Class 10th-Quadratic Equation- Exercise 4.3.Full notes

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

(i) 2x²-7x+3=0

(ii) 2x²+x-4=0

(iii)4x²+4√3x+3=0

(iv) 2x²+x+4=0

 Ans. (i) 2x²-7x+3=0

First we divide equation by 2 to make co-efficient of x² equal to 1,

x²-(7/2)x+3/2=0

Divide middle term of the equation by 2x, we get

(7/2)x*(1/2x)=7/4 

Add and subtract square of 7/4  from the equation 

x²-(7/2)x+3/2=0

x²-(7/2)x+3/2+(7/4)²-(7/4)²=0

x²+(7/4)²-7/2 x+3/2+(-7/4)²=0

{(a-b)²=a²+b²-2ab}

(x-7/4)²-(24-49)/16=0

(x-7/4)²=(49-24)/16

Taking Square root on both sides,

x-7/4= 士5/4

X=5/4+7/4=12/4=3 

or 

x=-5/4+7/4=2/4=1/2

Therefore, x=1/2 or 3  

(ii) 2x²+x-4=0

First we divide equation by 2 to make co-efficient of x² equal to 1,

x²+x/2-2=0

Divide middle term of the equation by 2x, we get

x/2*1/2x=1/4

Add and subtract square of 1/4  from the equation 

x²+x/2-2=0

x²-x/2-2+(1/4)²-(1/4)²=0

x²+x/2+(1/4)²-2-(1/16)=0

{(a+b)²=a²+b²+2ab}

(x+1/4)²-(33/16)=0

(x+1/4)²=33/16

Taking square root on the both sides,

⇒ x+1/4=士√33/4
x=√33/4-1/4=(√33-1)/4

and x= (-√33/4)-1/4 =(-√33-1)/4

Therefore, x=(√33-1)/4 , or (-√33-1)/4

(iii)4x²+4√3x+3=0

Dividing the equation by “4”

x²+√3x +3/4=0

Now According to the procedure of completing the square,

x²+√3x +3/4 +(√3/2)²-(√3/2)² =0

x²+(√3/2)²+√3x+3/4-3/4 =0

{(a+b)²=a²+b²+2ab}

(x+√3/2)²=0

(x+√3/2)(x+√3/2)=0

Taking square root on the both sides,

(x+√3/2)=0 , (x+√3/2)=0

 x=-√3/2 , x=-√3/2

(iv) 2x²+x+4=0

Dividing the equation by “2”

x²+x/2+2=0

Now According to the procedure of completing the square,

x²+x/2+2 +(1/4)²-(1/4)² =0

x²+(1/4)²+x/2+2-(1/4)² =0

{(a+b)²=a²+b²+2ab}

(x+1/4)²+2-(1/16)=0

(x+1/4)²=(1/16)-2 =(1-32)/16

Taking square root on both sides. Right hand side does not exist because square root of negative number does not exist.

Therefore, there is no solution for Quadratic Equation 2x²+x+4=0.

2. Find the roots of the following Quadratic Equations by applying quadratic formula.

(i)  2x²-7x+3=0

(ii) 2x²+x-4=0

(iii)4x²+4√3x+3=0

(iv) 2x²+x+4=0

Ans. (i)  2x²-7x+3=0

Comparing quadratic equation 2x²-7x+3=0 with general form ax²+bx+c=0 , we get a=2,b=-7 and c=3

Putting these values in quadratic formula  

⇒x= [-b 士√b²-4ac]/ 2a

⇒x= [7 士√(-7)²-4*2*3)]/ 2*2 

⇒x= [7 士√(49-24)]/ 4 

⇒x= (7 士√25)/4 

⇒x= (7 士5)/4

⇒x= (7+5)/4 or (7-5)/4

⇒ x = 3 or ½

(ii) 2x²+x-4=0

Comparing quadratic equation 2x²+x-4=0 with general form ax²+bx+c=0 , we get a=2,b=1 and c=-4

Putting these values in quadratic formula  

⇒x= [-b 士√b²-4ac]/ 2a

⇒x= [-1 士√(1)²-4*2*(-4))]/ 2*2 

⇒x= (-1 士√33)/4 

⇒x= (-1 +√33)/4 or x= (-1-√33)/4 

(iii) 4x²+4√3x+3=0

Comparing quadratic equation 4x²+4√3x+3=0  with general form ax²+bx+c=0, we get a=4,b=4√3  and c=3

Putting these values in quadratic formula  

⇒x= [-b 士√b²-4ac]/ 2a

⇒x= [4√3  士√(4√3)²-4*4*3)]/ 2*4 

⇒x= [4√3 士√(0)]/ 8 

⇒x= -4√3)/8 

⇒x= -√3/2

⇒x= -√3/2   , -√3/2

A quadratic equation has two roots. Here, both the roots are equal.

(iv) 2x²+x+4=0

Comparing quadratic equation 2x²+x+4=0 with general form ax²+bx+c=0 , we get a=2,b=1 and c=4

Putting these values in quadratic formula  

⇒x= [-b 士√b²-4*a*c]/ 2*a

⇒x= [-1 士√(1)²-4*2*4)]/ 2*2 

⇒x= [-1 士√(-31)]/ 4 

But, the square root of negative number is not defined.

Therefore,

Quadratic equation 2x²+x+4=0 has no solution.

 3. Find the roots of the following equations:-

(i) (x-1)/x=3, x is not equal to 0

(ii) 1/(x+4)- 1/(x-7) =11/30, x is not equal to -4 and 7

Ans. (i) x-1/x=3, x is not equal to 0

⇒ (x² -1)/x=3

⇒ (x² -1)=3x

⇒ x^2-3x-1=0

Comparing quadratic equation x²-3x-1=0 with general form ax²+bx+c=0 , we get a=1,b=-3 and c=-1.

Putting these values in quadratic formula  

 ⇒x= [-b 士√b²-4ac]/ 2a

⇒x= [3士√(3)²-4*1*(-1))]/ 2*1

⇒x= [3 士√(13)]/2

⇒x= (3 士√13)/2

⇒x= (3 +√13)/2 or (3 -√13)/2

Hence, proved.

(ii) 1/(x+4)- 1/(x-7) =11/30, x is not equal to -4 and 7 

 1/(x+4)- 1/(x-7) =11/30, x is not equal to -4 and 7

⇒ (x-7)-(x+4)/(x-4)(x-7) =11/30

[LCM]

⇒-11/(x-4)(x-7) =11/30

⇒ -30=x² -7x+4x-28

⇒ x² -3x+2=0

Comparing quadratic equation x²-3x+2=0 with general form ax²+bx+c=0 , we get a=1,b=-3 and c=2.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x= [3士√(3)²-4*1*(2))]/ 2*1

⇒x= [3 士√(1)]/2

⇒x= (3 +√1)/2 or (3 -√1)/2

⇒ x=2,1

Hence, proved.

4. The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Ans. Let present age of Rehman = x yrs.

3 years Ago, Age of Rehman= (x-3) years

After 3 years , Age of Rehman= (x+5) years

Acc. To the given condition:-

1/(x-3)+1/(x+5) =1/3

⇒ [(x+5)+(x-3)]/[(x-3)(x+5)]=1/3

⇒ 3(2x+2)=(x-3)(x+5)

⇒ 6x+6= x²-3x+5x-15

⇒ x²-4x-15-6=0

⇒ x²-4x-21=0

Comparing quadratic equation x²-4x-21=0 with general form ax²+bx+c=0 , we get a=1,b=-4 and c=-21.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x= [4士√(4)²-4*1*(-21))]/ 2*1

⇒x= [4士√(16+84)]/2

⇒x= [4士√(100)]/2

⇒x= (4 +10)/2 , (4 -10)/2

⇒ x=7,-3

Age cannot be negative, so we discard x=-3

Therefore, present age of Rehman is 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.

Ans:Let Shefali’s marks in Mathematics = x

Let Shefali’s marks in English = 30 − x

If, she had got 2 marks more in Mathematics, than her marks would be = x + 2

If, she had got 3 marks less in English, then her marks in English would be = 30 – x − 3 = 27 − x

Acc. to given condition:-

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

⇒ 27x-x²+54-2x=210

⇒ x²-25x+156=0

Comparing quadratic equation x²-25x+156=0 with general form ax²+bx+c=0 , we get a=1,b=-25 and c=156.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x= [25士√(25)²-4*1*156]/ 2*1

⇒x= [25士√(625-624)]/2

⇒x= [25士√(1)]/2

⇒x= (25+1)/2 , (25 -1)/2

⇒ x= 13 or 12.

Therefore, Shefali’s marks in Mathematics = 13 or 12

Shefali’s marks in English = 30 – x = 30 – 13 = 17

Or Shefali’s marks in English = 30 – x = 30 – 12 = 18

Therefore, her marks in Mathematics and English are (13, 17) or (12, 1)

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.

Ans. Let shorter side of rectangle =x metres

Let diagonal of rectangle = (x + 60) metres

Let longer side of rectangle = (x + 30) metres

Acc. to Pythagoras theorem,

(x+60)²=(x+30)²+x²

⇒ x²+3600+120x=x²+900+60x+x²

⇒ x²-60x-2700=0

Comparing quadratic equation x²-60x-2700=0 with general form ax²+bx+c=0 , we get a=1,b=-60 and c=-2700.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x= [60士√(-60)²-4*1*(-2700)]/ 2*1

⇒x= [60士√(3600+10800)]/2

⇒x= [60士120]/2

⇒x= (60+120)/2 , (60-120)/2

⇒ x= 90 ,-30

We ignore -30. Since length cannot be in negative.

 Therefore, x=90 which means length of shorter side = 90 meters

And length of longer side= x+30 =90+30=120 metres

Therefore, length of sides are 90 and 120 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.

Ans. Let smaller number =x and

let larger number =y  

Acc. to condition:-

y²-x²=180…….(1)

Also, we are given that square of smaller number is 8 times the larger number.

x² = 8y….(2)

Putting equation (2) in (1), we get

y²-(8y)=180

⇒  y²-8y-180=0

Comparing quadratic equation y²-8y-180=0 with general form ax²+bx+c=0 , we get a=1,b=-8 and c=-180.

Putting these values in quadratic formula 

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒y= [8士√(-8)²-4*1*(-180)]/ 2*1

⇒y= [8士√(64+720)]/2

⇒y= [8士√(784)]/2

⇒y= (8+28)/2 , (8-28)/2

⇒ y=18 or 10

Using equation (2) to find smaller number.

⇒ x²=8y =8*18=144

⇒ x = ±12

And, 

x²=8y =8*(-10)=144  {No real solution for x}

Therefore, two number are (12,18) (-12,18)

8. A train travels 360 km at a uniform speed. If, the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Ans. Let the speed of the train =x km/hr

If, speed had been 5 km/hr more, train would have taken 1 hr less.

Acc. to this condition

360/x=360/(x+5)=1

⇒ 360[(1/x)-1/(x+5)] =1

⇒ 360 [(x+5)-x]/x(x+5)=1

⇒ 360*5 = ⇒ x² +5x

⇒ x² +5x-1800=0

Comparing quadratic equation x²+5x-1800=0. x with general form ax²+bx+c=0 , we get a=1,b=5 and c=-1800.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x = [-5士√(5)²-4*1*(-1800)]/ 2*1

⇒x = [-5士√(25+7200)]/2

⇒x = [-5士√(7225)]/2

⇒x = [-5士85]/2

⇒x= (-5+85)/2 , (-5-85)/2

⇒ x=40 , -45

Since speed of train cannot be in negative. Therefore, we discard x= -45

Therefore, speed of train =40 km/hr

9. Two water taps together can fill a tank in 75/8 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.

Ans. Let time taken by tap of smaller diameter to fill the tank = x hours

Let time taken by tap of larger diameter to fill the tank = (x-10) hours

It means that tap of smaller diameter fills (1/x)^th part of tank in 1 hour…..(1)

And, tap of larger diameter fills (1/x-10)^th part of tank in 1 hour         …….(2)

When two taps are used together, they fill tank in 75/8 hours.

In 1 hour, they fill (8/75)^th  part of tank 1/(75/8)= 8/75                       ……..(3)

From (1), (2) and (3),

(1/x+1)/(x-10) =8/75

⇒ (x-10+x)/ x(x-10) = 8/75

⇒ 75 (2x − 10) = 8(x² -10x)

⇒ 150x – 750 =8x² -80x

⇒ 8x² -80x-150x+750=0

⇒ 4x² -115x+375=0 

Comparing quadratic equation 4x²-115x+375=0 with general form ax²+bx+c=0 ,we get a=4,b=-115 and c=-375.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x = [115士√(-115)²-4*3*(375)]/ 2*4

⇒x = [115士√(13225-6000)]/8

⇒x = [115士√(7225)]/8

⇒x = [115士85]/8

⇒x= (115+85)/8 , (115-85)/8

⇒ x=25, 3.75

Time taken by larger tap = x-10 =3.75 -10= -6.25 hours

Time cannot be in negative. Therefore , we ignore this value.

Time taken by larger tap= x-10=25-10= 15 hours

Therefore, time taken by larger tap is 15 hour and time taken by smaller tap is 25 hours.

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 two trains.

Ans. Let average speed of passenger train = x km/hr

Let average speed of express train = (x+11) km/hr

Time taken by passenger train to cover 132 km= (132/x ) hours

Time taken by express train to cover 132 km =[132/(x+11)] hours

Acc. to the given condition,

132/x=[132/(x+11)]+1

⇒ 132 [1/x - 1/(x+11)] =1

⇒ 132[ (x+11-x)/ x(x+11)] =1

⇒ 132 *11=x(x+11)

⇒1452 = x²+11x

⇒ x²+11x-1452=0

Comparing quadratic equation x²+11x-1452=0 with general form ax²+bx+c=0 ,we get a=1,b=11 and c=-1452.

Putting these values in quadratic formula  

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x = [-11士√(11)²-4*1*(-1452)]/ 2*1

⇒x = [-11士√(121+5808)]/2

⇒x = [-11士√(5929)]/2

⇒x = [-11士77]/2

⇒x= (-11+77)/2 , (-5-77)/2

⇒x= 33,-44

As speed cannot be in negative. Therefore, speed of passenger train =33 km/hr

And, speed of express train = x+11 =33 +11 = 44 km/hr

11. Sum of areas of two squares is 468 m². If, the difference of their perimeters is 24 metres, find the sides of the two squares.

Ans. Let perimeter of first square = x metres

Let perimeter of second square = (x+24) metres

Length of side of first square= x/4 metres 

                                             { Perimeter of square = 4* length of side}

Length of side of first square = (x+24)/4 metres

Area of first square= side * side = x/4* x/4 = x²/16 m²

Area of second square = [(x+24)/4]²

Acc. to given condition:-

x²/16 + [(x+24)/4]²= 0

⇒ x²/16 + (x²+576+48x)/16 = 468

⇒  (x²+ x²+576+48x)/16 = 468

⇒2x²+576+48x=468*16

⇒ 2x²+48x+576=7488

⇒ 2x²+48x-6912=0

⇒ x²+24x-3456=0

Comparing quadratic equation x²+24x-3456=0 with general form ax²+bx+c=0 ,we get a=1,b=24 and c=-3456.

Putting these values in quadratic formula 

⇒x= [-b 士√(b²-4ac)]/ 2a

⇒x = [-24士√(24)²-4*1*(-3456)]/ 2*1

⇒x = [-24士√(576+13824)]/2

⇒x = [-24士√(14400)]/2

⇒x = [-24士120]/2

⇒x= (-24+120)/2 , (-24-120)/2

⇒ x=48 or -72

Perimeter of square cannot be in negative. Therefore, we discard x=-72.

Therefore, perimeter of first square= 48 metres

And, perimeter of second square = x+24 = 48 +24 = 72 metres

⇒ Side of first square = perimeter / 4 =48/4 =12 m

And , side of second square