NCERT full solutions here you can find full detailed solutions for all the questions with proper format. class 10 maths ncert solutions chapter 4 exercise 4.3.
You can see the solution for complete chapter here –
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Complete 10th Class Maths Solution
- NCERT Class 10th – Chapter 3- Exercise 3.6 Pair of Two Equations in Two Variables
- NCERT Class 10th – Chapter 4 – Exercise 4.1 Quadratic Equations
- NCERT Class 10th – Chapter 4 – Exercise 4.2 Quadratic Equations
EXERCISE :-4.3
QUE:- 1. Find the roots of the following quadratic equations, if they exist, by the method of
completing the square:
(i) 2x² – 7x + 3 = 0 (ii) 2x² + x – 4 = 0
(iii) 4x ²+ 4√3x + 3 = 0 (iv) 2x² + x + 4 = 0
SOL:-
(i) 2x² – 7x + 3 = 0
⇒ 2x² -7x =-3
on dividing both side by 2
⇒ x² -7x/2 =-3/2
⇒ x² -2×x× 7/4 =-3/2
adding both side( 7/4)²
⇒ x² – 2×x× 7/4 +( 7/4)² =( 7/4)² -3/2
⇒ (x -7/4)² = 49/16-3/2
⇒ (x -7/4)² =25/16
⇒(x -7/4)² =5²/4²
⇒(x-7/4) =±5/4
⇒ x =5/4 ±7/4
x =7/4+ 5/4 ; x=12/4 =3
x= 7/4 – 5/4 ; x=2/4 =1/2
x= 3 or 1/2
(ii) 2x² + x – 4 = 0
⇒2x² +x =4
dividing by 2 the equation
⇒ x² +x/2 =2
on adding both side (1/4)²
⇒ x² +x/2 +(1/4)² =2 +(1/4)²
⇒ x² +2 ×1/4×x +(1/4)² =2+1/16
⇒( x +1/4)² = 33/16
x +1/4 =√33/4
x =±(√33-1)/4
x= (√33-1)/4 or x=-(√33-1)/4
(iii) 4x ²+ 4√3x + 3 = 0
(2x)² +2 × 2x ×√3 +(√3)²=0
(2x +√3)²=0
(2x +√3) =0
x =-√3/2
(iv) 2x² + x + 4 = 0
2x² + x = – 4
dividing by 2 both side
x² +x/2 =-2
adding (1/4)² both side
x² +x/2 +(1/4)²=-2 +(1/4)²
x² + 2× x ×1/4+(1/4)² =1/16-2
(x+1/4)² =-31/16
the square of a number is never a negetive therefor the equation has no real root.
QUE:-2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic
formula.
(i) 2x² – 7x + 3 = 0 (ii) 2x² + x – 4 = 0
(iii) 4x ²+ 4√3x + 3 = 0 (iv) 2x² + x + 4 = 0
SOL :-
(i) 2x² – 7x + 3 = 0
on compairing this equation with ax² + bx +c =0 we obtain
a =2 ,b=-7 c=3 by using quadratic formula
x=(-b ±√b²-4ac)/2a
x= (-(-7) ±√(-7)²-4×2×3)/2×2
x= (7 ±√49-24)/4
x= (7 ± 5)/4
x = 12/4; x=2/4
x=3 or 1/2
(ii) 2x² + x – 4 = 0
on compairing this equation with ax² + bx +c =0 we obtain
a =2 ,b=1 c=-4by using quadratic formula
x=(-b ±√b²-4ac)/2a
x = (-1 ±√ 1 -4 ×2×(-4))/2×2
x = (-1 ± √ 1+32)/4
x = (-1 ± √33)/4
x =(-1 +√33)/4 or x =(-1 -√33)/4
(iii) 4x ²+ 4√3x + 3 = 0
on compairing this equation with ax² + bx +c =0 we obtain
a =4 ,b=4√3 c=3 by using quadratic formula
x=(-b ±√b²-4ac)/2a
x = (-4√3 ±√(4√3)²-4×4×3)/2×4
x= (-4√3 ± √48 -48)/8
x= (-4√3 ± 0)/8
x =-√3 /2
(iv) 2x² + x + 4 = 0
on compairing this equation with ax² + bx +c =0 we obtain
a =2 ,b=1 c=4 by using quadratic formula
x=(-b ±√b²-4ac)/2a
x =(-1 ±√1²-4×2×4)/2×2
x = (-1±√1-32)/4
x =(-1±√-31/4
square of anumber never negetive therefore for the quadratic equatio there are not any real root
QUE:-3. Find the roots of the following equations:
(i)x -1/x =3 ;x ≠0 (ii) 1/(x+4) -1/(x-7)=11/30 ;x ≠-4 ,7
SOL:-
(i)x -1/x =3 ;x ≠0
x² -1=3x
x² -3x -1=0
on compairing this equation with ax² + bx +c =0 we obtain
a =1 ,b=-3 c=-1 by using quadratic formula
x=(-b ±√b²-4ac)/2a
x = ( -(-3) ±√(-3)²-(4×1×-1))/2×1
x = (3 ±√9 -4)/2
x = (3 ±√5)/2
therefore roots for the quadratic equation (3 -√5)/2 and (3 +√5)/2
(ii) 1/(x+4) -1/(x-7)=11/30 ;x ≠-4 ,7
(x-7-x-4)/(x+4)(x-7) =11/30
-11/(x+4)(x-7) =11/30
(x+4)(x-7) =-30
x² -7x +4x -28 =-30
x² -3x -28 +30=0
x² -3x +2 =0
on compairing this equation with ax² + bx +c =0 we obtain
a =1 ,b=-3 c=2 by using quadratic formula
x=(-b ±√b²-4ac)/2a
x =(-(-3) ±√-3²-4×1×2)/2×1
x= (3 ±√ 9 -8)/2
x =(3 ±√1)/2
x =(3 ±1)/2
x =(3-1)/2=2/2=1
or x =(3+1)/2 =4/2 =2
therefore roots for the quadratic equation 1 and 2
QUE:-4. The sum of the reciprocals of Rehman’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 is x
3 year ago (x-3)
and 5 year hence (x +5)
sum of the reciprocals 1/(x-3)+1/(x +5)=1/3
(x +5 +x -3)/(x-3)(x +5)=1/3
( 2x +2) 3 =(x-3)(x +5)
6x +6 = x² +5x -3x -15
x² +5x -3x -15-6x -6 =0
x² -4x -21 =0
on compairing this equation with ax² + bx +c =0 we obtain
a =1 ,b=-4 c=-21 by using quadratic formula
x=(-b ±√b²-4ac)/2a
x=(-(-4) ±√-4²-4×1×(-21))/2×1
x =( 4 ±√16 +84)/2
x= (4 ± 10)2
x =(4+10)/2 =7
or x = (4-10)/2 =-6/2=-3 age never negetive
therefore present age of Rehman is 7 year
QUE:-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 the marks in maths x
and in english 30-x
according to the quetion
in maths x +2
and in english 30-x -3 =27-x
product of marks (x +2)(27-x )=210
27x -x² +54 -2x =210
x² -25x +156 =0
x² -12x -13x +156=0
x(x-12) -13(x-12)=0
(x-12)(x-13) =0
x = 12 ,13
if the marks in maths 12
and in english 30-12 =18
or the marks in maths 13
and in english 30-13 =17
QUE:-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 sorter side of rectangle is x
then the larger side of rectangle x +30
diagonal of rectangle =√ x² +(x +30)²
it is given that diagonal of rectangale is 60 meter more than shorter side =x +60
√ x² +(x +30)² = x +60
x² +(x +30)² = (x +60)²
x² + x² +900 +60x =x² +3600+120x
x² -60x -2700=0
x² -90x +30x -2700=0
x(x -90) +30(x -90)=0
(x-90) (x +30) =0
x =90 ,-30
shorter side of rectangel is 90 meter becuase any side never be negetive
the sorter side of rectangle is 90
then the larger side of rectangle 90 +30 =120
diagonal of rectangale is 60 meter more than shorter side =90+60 =150
QUE:-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 lerger number is x
square of smaller number is =8x
according to the question
x² -8x =180
x² -8x -180=0
x² -18x +10x -180=0
x(x -18) +10(x -18)=0
(x-18)(x+10)=0
x =18 ; -10
if the lerger number is 18
smaller number is =√8×18=±12
QUE:-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 speed of train is x
time taken by train for 360 km =360/x
according to given information
(x +5) (360/x- 1) =360
360 -x +1800/x -5=360
-x +1800/x -5 =0
-x² +1800 -5x =0
x² +5x -1800=0
x² +45x -40x -1800=0
x(x +45)-40(x+45)=0
(x +45)(x-40)=0
x = 40 ;-45
speed is never negetive so speed is 40km/h
QUE:-9. Two water taps together can fill a tank in 9 3/8hours( 75/8h) . 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 time taken by the smaller pipe is x hours
and the time taken by large pipe is (x -10)
that the time taken by both pipe is 9 3/8hours( 75/8h)
part fill by smaller pipe in 1 hour 1/x
part fill by large pipe in 1 hour 1/x -10
according to the question
1/x + 1/(x-10) =8/75
(x-10+x)/x (x-10) =8/75
(2x -10 )75 =8(x² -10x)
150x -750 =8x²-80x
8x² -230x +750=0
8x(x -25) -30(x-250)=0
(x -25)(8x-30) =0
x =25 ; 30/8 =3.75 hour
so the small pipe taken 25 hour
and large pipe taken 25-10=15 hour
QUE:-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 avrege speed of passenger train is x km/h
avrage speed of express train is (x +11) km/h
according to the question
express train taken 1 hour less then passenger train to cover 132km
132/x -132/x+11=1
132 ((x +11 -x)/x(x+11))=1
132×11 =x² +11x
x² +11x -1452=0
x² +44x -33x -1452 =0
x(x +44) -33(x -44)=0
(x+44)(x-33) =0
x =-44 ; 33
so speed of passenger train is 33 km/h because speed is never negetive
and speed of express train is 33+11=44 km/h .
QUE:-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 two square x and y
parimeter of squares 4x and 4y
accordinq to the question
4x -4y =24
4x =24+4y
x = 6+y
x ² +y² =468
(6+y )² +y² =468
36 +12y +y²+y²=468
2y² +12y -432 =0
y² +6y -216=0
y² +18y -12y -216 =0
y(y+18) -12(y+18)=0
(y+18)(y-12) =0
y =-18 ;12
x = 6+12 =18
so the side of square is 12 ,18.
You can see the solution for complete chapter here –
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