Exercise 4.3

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

1)

$$2x^{2}-7x+3$$ =0

Sol.The given quadratic equation is

$$2x^{2}-7x+3$$ =0

Shifting constant term to R.H.S.,

$$2x^{2}-7x=-3$$

$$X^{2}-\frac{7}{2}x=-\frac{3}{2}$$

| dividing throughout by 2

Add

$$(\frac{1}{2} \:\: coeff. \:\: of \:\: x)^{2}$$    

i.e

$$(\frac{1}{2} \:\: X \:\: \frac{7}{2})^{2}$$

i.e.,

$$(\frac{7}{4})^{2}$$ to both sides

$$x^{2}-\frac{7}{2}x+(\frac{7}{4})^{2}=(\frac{7}{4})^{2}-\frac{3}{2}$$

=

$$\frac{49}{16}-\frac{3}{2}$$

=

$$\frac{49-24}{16}$$

$$\Rightarrow (x-\frac{7}{4})^{2}=\frac{25}{16}$$

Taking square roots of both sides,

$$x - \frac{7}{4} = ± \sqrt{\frac{25}{16}}$$

$$x-\frac{7}{4}=±\frac{5}{4}$$

x=

$$\frac{7}{4}±\frac{5}{4}$$

a=

$$\frac{7}{4}+\frac{5}{4}$$    and $$\frac{7}{4}-\frac{5}{4}$$

x=

$$\frac{12}{4}$$    and  $$\frac{2}{4}$$

x=3   and

$$\frac{1}{2}$$

hence, the roots of the quadratic equation

$$2x^{2}-7x+3$$ =0   are 3 and $$\frac{1}{2}$$ .

ii)

$$2x^{2}+x-4=0$$

Sol.The given quadratic equation is

$$2x^{2}$$ +x-4=0shifting constant term to R.H.S.,

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

Dividing every term by a=2 to make coefficient of

$$x^{2}$$ unity,

$$X^{2}+\frac{1}{2}$$ x=2

Add

$$(\frac{1}{2} \:\: coefficient \:\: of \:\: x)^{2}$$

i.e.,

$$(\frac{1}{2} \:\: X \:\: \frac{1}{2})^{2}=\frac{1}{4}^{2}$$ to both sides.

$$X^{2}+\frac{1}{2}x+(\frac{1}{2})^{2}=2+(\frac{1}{4})^{2}=2+\frac{1}{16}$$

$$(x+\frac{1}{4})^{2}=\frac{32+1}{16}=\frac{33}{16}$$

Taking  square roots of both sides

$$X + \frac{1}{4} = ±\sqrt{\frac{33}{16}}$$

X=

$$\frac{-1}{4}±\frac{\sqrt{33}}{4}=\frac{-1±\sqrt{33}}{4}$$

Hence the roots of the given equation are

X=

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

iii)

$$4x^{2}+4\sqrt{3}x+3$$ =0

Sol.The given quadratic equation is

$$4x^{2}+4\sqrt{3}x+3$$ =0

Shifting constant term to R.H.S.,

$$4x^{2}+4\sqrt{3}x$$ =-3

Dividing every term by a=4 to make coefficient of

$$x^{2}$$ unity,

$$X^{2}+\sqrt{3}x=-\frac{3}{4}$$

Adding

$$(\frac{1}{2}coeff. Of x)^{2}  i.e., (\frac{\sqrt{3}}{2})^{2}$$   to both sides,

$$X^{2}+\frac{2\sqrt{3}}{2}x+(\frac{\sqrt{3}}{2})^{2}=(\frac{\sqrt{3}}{2})^{2}-\frac{3}{4}$$

$$\Rightarrow (x+\frac{\sqrt{3}}{2})^{2}=\frac{3}{4}-\frac{3}{4}$$ =0

Taking square roots on both sides,

$$X+\frac{\sqrt{3}}{2}$$ =±0

X=

$$-\frac{\sqrt{3}}{2}±0=-\frac{\sqrt{3}}{2}+0, -\frac{\sqrt{3}}{2}-0$$

X=

$$-\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}$$

i.e., roots are equal.

Remark . here root

$$-\frac{\sqrt{3}}{2}$$ is repeated twice

iv)

$$2x^{2}+x+4$$ =0

Sol.The given quadratic equation is

$$2x^{2}+x+4$$ =0

Shifting constant term to R.H.S.,

$$2x^{2}+x=-4$$

$$X^{2}+\frac{1}{2}x=-2$$

|Dividing throughout by 2

Ass

$$(\frac{1}{2}coeff. Of x)^{2}  i.e. (\frac{1}{2} x \frac{1}{2})^{2}$$

i.e.

$$(\frac{1}{4})^{2}$$ to both sides,

$$x^{2}+2\frac{1}{4}x+(\frac{1}{4})^{2}=\frac{1}{16}-2$$

$$(x+\frac{1}{4})^{2}=\frac{1}{16}-2$$

=

$$\frac{1-32}{16}$$

$$(x+\frac{1}{4})^{2}=-\frac{31}{16}$$

Clearly R.H.S is negative

But

$$(x+\frac{1}{4})^{2}$$ cannot be negative for

Any real value of x because the square of a real number cannot be negative . Hence the given equation has no real roots i.e., solutions of the given equation do not exist.

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

i)

$$2x^{2}-7x+3=0$$

Sol.The given quadratic equation is

$$2x^{2}-7x+3=0$$              

|

$$ax^{2}+bx+c=0$$

Here , a=2  b=-7   c=3

So,

$$D=b^{2}-4ac=(-7)^{2}-4(2)(3)$$

=49-24

25≥0

Therefore,

$$x=\frac{-b±\sqrt{D}}{2a}$$

$$x=\frac{-(-7)±\sqrt{25}}{2(2)}=\frac{7±5}{4}$$

=

$$\frac{7+5}{4},  \frac{7-5}{4}$$

=3, 

$$\frac{1}{2}$$

ii)

$$2x^{2}+x-4=0$$

Sol.The given quadratic equation is

$$2x^{2}+x-4=0$$        | $$ax^{2}+bx+c=0$$

Here, a=2    b=1    c=-4

D=

$$b^{2}-4ac$$

=

$$(1)^{2}-4(2)(-4)$$

=1+32

=33≥0

Therefore,

$$x=\frac{-b±\sqrt{D}}{2a}$$

=

$$\frac{-1±\sqrt{33}}{2(2)}=\frac{-1±\sqrt{33}}{4}$$

=

$$\frac{-1+\sqrt{33}}{4},  \frac{-1-\sqrt{33}}{4}$$

So, the roots are

$$\frac{-1+\sqrt{33}}{4}$$    and $$\frac{-1-\sqrt{33}}{4}$$

iii)

$$4x^{2}+4\sqrt{3}x+3$$ =0

Sol.The given quadratic equation is

$$4x^{2}+4\sqrt{3}x+3$$ =0     | $$ax^{2}+bx+c=0$$

Here, a=4    b=

$$4\sqrt{3}$$     c=3

D=

$$b^{2}-4ac$$

=

$$(4\sqrt{3})^{2}-4(4)(3)$$

=16 x 3-16 x3=0

Therefore,

X=

$$\frac{-b±\sqrt{D}}{2a}$$

|using the quadratic formula

=

$$\frac{-4\sqrt{3}±\sqrt{0}}{2x4}$$

=

$$\frac{-4\sqrt{3}+0}{8}, \frac{-4\sqrt{3}-0}{8}$$

=

$$-\frac{\sqrt{3}}{2},  -\frac{\sqrt{3}}{2}$$

So, both the roots are

$$-\frac{\sqrt{3}}{2}$$ .

iv)

$$2x^{2}$$ +x+4=0

Sol.The given quadratic equation is

$$\frac{-4\sqrt{3}±0}{8}$$ ,   

Here, a=2    b=1   c=4

So  D=

$$b^{2}-4ac$$

=

$$(1)^{2}-4(2)(4)$$

=-31<0

Therefore X=

$$\frac{-b±\sqrt{D}}{2a}$$

=

$$\frac{-1±\sqrt{31}}{4}$$

Now

$$\sqrt{-31}$$ cant be a real number as there is no real number whose square is negative.

So, there are no real roots of the given equation , i.e., the solutions of the given equation do not exist.

3.Find the roots of the following equations:

i)

$$x-\frac{1}{x}$$ =3, x≠0

Sol.The given equation is

$$x-\frac{1}{x}$$ =3

taking L.C.M

$$\frac{x^{2}-1}{x}$$ =3

Cross -multiplying

$$X^{2}-1=3x$$

$$X^{2}-3x-1=0$$

Which is quadratic equation in x.

Here,

A=1,  b=-3,  c=-1

So,

$$D=b^{2}-4ac=(-3)^{2}-4(1)(-1)$$

=9+4

=13≥0

Therefore

X=

$$\frac{-b±\sqrt{D}}{2a}$$ .    |using the quadratic formula

=

$$\frac{-(-3)±\sqrt{13}}{2(1)}=\frac{3±\sqrt{13}}{2}$$

=

$$\frac{3+\sqrt{13}}{2}, \frac{3-\sqrt{13}}{2}$$

So, the roots are

$$\frac{3+\sqrt{13}}{2}$$   and $$\frac{3 - \sqrt{13}}{2}$$ .

ii)

$$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$$ , x≠-4,7

Sol.The given equation is

$$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$$

Taking L.C.M

$$\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}$$

$$\frac{x-7-x+4}{(x+4)(x-7)}=\frac{11}{30}$$

$$\frac{11}{(x+4)(x-7)}=\frac{11}{30}$$

Dividing both sides by 11,

$$\frac{-1}{(x+4)(x-7)}=\frac{1}{30}$$

Cross-multiplying

(x+4)(x-7)=-30

(

$$x^{2}-7x+4x-28$$ )=-30

(

$$x^{2}-3x-28$$ )=-30

$$X^{2}$$ -3x+2=0

Which is a quadratic equation in x.

Here , a=1,  b=-3,  c=2

So, D=

$$b^{2}-4ac$$

=

$$(-3)^{2}-4(1)(2)$$ =9-8

=1≥0

Therefore,

X=

$$\frac{-b±\sqrt{D}}{2a}$$

|using the quadratic formula

X=

$$\frac{-(-3)±\sqrt{1}}{2x1}$$

X=

$$\frac{3±1}{2}$$

X=

$$\frac{3+1}{2}, \frac{3-1}{2}$$

X=2, 1

So, the roots are 2 and 1.

4.The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from

now is  

$$\frac{1}{3}$$ Find his present age.

Sol.Let the present age of Rehman be x years.

Then,

Rehmans age 3 years ago =(x-3)years and, Rehmans age 5 years from now

=(x+5) years

According to the question,

$$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}$$

Taking L.C.M

$$\frac{(x+5)+(x-3)}{(x-3)(x+5)}=\frac{1}{3}$$

$$\frac{2x+2}{(x-3)(x+5)}=\frac{1}{3}$$

Cross -multipying

(x-3)(x+5)=3(2x+2)

$$X^{2}+5x-3x-15=6x+6$$

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

$$x^{2}-4x-21$$ =0

which is a quadratic equation in x.

here, a=1   b=-4   c=-21

D=

$$b^{2}-4ac=(-4)^{2}-4(1)(-21)$$

=16+84=100

Using the quadratic formula, we get

X=

$$\frac{-b±\sqrt{D}}{2a}$$

X=

$$\frac{-4±\sqrt{100}}{2}$$

=

$$\frac{4±10}{2}$$

=

$$\frac{4+10}{2}, \frac{4-10}{2}$$

=7, -3

Since , x is the age , it cannot be negative, so, we reject the root   x=-3.

Therefore, x=7 gives the present age of Rehman as 7 years.

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.

Sol.Let shefali’s marks in Mathematics be x,

Then shefali’s marks in English

=(30-x)

|

$$\because$$ the sum of shefalis marks in mathematics and English is 30 according to the question,

(x+2){(30-x)}-3}=210

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

27x-

$$x^{2}$$ +54-2x=210

$$-x^{2}+25x+54-210$$ =0

$$-x^{2}+25x-156$$ =0

Dividing by -1

$$X^{2}-25x+156$$ =0

Which is  a quadratic equation in x

Here a=1   b=-25    c=156

D=

$$b^{2}-4ac$$

=

$$(-25)^{2}-4(1)(156)$$

=625-624=1

Using the quadratic formula , we get

X=

$$\frac{-b±\sqrt{D}}{2a}=\frac{25±\sqrt{1}}{2x1}=\frac{25±1}{2}$$

X=

$$\frac{25+1}{2},  \frac{25-1}{2}$$ =13, 12

30-x=30-13   or  30-12

30-x=17, 18

Therefore , either shefalis marks in mathematics are 13 and in English 17 or her marks in mathematics are 12 and in English 18.

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 the rectangular field be x meteres.

Then the longer side of the rectangular field

=

$$\sqrt{(length \:\: of \:\: the \:\: shorter \:\: side})^{2}+(length \:\: of \:\: the \:\: longer \:\: side)^{2}$$ (By Pythagoras theorem)

=

$$\sqrt{x^{2}+(x+30)^{2}}$$ metres

According to the question

=

$$\sqrt{x^{2}+(x+30)^{2}}$$ =x+60

Squaring both sides , we get

$$X^{2}+(x+30)^{2}=(x+60)^{2}$$

$$X^{2}+x^{2}$$ +60x+900

=

$$x^{2}$$ +120x+3600

$$X^{2}$$ -60x-2700=0

Which is  a quadratic equation in x

Here a=1   b=-60  c=-2700

So, D=

$$b^{2}-4ac$$

=

$$(-60)^{2}-4(1)(-2700)$$

=3600+10800=14400

Using the quadratic formula we have

X=

$$\frac{-b±\sqrt{D}}{2a}$$

=

$$\frac{-(-60)±\sqrt{14400}}{2x1}=\frac{60±120}{2}$$

=

$$\frac{60+120}{2}, \frac{60-120}{2}$$

X=90, -30

Since x cannot be negative , being a dimension , therefore the length of the shorter side of the rectangular field is 90 metres.

The length of the longer side

=x+30

=90+30

=120 metres.

Hence, the sides of the field are 120 metres and 90 metres.

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

Then , A.T.Q.

$$(small \:\: number)^{2}=8(large \:\: number)$$

=8x………..(1)

Again according to situation of the question ,

$$(larger \:\: number)^{2}-(small \:\: number)^{2180}$$

$$X^{2}-8x$$ =180      by (1)

Shifting constant term to L.H.S.,

$$X^{2}-8x-180$$ =0

Here a=1   b=-8   c=-180

$$\therefore$$   D= $$b^{2}-4ac=(-8)^{2}-4(1)(-180)$$

=64+720=784

Using the quadratic formula, we have

X=

$$\frac{-b±\sqrt{D}}{2a}=\frac{8±\sqrt{784}}{2}$$

X=

$$\frac{8±28}{2}=\frac{8+28}{2}, \frac{8-28}{2}$$

X=18, -10

X=-10 is inadmissible as then smaller number

=

$$\sqrt{8x}$$           by (1)

=

$$\sqrt{8 x (-10)}=\sqrt{-80}$$

Which does not exist because -80 is negative

$$\therefore x$$ =18

$$\therefore \sqrt{8x}=\sqrt{8 x 18}=\sqrt{144}$$ =±12

Hence, the two numbers are 18, 12 or 18, -12.

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/h.

Then the time taken by the train to travel 1360 km=

$$\frac{360}{x}$$ hours

|time =

$$\frac{Distance}{speed}$$

If the speed had been 5 km/h more i.e., speed is (x+5) km/h, then the time by the train to travel 360km  =

$$\frac{360}{x+5}$$ hours

|time =

$$\frac{Distance}{speed}$$

According to the question,

$$\frac{360}{x+5}=\frac{360}{x}$$ -1

$$\frac{360}{x}-\frac{360}{x+5}$$ =1

$$360(\frac{1}{x}-\frac{1}{x+5})$$ =1

Dividing by 360,

$$\frac{1}{x}-\frac{1}{x+5}=\frac{1}{360}$$

Taking L.C.M

$$\frac{x+5-x}{x(x+5)}=\frac{1}{360}$$

$$\frac{5}{x(x+5)}=\frac{1}{360}$$

Cross-multiplying

X(x+5)=(5)(360)

X(x+5)=1800

$$X^{2}+5x-1800$$ =0

Here a=1  b=5   c=-1800

So D=

$$b^{2}-4ac=5^{2}-4(1)(-1800)$$

=25+7200=7225

Using the quadratic formula, we have

X=

$$\frac{-b±\sqrt{D}}{2a}$$

X=

$$\frac{-5±\sqrt{7225}}{2}$$

X=

$$\frac{-5±85}{2}=\frac{-5+85}{2}, \frac{-5-85}{2}$$

X=40, -45

Since , x is the speed of the train it can not be negative . so we reject the root x=-45.

Therefore , x=40 gives the speed of the train as 40 km/h.

9.Two water taps together can fill a tank in

$$9\frac{3}{8}$$ 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.

Sol.Let the smaller tap fill the tank in x hours.

Then , the larger tap will fill the tank in (x-10)hours.

$$\therefore$$ part of the filled by the smaller tap in 1 hour = $$\frac{1}{x}$$

And, part pf the tank filled by the larger tap in 1 hour =

$$\frac{1}{x-10}$$ .

$$\therefore$$ part of tank filled by the two taps together in 1 hour = $$(\frac{1}{x}+\frac{1}{x-10})$$

$$\therefore$$   part of tank filled by two taps together in $$9\frac{3}{8}$$ hours = $$\frac{75}{8}(\frac{1}{x}+\frac{1}{x-10})$$

|

$$\because 9\frac{3}{8}=9+\frac{3}{8}=\frac{75}{8}$$

According to the question

$$\frac{75}{8}(\frac{1}{x}+\frac{1}{x-10})$$ =1

| whole tank is filled in 9\frac{3}{8}hours (given)

|

$$\because$$ the whole tank has been filled.

Dividing by \frac{75}{8},

$$\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}$$

Taking l.C.M

$$\frac{x-10+x}{x(x-10)}=\frac{8}{75}$$

$$\frac{2x-10}{x(x-10)}=\frac{8}{75}$$

Cross multiplying

$$8x(x-10)=75(2x-10)$$

$$8x^{2}-80x=150x-750$$

$$8x^{2}-230x+750=0$$

Comparing eith ax^{2}+bx+c=0, we get a=8, b=-230, c=750

So, D=

$$b^{2}-4ac$$

=

$$(-230)^{2}-4(8)(750)$$

=52900-8 x 3000

52900 – 24000=28900

Using the quadratic formula, we get

X=

$$\frac{-b±\sqrt{D}}{2a}=\frac{230±\sqrt{28900}}{2x8}$$

X=

$$\frac{230±\sqrt{28900}}{16}=\frac{230±170}{16}$$

=

$$\frac{230+170}{16}, \frac{230-170}{16}$$

=

$$\frac{400}{16}, \frac{60}{16}$$

=25, 

$$\frac{15}{4}$$

X=

$$\frac{15}{4}$$ is inadmissible as then x-10

=

$$\frac{15}{4}-10=-\frac{25}{4}$$ which is negative and the time cannot be negative 

X=25

x-10=25-10=15

hence the time in which the smaller tap and the larger tap fill the tank separately are 25 hours and 15 hours respectively.

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.

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

Then, the average speed of the express train =(x+11)km/h.

Time taken by the passenger train to travel 132km between Mysore and Bangalore

=

$$\frac{132}{x}h$$

Time taken by the express train to travel 132 km between Mysore and Bangalore

=

$$\frac{132}{x+11}h$$

According to the question,

$$\frac{132}{x+11}=\frac{132}{x}-1$$

1=

$$\frac{132}{x}-\frac{132}{x+11}$$

$$\frac{132}{x}-\frac{132}{x+11}$$ =1

$$132(\frac{1}{x}-\frac{1}{x+11})$$ =1

Dividing by 132,

$$\frac{1}{x}-\frac{1}{x+11}=\frac{1}{132}$$

Taking L.C.M.,

$$\frac{(x+11)-x}{x(x+11)}=\frac{1}{132}$$

$$\frac{11}{x(x+11)}=\frac{1}{132}$$

Cross multiplying

X(x+11)=11 x 132

X(x+11)=(11)(132)

X(x+11)=1452

$$X^{2}+11x-1452$$ =0

This is of the from

$$ax^{2}+bx+c=0$$

Comparing , we get

a=1, b=11, c=-1452

So, D=

$$b^{2}-4ac$$

=

$$(11)^{2}-4(1)(-1452)$$

=121+4(1452)

=121+5808=5929

Using the quadratic formula, we have

X=

$$\frac{-b±\sqrt{D}}{2a}$$

X=

$$\frac{-11±\sqrt{5929}}{2}=\frac{-11±77}{2}$$

X=

$$\frac{-11+77}{2}, \frac{-11-77}{2}$$

X=

$$\frac{66}{2}, \frac{-88}{2}$$

X=33, -44

Since x is the average speed of the passenger train , it cannot be negative. So , we reject the root x=-44

Therefore, x=33 gives the average speed of the passenger train as 33 km/h.

$$\therefore$$ Average speed of the express train

=x+11

=33+xx

44 km/h

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.

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

Then perimeter of the smaller square =4xm

Therefore,  perimeter of the larger square

=(4x+24)m

=4x side of larger square

Therefore, side of the larger square

=

$$\frac{4x+24}{4} m =\frac{4(x+6)}{4}$$ cm

=(x+6)m

Again

Area of the smaller square =

$$x^{2}cm^{2}$$

Area of the larger square =

$$(x+6)^{2}cm^{2}$$

According to the question ,

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

$$X^{2}+x^{2}+12x+36=468$$

$$2x^{2}+12x-432=0$$

$$X^{2}+6x-216$$ =0

|Dividing throughout by 2

Which is a quadratic equation in x.

Comparing with

$$ax^{2}$$ +bx+c=0, we get

a=1, b=6, c=-216

therefore

D=

$$b^{2}-4ac=6^{2}-4(1)(-216)$$

=36+564

900≥0

So, the given equation can be solved for x.

Using the quadratic formula, we get

X=

$$\frac{-b±\sqrt{D}}{2a}$$

=

$$\frac{-6±\sqrt{900}}{2}=\frac{-6±30}{2}$$

$$\frac{-6+30}{2},  \frac{-6-30}{2}$$

=12,  -18

Since x cannot be negative, being the length of side of the smaller square, therefore the length of the side of the smaller square is 12m.

$$\therefore$$ the length of the side of the larger square =x+6

=12+6=18m.

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