NCERT Solutions For Class 10 MATH Chapter – 4 Exercise – 4.3
Q1. Find the roots of the following quadratic equations, if they exist , by the method of completing the square.
- 2x2– 7x + 3 = 0
- 2x2+ x – 4 = 0
- 4x2+ 4√3x + 3 = 0
- 2x2+ x+ 4 = 0
Solution:-
- 2x2– 7x + 3= 0
2x2 – 7x = -3
On dividing both sides of the equation by 2 , we get
= x2 – 7x/2 = -3/2
= x2 – 2 x x x 7/4 = -3/2
On adding (7/4)2 to both sides of equation we get
= (x)2 – 2 x x x 7/4 + (7/4)2
= x = 12/4 or x = 2/4
- 2x2+ x – 4 = 0
= 2x2 + x = 4
On dividing both sides of the equation we get
= x2 + x/2 = 2
On adding (1/4)2 to both sides of the equation we get
= (x)2 + 2 x x x (1/4) + (1/4)2
= x = √33 – 1/4 or x = -√33 – 1/4
- 4x2+ 4√3x + 3 = 0
= (2x)2 + 2 x 2x x √3 + (√3)2 = 0
= x = -√3/2 or x = -√3/2
- 2x2+ x + 4 = 0
= 2x2 + x= -4
On dividing both sides of the equation we get
= x2 + 1/2x = 2
= (x +1/4)2
= -31/6
Q2. Find the roots of the quadratic equations given Q1. above by applying the quadratic formula.
Solution:-
- 2x2– 7x + 3 = 0
On comparing this equation with ax2 + bx + c = 0 we get
A= 2 , b= -7 and c = 3
By using quadratic formula we get
X = -b + √b2 – 4ac/2a
X = 7 + √49 – 24/4
X = 12/4 or x = 2/4
- 2x2+ x – 4 = 0
On comparing this equation with ax2 + bx + c = 0 we get
A = 2 , b= 1 and c = 4
X = -b + √b2 – 4ac/2a
X = -1 + √1 + 32/4
X = -1 + √33/4
X = -1 + √33/4 or x = -1-√33/4
- 4x2+ 4√3 + 3 = 0
On comparing this equation with ax2 + bx + c = 0 we get
A = 2 , b = 1 and c = -4
By using quadratic formula we get
x = -b+ √b2 – 4ac/2a
X = -√3/2 or x = -√3/2
- 2x2 + x + 4 = 0
A = 2 , b = 1 and c = 4
By using quadratic equation formula we get
X = -b+√b2 – 4ac/2a
X = -1+√1-32/4
X = -1+√-31/4
Q3. Find the roots of the following equations:
- X-1/x = 3 x ≠0
- 1/x +4 -1/x – 7= 11/30 , x= -4 , 7
Solution:-
- X -1/x = 3
X2 -3x – 1 = 0
A = 1 , b = -3 and c = -1
X = -b + √b2 – 4ac/2a
X = 3 + √9 + 4/2
X = 3 + √13/2 or x = 3 – √13/2
- 1/x+4 -1/x – 7 = 11/30
= x-7-x-4/(x +4)(x-7)= 11/30
= x = 1 or x = 2
Q4. The sum of the reciprocal of Rehman age (in years) 3 years ago and 5 years from now is 1/3 . Find his present age.
Solution:-
1/x-3 +1/x-5 = 1/3
X + 5 + x – 3/(x-3)(x+5) = 1/3
2x + 2/(x-3)(x +5) = 1/3
X = 7 , -3
Q5. In a class test , the sum of Shefali 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.
Solution:-
(x+2)(30 – x – 3) = 210
(x+2)(27-x)=210
= -x2 – 25x + 54 = 210
= x2 – 25x + 156 = 0
= x2 – 12x – 13x + 156 = 0
= (x-12)(x-13)=0
= x = 12 , 13
Q6. The diagonal of a rectangular field is 60 meters more than the shorter side . If the longer side is 30 meters more than the shorter side , find the sides of the field.
Solution:-
√x2 + (x + 30)2 = x + 60
= x2 + (x + 30)2 = (x + 60)2
= x2 + x2 + 900 + 60x = x2 + 3600 + 120x
= (x-90)(x+30) = 0
= x = 90 , -30
Q7. The difference of square of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:-
x2-8x = 180
= x2-8x-180=0
X2-18x + 10x – 180 =0
(x-18)(x+10) = 0
X = 18 , -10
Q8. A train travels 360km at a uniform speed If the speed had been 5 km/h more, it would have been taken 1 hour less for the same journey. Find the speed of the train.
Solution:-
(x+5)(360-1/x)= 360
= 360 – x + 1800 – 5/x = 360
= x2 + 5x + 10x – 1800 = 0
= x(x+45)-40(x+45)=0
= x = 40 , -45
Q9. Two water taps together can fill in 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 top can separately fill the tank.
Solution:-
1/x + 1/x-10 = 8/75
X-10+x/x(x-10) = 8/75
2x – 10/x(x-10) = 8/75
8x2 – 230x + 750 = 0
(x-25)(8x – 30)= 0
X = 25 , 30/8
Q10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore
and Bangalore If the average speeds of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.
Solution:-
= 132 x 11 = x(x+11)
= x2 + 11x – 1452 = 0
= x2 + 44x – 33x – 1452 = 0
= x(x+44) -33(x+44) = 0
= x = -44 , 33
Q11. Sum of the areas of two square is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:-
= (6 +y2) + y2 = 468
= y2 + 18y – 12y – 216 = 0
= (y+18)(y-12) = 0
= y = -18 , 12