NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.3
In this page we have NCERT Solutions for Class 10th Maths: Chapter 4 - Quadratic Equations for
EXERCISE 4.3 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.
Formula and Method Used
NCERT Solutions Quadratic Equations Exercise 4.3
Question 1
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x^{2}– 7x+3 = 0
(ii)2x^{2}+x– 4 = 0
(iii)4x^{2}+ 4√3x+ 3 = 0
(iv)2x^{2}+x+ 4 = 0
Answer
(i) 2x^{2}–7x+ 3 = 0
On dividing both sides of the equation by 2, we get x^{2}– 7x/2 +3/2=0 x^{2}– 2 ×x× 7/4 +3/2=0
adding (7/4)^{2} and subtracting on LHS, we get
(x)^{2}- 2 ×x× 7/4 + (7/4)^{2}- (7/4)^{2}+ 3/2=0
(x- 7/4)^{2}= 49/16 - 3/2
(x- 7/4)^{2}= 25/16
(x- 7/4) =± 5/4 x= 7/4± 5/4 x= 7/4+ 5/4 orx= 7/4 - 5/4 x= 12/4 orx= 2/4 x= 3 or 1/2
(ii) 2x^{2}+x– 4 = 0
On dividing both sides of the equation, we get x^{2}+x/2 – 2=0
adding (1/4)^{2}and Subtracting to LHS, we get
(x)^{2}+2 ×x× 1/4 + (1/4)^{2}- (1/4)^{2} -2 =0
(x+ 1/4)^{2}= 33/16
⇒x+ 1/4 = ± √33/4
⇒x= ± √33/4 - 1/4
⇒x= ± √33-1/4
⇒x= √33-1/4 orx= -√33-1/4
(iv) 2x^{2}+x+ 4 = 0
On dividing both sides of the equation, we get x^{2}+ 1/2x+2=0
Adding and Subtracting (1/4)^{2}to LHS, we get
(x)^{2}+2 ×x× 1/4 + (1/4)^{2}- (1/4)^{2}+ 2=0
(x+ 1/4)^{2}= 1/16 - 2
(x+ 1/4)^{2}= -31/16
However, the square of number cannot be negative.
Therefore, there is no real root for the given equation.
Question 2
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Answer
(i) 2x^{2}–7x+ 3 = 0
On comparing this equation withax^{2}+bx+c= 0, we get a= 2,b= -7 andc= 3
By using quadratic formula
$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
Substituting the values, we get
$x = \frac {7\pm \sqrt {49-24}}{4}$
$x = \frac {7\pm 5}{4}$
Taking + and – separately we get x=3 or 1/2
(ii) 2x^{2}+x- 4 = 0
On comparing this equation withax^{2}+bx+c= 0, we get
a = 2, b = 1 andc= -4
By using quadratic formula
$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
Substituting the values, we get x= -1+√33/4 orx= -1-√33/4
(iii) 4x^{2}+4√3x+ 3 = 0
On comparing this equation withax^{2}+bx+c= 0, we get a=4,b=4√3 and c = 3
By using quadratic formula
$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
Substituting the values, we get x= √3/2 orx= -√3/2
(iv)2x^{2}+x+ 4 = 0
On comparing this equation withax^{2}+bx+c= 0, we get a= 2,b= 1 andc= 4
By using quadratic formula
$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$ x= -1±√1-32/4 x= -1±√-31/4
The square of a number can never be negative.
∴There is no real solution of this equation.
Question 2
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
Answer
(i)x-1/x= 3
⇒x^{2}- 3x-1 = 0
On comparing this equation withax^{2}+bx+c= 0, we get a= 1,b= -3 andc= -1
By using quadratic formula, we get
$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$ x=3±√9+4/2 x= 3±√13/2 x= 3+√13/2 orx= 3-√13/2
(ii) 1/x+4 - 1/x-7 = 11/30 x-7-x-4/(x+4)(x-7) = 11/30
-11/(x+4)(x-7) = 11/30
(x+4)(x-7) = -30 x^{2}- 3x- 28 = 30 x^{2}- 3x+ 2 = 0
By using quadratic formula, we get
$x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
Substituting the values x= 1 or 2
Question 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.
Answer
Let the present age of Rehman bexyears.
Three years ago, his age was (x- 3) years.
Five years hence, his age will be (x+ 5) years.
As per question, the sum of the reciprocals of Rahman’s ages 3 years ago and 5 years from now is 1/3.
∴ 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) x^{2}-4x- 21 = 0
Solving as per Factoring method x^{2}- 7x+ 3x- 21 = 0 x(x- 7)+ 3(x- 7) = 0
(x- 7)(x+ 3) = 0 x= 7, -3
However, age cannot be negative.
Therefore, Rehman's present age is 7 years.
Question 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.
Answer
Let the marks in Mathematics bex.
Then, the marks in English will be 30 -x.
According to the question,
(x+ 2) (30 -x- 3) = 210
(x+ 2) (27 -x) = 210
-x^{2}+ 25x+ 54 = 210 x^{2}- 25x+ 156 = 0 x^{2}-12x- 13x+ 156 = 0 x (x- 12) -13(x- 12) = 0
(x- 12) (x- 13) = 0 x= 12, 13
If the marks in Mathematics are 12, then marks in English will be 30 - 12 = 18
If the marks in Mathematics are 13, then marks in English will be 30 - 13 = 17
Question 6
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. Answer
Let the shorter side of the rectangle bexm.
Then, larger side of the rectangle = (x+ 30) m
Now Diagonal of the rectangle is given by
$D= \sqrt {l^2 + b^2}$
$D= \sqrt {x^2 + (x+30)^2}$
Now D=(x+30)
So
$x + 60 = \sqrt {x^2 + (x+30)^2}$
Squaring both the sides x^{2}+ (x+ 30)^{2}= (x+ 60)^{2} x^{2}- 60x- 2700 = 0\ x^{2}- 90x+ 30x- 2700 = 0 x (x- 90)+ 30(x-90)
(x- 90) (x+ 30) = 0 x= 90, -30
However, side cannot be negative. Therefore, the length of the shorter side will be90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m.
Question 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. Answer
Let the larger and smaller number bexandyrespectively.
According to the question, x^{2}-y^{2}= 180 ---(A) y^{2}= 8x ---(B)
From equation A and B x^{2}- 8x= 180 x^{2}-8x- 180 = 0 x= 18, -10
However, the larger number cannot be negative as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.
Therefore, the larger number will be 18 only. x= 18
∴y^{2}= 8x = 8 × 18 = 144
⇒y= ±√44= ±12
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and - 12.
Question 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.
Answer
Let the speed of the train bexkm/hr.
Time taken to cover 360 km = 360/xhr.
According to the question,
(x+ 5) [360-(1/x)] = 360
360 -x+ 1800-(5/x)= 360 x^{2}+5x+ 10x- 1800 = 0 x= 40, -45
However, speed cannot be negative.
Therefore, the speed of train is 40 km/h.
Question 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.
Answer
Let the time taken by the smaller pipe to fill the tank be x hr.
Time taken by the larger pipe = (x- 10) hr.
Part of tank filled by smaller pipe in 1 hour = 1/x
Part of tank filled by larger pipe in 1 hour = 1/x- 10
It is given that the tank can be filled in 75/8 hours by both the pipes together. Therefore,
1/x+ 1/(x-10) = 8/75 (x-10+x)/x(x-10) = 8/75
2x-10/x(x-10) = 8/75
75(2x- 10) = 8x^{2}- 80x
8x^{2}- 230x+750 = 0 x= 25, 30/8
Time taken by the smaller pipe cannot be 30/8= 3.75 hours. As in this case, the time taken by the larger pipe will be negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 - 10 =15 hours respectively.
Question 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 the two trains.
Answer
Let the average speed of passenger train bexkm/h.
Average speed of express train = (x+ 11) km/h
It is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.
$ \frac {132}{x} - \frac {132}{x+ 11} =1$
$132 [\frac {x+11 -x}{x(x+11)}] =1$
132 × 11 =x (x+ 11) x^{2}+ 11x- 1452 = 0 x= - 44, 33
Speed cannot be negative.
Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.
Question 11
Sum of the areas of two squares is 468 m^{2}. If the difference of their perimeters is 24 m, find the sides of the two squares.
Answer
Let the sides of the two squares bexm andym. Therefore, their perimeter will be 4xand 4y respectively and their areas will bex^{2}andy^{2}respectively.
As per question
4x- 4y= 24 x-y= 6 x=y+ 6
Also,x^{2}+y^{2}= 468
So (6+y^{2})+y^{2}= 468
2y^{2}+ 12y+ 432 = 0 y^{2}+ 6y - 216 = 0
Solving by factor method y= -18, 12
However, side of a square cannot be negative.
Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.
Summary
NCERT Solutions for Class 10th Maths: Chapter 4 - Quadratic Equations Ex 4.3 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail.You can download this as pdf Download NCERT solution Quadratic Equation Exercise 4.3 as pdf
This chapter 4 has total 4 Exercise 4.1 ,4.2,4.3 and 4.4. This is the Third exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below