EXERCISE 4.1
1. Check whether the following are quadratic equation?
(i) (x+1)2 = 2(x-3)
(ii) x2-2x = (-2)(3-x)
(iii) (x − 2) (x + 1) = (x − 1) (x + 3)
(iv) (x − 3) (2x + 1) = x (x + 5)
(v) (2x − 1) (x − 3) = (x + 5) (x − 1)
(vi) x2+3x+1 = (x-2)2
(vii) (x+2)3 = 2x(x2-1)
(viii) x3-4x2-x+1 = (x-2)3
Answer:
An equation of the form ax2+bx+c = 0, a ≠ 0 and a, b, c are real numbers is called quadratic equation.
(i) (x+1)2 = 2(x-3)
x2+2x+1 = 2x-6
x2+2x+1 -2x+6 = 0
x2+7 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 1≠ 0, b = 0, c = 7
Hence the given equation is quadratic equation.
(ii) x2-2x = (-2)(3-x)
x2-2x = -6+2x
x2-2x -2x+6 = 0
x2-4x+6 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 1≠ 0, b = -4, c = 6
Hence the given equation is quadratic equation.
(iii) (x-2)(x+1) = (x-1)(x+3)
x2-2x+x-2 = x2-x+3x-3
x2-2x+x-2 -x2+x-3x+3= 0
-3x+1 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 0, b = -3, c = 1
Hence the given equation is not quadratic equation.
(iv) (x-3)(2x+1) = x(x+5)
x2-6x+x-3 = x2+5x
x2-6x+x-3 -x2-5x = 0
-10x-3 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 0, b = -10, c = -3
Hence the given equation is not quadratic equation.
(v) (2x-1)(x-3) = (x+5)(x-1)
2x2-6x-x+3 = x2+5x-x-5
2x2-6x-x+3 -x2-5x+x+5 = 0
x2-11x+8 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 1 ≠ 0, b = -11, c = 8
Hence the given equation is quadratic equation.
(vi) x2+3x+1 = (x-2)2
x2+3x+1 = x2-4x+4
x2+3x+1 -x2+4x-4= 0
7x-3 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 0, b = 7, c = -3
Hence the given equation is not quadratic equation.
(vii) (x+2)3 = 2x(x2-1)
x3+6x2+12x+8 = 2x3-2x
x3+6x2+12x+8 -2x3+2x = 0
x3 – 6x2-14x-8 = 0
It is not in the form of ax2+bx+c = 0s
Hence the given equation is not a quadratic equation.
(viii) x3-4x2-x+1 = (x-2)3
x3-4x2-x+1 = x3-6x2+12x-8
x3-4x2-x+1 -x3+6x2-12x+8= 0
2x2-13x+9 = 0
Comparing above equation with ax2+bx+c = 0, we have a = 2 ≠ 0, b = -13, c = 9
Hence the given equation is quadratic equation.
Represent the following situations in the form of quadratic equations:
2. (i) The area of rectangular plot is 528 m2. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot .
Answer:
Let the breadth of the plot is x meters
Then, the length is (2x + 1) meters.
Area = length × breadth
528 = x × (2x + 1)
528 = 2x2 + x
2x2 + x – 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let x and x+1 be two consecutive integers.
Then, x(x + 1) = 306
x2 + x – 306 = 0
(iii) Rohan’s mother is 26 years older than him. The products of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Answer:
Let x be the Rohan’s present age. Then, his mother’s present age is x+26
After 3 years theirs ages will be x+3 and x+26+3 = x+29 respectively.
Hence,
(x+3)(x+29) = 360
x2+3x+29x+87 = 360
x2+32x- 273 = 0
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer:
Let speed of train be x km/h
Time taken by train to cover 480 km = 480xhours
If, speed had been 8km/h less then time taken would be (480x−8) hours
According to given condition, if speed had been 8km/h less then time taken is 3 hours less.
Therefore, 480x – 8 = 480x + 3
⇒480 (1x – 8 − 1x) = 3
⇒480 (x – x + 8) (x) (x − 8) = 3
⇒480 × 8 = 3 (x) (x − 8)
This is a Quadratic Equation.
EXERCISE 4.2
1. Find the roots of the following Quadratic Equations by factorization.
(i) x2-3x-10 = 0
Answer:
x2-3x-10 = 0
x2-5x+2x-10 = 0
x(x-5)+2(x-5) = 0
(x-5)(x+2) = 0
⇒ x-5 = 0 or, x+2 = 0
x = 5 or x =-2
Hence the required roots are 5, -2.
(ii) 2x2+x-6=0
Answer:
2x2+x-6 = 0
2x2+4x-3x-6 = 0
2x(x+2)-3(x+2) = 0
(2x-3)(x+2) = 0
⇒ 2x-3 = 0 or, x+2 = 0
x = 3/2 or x = -2
Hence the required roots are 3/2 , -2.
(iii) √2×2+7x+5√2 = 0
Answer:
√2x2+7x+5√2 = 0
√2x2+2x+5x+ 5√2 = 0
√2x(x+√2)+5(x+√2) = 0
(x+√2)(√2x +5) = 0
⇒ x+√2 = 0 or, √2x +5 = 0
x = -√2 or x = –5/√2
Hence the required roots are -√2, –5/√2.
(iv) 2x2-x+1/8=0
Answer:
2x2-x+1/8=0
16x2-8x+1 = 0
(4x)2-2.4x.1+12 = 0
(4x-1)2= 0
⇒ 4x-1 = 0 or 4x-1 = 0
x = 1/4 or x = 1/4
Hence the required roots are 1/4 , 1/4
(v) 100x2-20x+1 = 0
Answer:
100x2-20x+1 = 0
(10x)2-2×10x.1+12= 0
(10x-1)2 = 0
10x-1 = 0 or 10x-1 = 0
x = 1/10 or x = 1/10
Hence the required roots are 1/10 or 1/10
Solve the following problems given :
2. (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each others, and product of the marbles they now have is 124. We would like to find how many marbles they had to start with?
Answer:
Let the number of marbles John had be x.
Then the number of marbles Jivanti had be 45-x
They lost 5 marbles to each others.
Hence now the number of marbles of John is x-5 and that of Jivanti is 45-x-5 = 40-x.
Then by given condition we have
(x-5)(40-x) = 124
40x-200-x2+5x = 124
x2-45x+324 = 0
x2-36x-9x+324 = 0
x(x-36)-9(x-36) = 0
(x-36)(x-9) = 0
x-36 = 0
x = 36
or x-9 = 0
x = 9
Therefore, they had 36 and 9 marvels respectively.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.
Answer:
Let the number of toys produced on that day is x.
Therefore, the cost of production of each toy on that day is (55-x)
So, the total cost of production on that day is = x(55-x)
Then using the given condition, we have,
x(55-x) = 750
55x-x2 =750
x2-55x+750 = 0
x2-25x-30x+750 = 0
x(x-25)-30(x-25) = 0
(x-25)(x-30) = 0
x-25 = 0
x =25
or x-30 =0
x = 30
Therefore, the number of toys produced on that day is 25 or 30.
3. Find the two numbers whose sum is 27 and product is 182.
Answer:
Let one number is x. Then another number is 27-x.
Then by the given condition,
x(27-x) = 182
27x- x2-182 = 0
x2-27x+182 = 0
x2-14x-13x+182 = 0
x(x-14)-13(x-14) = 0
(x-14)(x-13) = 0
x-14 = 0
x =14
or, x-13 = 0
x =13
Hence the required two numbers are 13 and 14.
4. Find the two consecutive positive integers, sum of whose squares is 365.
Answer:
Let the two consecutive positive integers be x and x+1
Then by the given condition,
x2+(x+1)2 = 365
x2+ x2+2x+1 = 365
2x2+2x-364 = 0
x2+x-182 = 0
x2+14x-13x-182 = 0
x(x+14)-13(x+14) = 0
(x+14)(x-13) = 0
x = -14 or 13
Since the numbers are positive so x = -14 not possible.
So, the numbers are 13 and 13+1 = 14.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer:
Let base of the right triangle is x cm. Then the altitude is x-7 cm.
Then by the property of right triangle and using the given property we have
x2+(x-7)2 = (13)2
x2+ x2-14x+49 = 169
2x2-14x-120 = 0
x2-7x-60 = 0
x2-12x+5x-60 = 0
x(x-12)+5(x-12) = 0
(x-12)(x+5) = 0
x-12 = 0
x =12
or, x+5 = 0
x = -5
Since length can’t be negative, so x = -5 is not possible.
Hence the base is 12 cm and the altitude is (12-7)cm = 5 cm
6. A cottage industry produces a certain number of pottery articles in a day. It was observed that on a particular day that the cost of production of each articles (in rupees) was 3 more than twice the number of articles produced on that day. If the total number of production on that day was ₹90, find the number of articles produced and the cost of each article.
Answer:
Let the number of articles produced on that day is x. then the price on that day is 2x+3
Then using the given condition we have,
x(2x+3) = 90
2x2+3x-90 = 0
2x2+15x-12x-90 = 0
x(2x+15)-6(2x+15)=0
(2x+15)(x-6) = 0
2x+15 = 0
x = – 15/2
or, x-6 = 0
x = 6
Since Articles can’t be negative then x = – 15/2 is not possible.
Therefore, the number of articles produced on that day = 6
Cost of each article = 2×6+3 = 15
EXERCISE 4.3
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
Answer:
(i)2x2 – 7x + 3 = 0
First we divide equation by 2,
(ii) 2x2+ x – 4 = 0
Dividing equation by 2,
(iii) 4x2+4√3x+3 = 0
Dividing equation by 4,
(iv) 2x2+ x + 4 = 0
Dividing equation by 2,
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
Answer:
(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
(ii) 2x2+ x – 4 = 0
Comparing quadratic equation 2x2+ x – 4 = 0 with the general form ax2 + bx + c = 0 , we get a = 2, b = 1 and c = −4
Putting these values in quadratic formula ,
(iii) 4x2+4√3x+3 = 0
Comparing quadratic equation 4x2+4√3x+3 = 0 with the general form ax2 + bx + c = 0, we get b = 
and c = 3
Putting these values in quadratic formula ,
(iv) 2x2+ x + 4 = 0
Comparing quadratic equation 2x2+ x + 4 = 0 with the general form ax2 + bx + c = 0, we get a = 2, b = 1 and c = 4
Putting these values in quadratic formula
3. Find the roots of the following equations:
(i) .png)
(ii)
Answer:
(i)
Comparing equation with general form ax2 + bx + c = 0,
We get a = 1, b = −3 and c = −1
Using quadratic formula
(ii).png)
⇒ x(x -2) -1 (x – 2) = 0
⇒ (x -1)(x – 2) = 0
⇒ x = 2, 1
4. The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is 13. Find his present age.
Answer:
Let present age of Rehman = x years
Age of Rehman 3 years ago = (x − 3) years.
Age of Rehman after 5 years = (x + 5) years
According to the given condition:
It is given that the sum of the reciprocals of Rehman’s ages 3 years ago and 5 years from now is 1/3.
⇒ 3 (2x + 2) = (x − 3) (x + 5)
⇒ 6x + 6 = x2 + 2x – 15
⇒x2 – 4x -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.
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 Shefali’s marks in Mathematics = x
Let Shefali’s marks in English = 30 − x
If, she had got 2 marks more in Mathematics, her marks would be = x + 2
If, she had got 3 marks less in English, her marks in English would be = 30 – x − 3 = 27 − x
According to given condition:
(x + 2) (27 − x) = 210
⇒ – x2 + 25x + 156 = 0
⇒ x(x – 12) -13(x – 12) = 0
⇒ (x – 12)(x – 13) = 0
⇒ x = 12, 13
If the marks in Maths are 12, then marks in English will be 30 – 12 = 18
If the marks in Maths are 13, then marks in English will be 30 – 13 = 17
Therefore, her marks in Mathematics and English are (13, 17) or (12, 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.
Answer:
Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
Diagonal of rectangle =
It is given that the diagonal of the rectangle = (x+30)m
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 – 60x – 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 be 90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m.
Therefore, length of sides are 90 and 120 in 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.
Answer:
Let the larger and smaller number be x and y respectively.
According to the question,
x2 – y2 = 180 and y2 = 8x
⇒ 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
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
∴ y2 = 8x = 8 × 18 = 144
⇒ y = 
∴ Smaller number = ±12
Therefore, the numbers are 18 and 12 or 18 and – 12.
Therefore, two numbers are (12, 18) or (−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.
Answer:
Let the original speed of train be x km/hr. Then,
Increased speed of the train = (x + 5)km/hr
Time taken by the train under usual speed to cover 360 km =
Time taken by the train under increased speed to cover 360 km = 
⇒ 1800 = x2 + 5x
⇒ x2 + 5x – 1800 = 0
⇒ x2 – 40x + 45x -1800 = 0
⇒ x(x – 40) + 45(x – 40) = 0
⇒(x – 40)(x + 45) = 0
So, either x – 40 = 0
x = 40
Or
x + 45 = 0
x = -45
But, the speed of the train can never be negative.
Hence, the original speed of train is x = 40 km/hr
9. Two water taps together can fill a tank in 
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 = 
It is given that the tank can be filled in hours by both the pipes together. Therefore,
⇒75(2x – 10) = 8x2 – 80x
⇒ 150x -750 = 8x2 – 80x
⇒ 8x2 – 230x +750 = 0
⇒ 8x2 – 200x – 30x +750 = 0
⇒ 8x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(8x -30) = 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.
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.
Answer:
Let the speed of the passenger train be x km/hr. Then,
Speed of the express train = (x + 11)km/hr
Time taken by the passenger train to cover 132 km between Mysore to Bangalore = 
hours
Time taken by express train to cover 132 km = 
According to the given condition,
⇒ 1452 = x2 + 11x
⇒ x2 – 33x + 44x -1452 = 0
x(x – 33) + 44(x – 33) = 0
(x – 33)(x + 44) = 0
So, either
x – 33 = 0
x = 33
Or
x + 44 = 0
x = -44
But, the speed of the train can never be negative.
Thus, when x = 33 then speed of express train
= x + 11
= 33 + 11
= 44
Hence, the speed of the passenger train is x = 33 km/hr
and the speed of the express train is x = 44 km/hr respectively.
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.
Answer:
Let the side of the first square be ‘a’ m and that of the second be ′A′ m.
Area of the first square = a2 sq m.
Area of the second square = A2 sq m.
Their perimeters would be 4a and 4A respectively.
Given 4A – 4a = 24
A – a = 6 ……(1)
A2 + a2 = 468 ……….(2)
From (1), A = a + 6
Substituting for A in (2), we get
(a + 6)2 + a2 = 468
a2 + 12a + 36 + a2 = 468
2a2 + 12a + 36 = 468
a2 + 6a + 18 = 234
a2 + 6a – 216 = 0
a2 + 18a – 12a – 216 = 0
a(a + 18) – 12(a + 18) = 0
(a – 12)(a + 18) = 0
a = 12, – 18
So, the side of the first square is 12 m. and the side of the second square is 18 m.
Exercise 4.4
1. Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
Answer:
We know that the quadratic equation ax2+bx+c = 0 has
(a) Two distinct real roots, if b2-4ac > 0,
(b) Two equal real roots, if b2-4ac = 0,
(c) No real roots, if b2-4ac < 0.
(i) 2x2-3x+5 = 0
Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 2, b = -3 and c = 5
Then, b2-4ac = (-3)2-4×2×5 = 9-40 = -31 < 0
Hence the given quadratic equation has no real roots.
(ii) 3x2-4√3x+4 = 0
Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 3, b = -4√3 and c = 4
Then, b2-4ac = (-4√3)2-4×3×4 = 48-48 = 0
Therefore, real roots exist for the given equation and they are equal to each other.
And the roots will be
Therefore, the roots are 
(iii) 2x2-6x+3 = 0
Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 2, b = -6 and c = 3
then, b2-4ac = (-6)2-4×2×3 = 36-24 = 12 > 0
Hence the given quadratic equation has two distinct real roots.
Applying quadratic formula
to find roots,
2. Find the value of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2+kx+3 = 0
Answer:
Comparing the given quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 2, b = k and c = 3
Now the given quadratic equation have two equal roots if
b2-4ac = 0
(k)2-4×2×3 = 0
k2-24 = 0
k2 = 24
k = ±√24
k = ± 2√6
Therefore, the required value of k is ± 2√6.
(ii) kx(x-2)+6 = 0
Answer:
The given quadratic equation can be written as
kx2-2kx+6 = 0….(1)
Comparing the quadratic equation (1) with the general form of quadratic equation ax2+bx+c = 0 we get
a = k, b = -2k and c = 6
Now the given quadratic equation have two equal roots if
b2-4ac = 0
(-2k)2-4×k×6 = 0
4k2-24k = 0
4k(k-6) = 0
k(k-6) = 0
k = 0
Or, k-6 = 0
k = 6
If k = 0, then equation will not have x2 and x, which is not possible because the given equation is quadratic equation.
Therefore, the required value of k is 6.
3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2. If so find its length and breadth.
Answer:
Let the breadth of the mango grove be x m and the length is 2x m.
Area = length × breadth
= x × 2x
= 2×2 m2
Then by the given condition,
2x2 = 800
x2 = 400
x = ±√400
x = ±20
Since, length cannot be negative, then x ≠ -20
Hence it is possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m2 and its breadth is 20 m and length is 20×2 = 40 m
4. Is the following situation possible? If so, determine their present ages .
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer:
Let the age of 1st friend is x.
Then the age of 2nd friend is (20-x)
Four year ago their age was (x-4) and (20-x-4)
Then using the given condition we have
(x-4)(16-x) = 48
16x – 64 – x2 +4x = 48
20x – x2 – 64 – 48 = 0
x2 – 20x + 112 = 0……(1)
Now comparing the above quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 1, b = -20 and c = 112
then, b2-4ac = (-20)2-4×1×112 = 400-448 = -48 > 0
Therefore, no real root is possible for this equation and hence this situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so find its length and breadth.
Answer:
Let the length and breadth of the park be “l” and “b”
Area of rectangle = 2(l + b)
2(l + b) = 80
l + b = 40
b = 40 – l
Then, Area = l(40 – l)
Then by the given condition,
l(40 – l) = 400
40l – l2 -400 = 0
l2 – 40l + 400 = 0….(1)
Now comparing the above quadratic equation with the general form of quadratic equation ax2+bx+c = 0 we get
a = 1, b = -40 and c = 400
Then, b2-4ac = (-40)2-4×1×400 = 1600-1600 = 0
As the quadratic equation has two equal roots then the given situation is possible.
Therefore, length of park, l = 20 m
And breadth of park, b = 40 – l = 40 – 20 = 20 m.