-(1)+\sqrt{(1)^{2}-4(2\times -4)}}{2\times 2}$ and

x=(1)(1)24(2×4)2×2$x=\frac{-(1)-\sqrt{(1)^{2}-4(2\times -4)}}{2\times 2}$

x=1+1+324$x=\frac{-1+\sqrt{1+32}}{4}$    and   x=11+324$x=\frac{-1-\sqrt{1+32}}{4}$

x=1+334$x=\frac{-1+\sqrt{33}}{4}$   and   x=1334$x= \frac{-1-\sqrt{33}}{4}$

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

x2 + 30x +3x +90 –1224 = 0

x2 + 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:

x=(33)+(33)24(1×1134)2×1$x=\frac{-(33)+\sqrt{(33)^{2}-4(1\times-1134)}}{2\times 1}$   and   x=(33)(33)24(1×1134)2×1$x=\frac{-(33)-\sqrt{(33)^{2}-4(1\times -1134)}}{2\times 1}$

x=33+1089+45362$x=\frac{-33+\sqrt{1089+4536}}{2}$   and   x=331089+45362$x=\frac{-33-\sqrt{1089+4536}}{2}$

x=33+56252$x=\frac{-33+\sqrt{5625}}{2}$   and   x=3356252$x= \frac{-33-\sqrt{5625}}{2}$

x=33+752$x=\frac{-33+75}{2}$   and   x=33752$x=\frac{-33-75}{2}$

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 = 860m2…… (2)

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

(3 + 2y) y = 860

2y2 + 3y = 860

2y2 + 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:

y=(3)+(3)24(2×860)2×2$y=\frac{-(3)+\sqrt{(3)^{2}-4(2\times-860)}}{2\times 2}$   and   y=(3)(3)24(2×860)2×2$y=\frac{-(3)-\sqrt{(3)^{2}-4(2\times -860)}}{2\times 2}$

y=3+9+68804$y=\frac{-3+\sqrt{9+6880}}{4}$   and   y=39+68804$y=\frac{-3-\sqrt{9+6880}}{4}$

y=3+68894$y=\frac{-3+\sqrt{6889}}{4}$   and   y=368894$y= \frac{-3-\sqrt{6889}}{4}$

y=3+834$y=\frac{-3+83}{4}$   and   y=3834$y=\frac{-3-83}{4}$

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 = 540x$\frac{540}{x}$ hrs

Now,

When speed is reduced to 6km/h then time taken to cover the same distance = 540x6$\frac{540}{x-6}$ hrs

Now, according to given condition:

540x6540x=1$\frac{540}{x-6}-\frac{540}{x} = 1$

540x – 540 ( x – 6 )  = x ( x – 6 )

x2 – 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:

x=(6)+(6)24(1×3240)2×1$x=\frac{-(-6)+\sqrt{(-6)^{2}-4(1\times-3240)}}{2\times 1}$   and   x=(6)(6)24(1×3240)2×1$x=\frac{-(-6)-\sqrt{(-6)^{2}-4(1\times -3240)}}{2\times 1}$

x=6+36+129602$x=\frac{6+\sqrt{36+12960}}{2}$   and   x=636+129602$x=\frac{6-\sqrt{36+12960}}{2}$

x=6+129962$x=\frac{6+\sqrt{12996}}{2}$  and   x=6129962$x= \frac{6-\sqrt{12996}}{2}$

x=6+1142$x=\frac{6+114}{2}$  and  x=61142$x=\frac{6-114}{2}$

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

3x2 + 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:

x=(5)+(5)24(3×750)2×3$x=\frac{-(-5)+\sqrt{(-5)^{2}-4(3\times-750)}}{2\times 3}$   and   x=(5)(5)24(3×750)2×3$x=\frac{-(-5)-\sqrt{(-5)^{2}-4(3\times -750)}}{2\times 3}$

x=5+25+90006$x=\frac{5+\sqrt{25+9000}}{6}$   and   x=525+90006$x=\frac{5-\sqrt{25+9000}}{6}$

x=5+90256$x=\frac{5+\sqrt{9025}}{6}$   and   x= 590256$\frac{5-\sqrt{9025}}{6}$

x=5+956$x=\frac{5+95}{6}$   and   x=5956$x=\frac{5-95}{6}$

Therefore    x=503$x=\frac{50}{3}$  and  x= -15

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

Hence number of chocolates produced = x = 503$\frac{50}{3}$ 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)       x2 + 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:

x=(2)+(2)24(1×168)2×1$x=\frac{-(2)+\sqrt{(2)^{2}-4(1\times-168)}}{2\times 1}$   and   x=(2)(2)24(1×168)2×1$x=\frac{-(2)-\sqrt{(2)^{2}-4(1\times -168)}}{2\times 1}$

x=2+4+6722$x=\frac{-2+\sqrt{4+672}}{2}$   and   x=24+6722$x=\frac{-2-\sqrt{4+672}}{2}$

x=2+262$x=\frac{-2+26}{2}$   and   x=2262$x= \frac{-2-26}{2}$

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:

x2 + (x+2)2 = 100

x2 + x2 + 4 + 4x = 100

2x2 + 4x – 96 = 0

x2 + 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:

x=(2)+(2)24(1×48)2×1$x=\frac{-(2)+\sqrt{(2)^{2}-4(1\times-48)}}{2\times 1}$   and   x=(2)(2)24(1×48)2×1$x=\frac{-(2)-\sqrt{(2)^{2}-4(1\times -48)}}{2\times 1}$

x=2+4+1922$x=\frac{-2+\sqrt{4+192}}{2}$   and   x=24+1922$x=\frac{-2-\sqrt{4+192}}{2}$

x=2+142$x=\frac{-2+14}{2}$   and   x=2142$x= \frac{-2-14}{2}$

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 = 10560$\frac{105}{60}$ hours

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 time=distancespeed$time =\frac{distance}{speed}$

Therefore time taken by motorboat to go upstream = 3521xhours$\frac{35}{21-x}\;hours$

And time taken by motorboat to go downstream       = 3521+x$\frac{35}{21+x}$hours

Now, according to the given condition:

3521x3521+x=10560$\frac{35}{21-x}-\frac{35}{21+x}=\frac{105}{60}$

35(21+x21+x212x2)=74$35(\frac{21+x-21+x}{21^{2}-x^{2}})=\frac{7}{4}$

2x441x2=74×35$\frac{2x}{441-x^{2}}=\frac{7}{4\times 35}$

40x=441x2$40x = 441-x^{2}$

x2+40x441=0$x^{2}+40x-441=0$

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:

x=(40)+(40)24(1×441)2×1$x=\frac{-(40)+\sqrt{(40)^{2}-4(1\times-441)}}{2\times 1}$   and   x=(40)(40)24(1×48)2×1$x=\frac{-(40)-\sqrt{(40)^{2}-4(1\times -48)}}{2\times 1}$

x=40+1600+17642$x=\frac{-40+\sqrt{1600+1764}}{2}$   and   x=401600+17642$x=\frac{-40-\sqrt{1600+1764}}{2}$

x=40+33642$x=\frac{-40+\sqrt{3364}}{2}$   and   x=4033642$x= \frac{-40-\sqrt{3364}}{2}$

x=40+582$x=\frac{-40+58}{2}$   and   x=40582$x=\frac{-40-58}{2}$

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 544m2. Find the sides of both the squares.

Sol.

Let x be the side of 1st square and y be the side of 2nd square.

Now, according to the given conditions: –

4x – 4y = 32

Or,                  x – y = 8 …..………………………….. (1)

And

x2 + y2 = 544 ……………………………  (2)

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

(8 – y)2 + y2 = 544

y2 + 64 – 16y + y2 = 544

2y2 -16y – 480 = 0

Or,     y2 – 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:

y=(8)+(8)24(1×240)2×1$y=\frac{-(-8)+\sqrt{(-8)^{2}-4(1\times-240)}}{2\times 1}$   and   y=(8)(8)24(1×240)2×1$y=\frac{-(-8)-\sqrt{(-8)^{2}-4(1\times -240)}}{2\times 1}$

y=+8+64+9602$y=\frac{+8+\sqrt{64+960}}{2}$   and   y=+864+9602$y=\frac{+8-\sqrt{64+960}}{2}$

y=8+10242$y=\frac{8+\sqrt{1024}}{2}$   and   y=810242$y= \frac{8-\sqrt{1024}}{2}$

y=8+322$y=\frac{8+32}{2}$           and          y=8322$y=\frac{8-32}{2}$

Therefore       y=20     and    y= -12

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

Therefore, The side of 2nd square    =       y     = 20m

And the side of 1st 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$\frac{9}{7} hours$  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 = 150x+15$\frac{150}{x+15}$ hours

And, time taken by a passenger train to cover the same distance = 150x$\frac{150}{x}$ hours

Now, according to the given condition:

150x$\frac{150}{x}$150x+15$\frac{150}{x+15}$= 97$\frac{9}{7}$

150(x+15x)x(x+15)=97$\frac{150(x+15-x)}{x(x+15)}=\frac{9}{7}$

150×15×7=9(x2+15x)$150\times 15\times 7=9(x^{2}+15x)$

50×5×7=x2+15x$50\times 5\times 7=x^{2}+15x$

Therefore x2+15x1750=0$x^{2}+15x-1750 = 0$

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:

x=(15)+(15)24(1×1750)2×1$x=\frac{-(15)+\sqrt{(15)^{2}-4(1\times-1750)}}{2\times 1}$   and   x=(15)(15)24(1×1750)2×1$x=\frac{-(15)-\sqrt{(15)^{2}-4(1\times-1750)}}{2\times 1}$

x=15+225+70002$x=\frac{-15+\sqrt{225+7000}}{2}$  and  x=15225+70002$x=\frac{-15-\sqrt{225+7000}}{2}$

x=15+72252$x=\frac{-15+\sqrt{7225}}{2}$   and   x=1572252$x= \frac{-15-\sqrt{7225}}{2}$

x=15+852$x=\frac{-15+85}{2}$     and     x=15852$x=\frac{-15-85}{2}$

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$\frac{24}{5} hours$.

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$\frac{1}{x}$ 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 1x4$\frac{1}{x-4}$ part of the tank.

Both the taps together can fill tank in 245hours$\frac{24}{5}hours$.

Therefore in 1hour both the taps can fill 524$\frac{5}{24}$  part of the tank.

Now, according to the given condition:

1x+1x4=524$\frac{1}{x}+\frac{1}{x-4}=\frac{5}{24}$ x4+xx(x4)=524$\frac{x-4+x}{x(x-4)}=\frac{5}{24}$ 48x965=x24x$\frac{48x-96}{5}=x^{2}-4x$

5x268x+96=0$5x^{2}-68x+96=0$

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=(68)+(68)24(5×96)2×5$x=\frac{-(-68)+\sqrt{(-68)^{2}-4(5\times96)}}{2\times 5}$  and  x= (68)(68)24(5×96)2×5$\frac{-(-68)-\sqrt{(-68)^{2}-4(5\times96)}}{2\times 5}$

x=68+4624192010$x=\frac{68+\sqrt{4624-1920}}{10}$  and  x=684624192010$x=\frac{68-\sqrt{4624-1920}}{10}$

x=68+270410$x=\frac{68+\sqrt{2704}}{10}$  and  x=68270410$x= \frac{68-\sqrt{2704}}{10}$

x= 68+5210$\frac{68+52}{10}$  and  x=685210$x= \frac{68-52}{10}$

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 575m2 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

y2 -50y +575=0

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

Since, D=b24(ac)$D=b^{2}-4(ac)$

Therefore, D=324(1×108)$D = 3^{2}-4(1\times -108)$

Therefore D=441

Since, D0$D\geq 0$,    Therefore the above quadratic equation will have real roots and hence the given situation is practically possible.

x=(3)+(3)24(1×108)2×1$\frac{-(3)+\sqrt{(3)^{2}-4(1\times-108)}}{2\times 1}$   and   x=(3)(3)24(1×108)2×1$\frac{-(3)-\sqrt{(3)^{2}-4(1\times-108)}}{2\times 1}$

x=3+9+4322$\frac{-3+\sqrt{9+432}}{2}$  and  x=39+4322$\frac{-3-\sqrt{9+432}}{2}$

x=3+4412$x=\frac{-3+\sqrt{441}}{2}$  and  x=34412$x= \frac{-3-\sqrt{441}}{2}$

x=3+212$\frac{-3+21}{2}$   and  x=3212$x=\frac{-3-21}{2}$

D=b24ac$b^{2}-4ac$

Therefore D = (50)24(1×575)$(-50)^{2}-4(1\times 575)$

Therefore D= 200

Since, D0$D\geq 0$ Therefore The above quadratic will have real roots and hence it is possible to design a park with given conditions.

Now,

y=(50)+(50)24(1×575)2×1$\frac{-(-50)+\sqrt{(-50)^{2}-4(1\times575)}}{2\times 1}$   and   y=(50)(50)24(1×575)2×1$\frac{-(-50)-\sqrt{(-50)^{2}-4(1\times575)}}{2\times 1}$

y=50+250023002$\frac{50+\sqrt{2500-2300}}{2}$  and   y=50250023002$y=\frac{50-\sqrt{2500-2300}}{2}$

y=50+2002$y=\frac{50+\sqrt{200}}{2}$   and   y= 502002$\frac{50-\sqrt{200}}{2}$

y=50+1022$\frac{50+10\sqrt{2}}{2}$   and   y=501022$\frac{50-10\sqrt{2}}{2}$

Therefore y=25+52$y=25+5\sqrt{2}$   and   y=2552$y=25-5\sqrt{2}$

Now,

If breadth of park =  y=25+52$y=25+5\sqrt{2}$m

Then, Length of park =  (50 – y) =x=2552$x = 25 – 5\sqrt{2}$m

And If breadth of park  = y=2552m$y=25-5\sqrt{2}m$

Then Length of park = (50 – y)    = x=25+52$x =25 + 5\sqrt{2}$m

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,     AC2 = AP2 + PC2             (Pythagoras Theorem)

152 = x2 + (x + 3)2

225 = x2+x2+9+6x

2x2 + 6x – 216 = 0

(or)  x2 + 3x – 108 = 0

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

Since,D=b24(ac)$D=b^{2}-4(ac)$

Therefore D=324(1×108)$D = 3^{2}-4(1\times -108)$

Therefore D=441

Since, D0$D\geq 0$, Therefore the above quadratic equation will have real roots and hence the given situation is practically possible.

x=(3)+(3)24(1×108)2×1$x=\frac{-(3)+\sqrt{(3)^{2}-4(1\times-108)}}{2\times 1}$   and   x=(3)(3)24(1×108)2×1$\frac{-(3)-\sqrt{(3)^{2}-4(1\times-108)}}{2\times 1}$

x=3+9+4322$\frac{-3+\sqrt{9+432}}{2}$  and  x=39+4322$\frac{-3-\sqrt{9+432}}{2}$

x=3+4412$x=\frac{-3+\sqrt{441}}{2}$  and  x=34412$x= \frac{-3-\sqrt{441}}{2}$

x=3+212$\frac{-3+21}{2}$   and  x=3212$x=\frac{-3-21}{2}$

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 k2x2 – 21x + k + 9 = 0 is 1.

Sol.

From the given quadratic equation: –    a= k2, b= -21 and c= (k + 9)

Let the other root of this quadratic equation be β

Therefore,    sum of their roots = 1 + β = ba$\frac{-b}{a}$ =21k2$\frac{21}{k^{2}}$

1+β=21k2$1+\beta =\frac{21}{k^{2}}$

Therefore β=21k2k2$\beta =\frac{21-k^{2}}{k^{2}}$…………(1)

And product of their roots = 1 × β = ca$\frac{c}{a}$ = k+9k2$\frac{k+9}{k^{2}}$

Therefore β=k+9k2(2)$\beta =\frac{k+9}{k^{2}}…………(2)$

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

k+9k2=21k2k2$\frac{k+9}{k^{2}} = \frac{21-k^{2}}{k^{2}}$

Therefore k2+k12=0$k^{2}+k-12=0$

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=(1)+(1)24(1×12)2×1$k=\frac{-(1)+\sqrt{(1)^{2}-4(1\times-12)}}{2\times 1}$  and k=(1)(1)24(1×12)2×1$k=\frac{-(1)-\sqrt{(1)^{2}-4(1\times-12)}}{2\times 1}$

k=1+1+482$k=\frac{-1+\sqrt{1+48}}{2}$ and k=11+482$k=\frac{-1-\sqrt{1+48}}{2}$

k=1+492$k=\frac{-1+\sqrt{49}}{2}$  and  k=1492$k= \frac{-1-\sqrt{49}}{2}$

k=1+72$k=\frac{-1+7}{2}$  and  k=172$k=\frac{-1-7}{2}$

k=3 and k= -4.

Therefore, the value of k is 3 or -4

Q.20 Determine the nature of roots of equation: 2x2 + 3x + 9 = 0

Sol.

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

Since,         D = b2 – 4ac

Therefore, D = 33 – 4(9×2)= -63

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

Q.21 If the ratio of roots of quadratic equation kx2 + 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   1st root will be α and 2nd root will be .

Now, sum of roots = α + 5α = ba$\frac{-b}{a}$= 18k$\frac{-18}{k}$

Therefore, 6α=18k$6\alpha =\frac{-18}{k}$

Therefore α=3k$\alpha =\frac{-3}{k}$ ……………….(1)

And product of roots = α × 5α = ca=15k$\frac{c}{a}=\frac{15}{k}$

Therefore α2=3k$\alpha^{2} =\frac{3}{k}$………… (2)

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

(3k)2=3k$(\frac{-3}{k})^{2}=\frac{3}{k}$

Therefore, k = 2