PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations
PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations
Ex 4.1
Question 1.
Check whether the following are quadratic equations:
(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)
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Solution:
(i) Given that
(x + 1)2 = 2(x – 3)
Or x2 + 1 + 2x = 2x – 6
Or x2 + 1 + 2x – 2x + 6 = 0
Or x2 + 7 = 0
Or x2 + 0x + 7 = 0
which is in the formof ax2 + bx + c = 0;
∴ It is a quadratic equation.
(ii) Given that
x2 – 2x = (-2) (3 – x)
Or x2 – 2x = -6 + 2x
Or x2 – 2x + 6 – 2x = 0
Or x2 – 4x + 6 = 0
which is the form of ax2 + bx + c = 0; a ≠ 0
∴ It is the quadratic equation.
(iii) Given that ,
(x – 2) (x + 1) = (x – 1) (x + 3)
Or x2 + x – 2x – 2 = x2 + 3x – x – 3
Or x2 – x – 2 = x2 + 2x – 3
Or x2 – x – 2 – x2 -2x + 3 = 0
Or -3x + 1 = 0 which have no term of x2.
So it is not a quadratic equation.
(iv) Given that
(x – 3)(2x + 1) = x(x + 5)
Or 2x2 + x – 6x – 3 = x2 + 5x
Or 2x2 – 5x – 3 – x2 – 5x = 0
Or x2 – 10x – 3 = 0
which is a form of ax2 + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.
(v) Given that ,
(2x – 1) (x – 3) = (x + 5) (x – 1)
0r2x2 – 6x – x + 3 = x2 – x + 5x – 5
Or 2x2 – 7x + 3 = x2 + 4x – 5
Or 2x2 – 7x + 3 – x2 – 4x + 5 = 0
Or x2 – 11x + 8 = 0
which is a form of ax2 + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.
(vi) Given that
x2+3x+1 = (x – 2)2
Or x2 + 3x + 1 = x2 + 4 – 4x
Or x2 + 3x + 1 – x2 – 4 + 4x = 0
Or 7x – 3 = 0
which have no term of x2.
So it is not a quadratic equation.
(vii) Given that
(x + 2)3 = 2x(x2 – 1)
Or x3 + (2)3 + 3 (x)2 2 + 3(x)(2)2 = 2x3 – 2x
Or x3 + 8 + 6x2 + 12x = 2x3 – 2x
Or x3 + 8 + 6x2 + 12x – 2x3 + 2x = 0
Or -x3 + 6x2 + 14x + 8 = 0
Here the highest degree of x is 3. which is a cubic equation.
∴ It is not a quadratic equation.
(viii) Given that
x3 – 4x2 – x+ 1= (x – 2)3
Or x3 – 4x2 – x + 1 = x3 – (2)3 + 3(x)2 (-2) + 3 (x) (-2)2
Or x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
Or x3 – 4x2 – x + 1 – x3 + 8 + 6x2 – 12x = 0
Or 2x2 – 13x + 9 = 0
which is in the form of ax2 + bx +c = 0; a ≠ 0
∴ It is a quadratic equation.
Question 2.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(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.
Solution:
(i) Let Breadth of rectangular plot = x m
Length of rectangular plot= (2x + 1) m
∴ Area of rectangular plot = [x (2x + 1)] m2 = (2x2 + x) m2
According to question,
2x2 + x = 528
S = 1
P = -528 × 2 = -1056
0r 2x2 + x – 528 = 0
Or 2x2 – 32x + 33x – 528 = 0
Or 2x(x – 16) + 33(x – 16) = 0
Or (x – 16) (2x + 33) = 0
Either x – 16 = 0 Or 2x + 33 = 0
x = 16 Or x = 2
∵ breadth of any rectangle cannot be negative, so we reject x = −33/2, x = 16
Hence, breadth of rectangular plot = 16 m
Length of rectangular plot = (2 ×16 + 1)m = 33m
and given problem in the form of Quadratic Equation are 2x2 + x – 528 = 0.
(ii) Let two consecutive positive integers are x and x + 1.
Product of Integers = x (x + 1) = x2 + x
According to question,
Or x2 + x – 306 = 0
S = 1, P = – 306
Or x2 + 18x – 17x – 306 = 0
Or x(x + 18) -17 (x + 18) = 0
Or (x + 18) (x – 17) = 0
Either x + 18 = 0 Or x – 17 = 0
x = -18 Or x = 17
∵ We are to study about the positive integers, so we reject x = – 18.
x = 17
Hence, two consecutive positive integers are 17, 17 + 1 = 18
and given problem in the form of Quadratic Equation is x2 + x – 306 = 0.
(iii) Let present age of Rohan = x years
Rohan’s mother’s age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
Rohan’s mother’s age = (x + 26 + 3) years = (x + 29) years
∴ Their product = (x + 3) (x + 29)
= x2 + 29x + 3x + 87
= x2 + 32x + 87
According to question,
x2 + 32x + 87 = 360
Or x2 + 32x + 87 – 360 = 0
Or x2 + 32x – 273 = 0
Or x2 + 39x – 7x – 273 = 0
S = 32, P = – 273
Or x(x + 39) – 7(x + 39) = 0
Or (x + 39) (x – 7) =
Either x + 39 = Or x – 7 = 0
x = -39 Or x = 7
∵ age of any person cannot be negative so, we reject x = -39
∴ x = 7
Hence, Rohans present age = 7 years
and given problem in the form of Quadratic Equation is x2 + 32x – 273 = 0.
or 3840 = 3 (u2 – 8u)
or u2 – 8u = 1280
or u2 – 8u – 1280=0
or u2 – 40u + 32u – 1280 = 0
S = -8, P = – 1280
or u(u – 40) + 32 (u – 40) = 0
or (u – 40)(u + 32) = 0
Either u – 40 = 0
or u + 32 = 0
u = 40 or u = -32
But, speed cannot be negative so we reject
u = – 32
∴ u = 40.
Hence speed of train is 40 km/hr Ans.
Ex 4.2
Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x2 – 3x – 10 = 0
(ii) 2x2 + x – 6 = 0
(iii) √2x2 + 7x + 5√2 = 0
(iv) 2 x2 – x + 1/8 = 0
(v) 100x2 – 20x + 1 = 0
Solution:
(i) Given quadratic
x2 – 3x – 10 = 0
Or x2 – 5x + 2x – 10 = 0
S = -3, p = -10
Or x (x – 5) + 2 (x – 5) = 0
Or (x – 5) (x + 2) = 0
Either x – 5 = 0 Or x + 2 = 0
x = 5 Or x = -2
Hence, 5 and -2 are roots of given Quadratic Equation.
(ii) Given quadratic equation
2x2 + x – 6 = 0 =1
0r 2x2 + 4x – 3x – 6 = 0
S = 1 P = -6 × 2 = -12
Or 2x (x + 2) -3 (x + 2) = 0
Or (x + 2) (2x – 3) = 0
Either x + 2 = 0 Or 2x – 3 = 0
x = -2 Or x = –3/2
Hence, – 2 and 3/2 are roots of given quadratic equation.
Question 2.
Solve the problems given in Example I. Statements of these problems are given below:
(i) John and Jivanti together have 45 marbles. Both of them lost S marbles each, and the product of the number of marbles they now have is 124. We would lfke to find out how many marbles they had to start with.
(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.
Solution:
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x
The number of marbles Íeft withJohn, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x2 – 200 + 5x
= -x2 + 45x – 200
According to question,
-x2 + 45x – 200 = 124
Or -x2 + 45x – 324 = 0
Or x2 – 45x + 324 =0
Or x2 – 36x – 9x + 324 = 0
S = -45, P = 324
Or x(x – 36) – 9(x – 36) = 0
Or (x – 36)(x – 9) = 0
Either x – 36 = 0, Or x – 9 = 0
x = 36 Or x = 9
∴ x = 36, 9
Hence, number of marbles they had to start with were 36 and 9 or 9 and 36.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
According to question.
x(55 – x) = 750
Or 55x – x2 = 750
Or -x2 + 55x – 750 = 0
Or x2 – 55x – 750 = 0
Or x2 – 30x – 25x + 750=0
S = -33, P = 750
Or x(x – 30) – 25(x – 30) = 0
Or (x – 30)(x – 25) = 0
Either x – 30 = 0 Or x – 25 = 0
x = 30 Or x = 25
∴ x = 30, 25
Hence, number of toys produced on that day were 30 and 25 or 25 and 30.
Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one number = x
2nd number = 27 – x
Their product = x (27 – x) = 27x – x2
According to question,
27x – x2 = 182
Or – x2 + 27x – 182 = 0
Or x2 – 27x + 182 = 0
S = -27, P = 182
Or x2 – 13x – 14x + 182 = 0
Or x(x – 13) – 14(x – 13) = 0
Or (x – 13) (x – 14) = 0
Either x – 13 = 0 Or x – 14 = 0
x = 13 Or x = 14
x = 13, 14
Hence, two numbers are 13 and 14 Or 14 and 13.
Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let one positive integer = x
2nd positive integer = x + 1
According to question,
(x)2 + (x + 1)2 = 365
Or x2 + x2 + 1 + 2x = 365
Or 2x2 + 2x + 365 = 0
Or 2x2 + 2x – 364 = 0
Or x2 + x – 182 = 0
Or x2 + 14x – 13x – 182 = 0
S = 1, P = -182
Or x(x + 14) – 13(x + 14) = 0
(x + 14)(x— 13) = O
Either x + 14 = 0
Or x = -14
Or
x – 13 = 0
x = 13
∵ We have positive integers.
So, we reject x = – 14.
∴ x = 13
∴ One positive integer = 13
and 2nd positive integer = 13 + 1 = 14
Hence, required consecutive positive integers are 13 and 14.
Question 5.
The altitude of a right triangle is 7 cm less than its base. 1f the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base of right triangle = x cm
Altitude of right triangle = (x – 7) cm
and hypotenuse of right triangle = 13 cm (Given)
According to Pythagoras Theorem,
(Base)2 + (Altitude)2 = (Hypotenuse)2
(x)2 ÷ (x – 7)2 = (13)2
Or x2 + x2 + 49 – 14x = 169
Or 2x2 – 14x + 49 – 169 = 0
Or 2x2 – 14x – 120 = 0
Or 2[x2 – 7x – 60] = 0
Or x2 – 7x – 60 = 0
Or x2 – 12x + 5x – 60 = 0
S = – 7 P = – 60
Or x(x – 12) + 5(x – 12) = 0
Or (x – 12) (x + 5) = 0
Either x – 12 = 0 Or x + 5 = 0
x = 12 Or x= – 5
∵ Length of any triangle cannot be negative.
So, we reject x = – 5
∴ x = 12
Hence, base of right triangle = 12 cm
Altitude of right triangle = (12 – 7) cm = 5 cm.
Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90, find the number of articles produced and the cost of each article.
Solution:
Let, number of pottery articles produced by industry in one day = x
Cost of production of each article = ₹ (2x + 3)
∴ Total cost of production in panicular day = ₹ [x(2x + 3)] = ₹ (2x2 + 3x)
According to question,
2x2 + 3x = 90
2x2 + 3x – 90 = 0
S = 3, P = 2 × -90 = -180
Or 2x2 – 12x + 15x – 90 = 0
Or 2x (x – 6) + 15 (x – 6) = 0
Or (x – 6) (2x + 15) = 0
Either x – 6 = 0 Or 2x + 15 = 0
x = 6 Or x = −15/2
∵ number of articles cannot be negative
So, we reject x = 2
∴ x = 6
Hence, number of articles produced on certain day = 6
and cost of production of each article = ₹ [2 × 6 + 3] = ₹ 15.
Ex 4.3
Question 1.
Find the roots of the following quadratic equations if they exist, by the method of completing the square:
(i) 2x2 + 7x + 3
(ii) 2x2 + x – 4 = 0
(ili) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is 13. F1nd his present age.
Solution:
Let Rehman’s present age = x years
3 years ago Rehman’s age (x – 3) years
5 years from now Rehman’s age =(x + 5) years
According to question,
Question 5.
In a class test, the sum of Shefall’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less ¡n English, the product of their marks would have been 210. Find her marks In the two subjects.
Solution:
Let Shefali get marks in Mathematics = x
Shefali’s marks in English = 30 – x
According to 1st condition,
Shefali’s marks in Mathematics = x + 2
and Shefali’s marks in English = 30 – x – 3 = 27 – x
∴ Their product = (x + 2) (27 – x)
= 27x – x2 + 54 – 2x
= x2 + 25x + 54
According to 2nd condition,
-x2+ 25x+ 54 = 210
Or -x2 + 25x + 54 – 210 = 0
Or -x2 + 25x – 156 = 0
Or x2 – 25x+ 156 = o
Compare it with ax2 + bx + c = O
a = 1, b = -25, c = 156
Now, b2 – 4ac = (-25)2 – 4 × 1 × 156
= 625 – 624 = 1 > 0
Case I:
When x = 13
then Shefaiis marks in Maths = 13
Shefali’s marks in English = 30 – 13 = 17.
Case II:
When x = 12
then Shefalis marks in Maths = 12
Shefali’s marks in English = 30 – 12 =18.
Hence, Shefalis marks in two subjects are 13 and 17 Or 12 and 18.
Question 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.
Solution:
Let shorter side of rectangular field = AD = x m
Longer side of rectangular field = AB = (x + 30) m
and diagonal of rectangular field = DB = (x + 60) m
In rectangle. the angle between the length and breadth is right angle.
∴ ∠DAB = 90°
Now, in right angled triangle DAB, using Pythagoras Theorem,
(DB)2 = (AD)2 + (AB)2
(x + 60)2 = (x)2 + (x + 30)2
Or x2 + 3600 + 120x = x2 + x2 + 900 + 60x
Or x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
Or -x2 + 60x + 2700 = 0
Or x2 – 60x – 2700 = 0
Compare it with ax2 + bx + e = O
∴ a = 1, b = -60, c = -2700
and b2 – 4ac = (-60)2 – 4. 1 . (-2700)
= 3600 + 10800 = 14400 > 0
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.
Solution:
Let larger number = x .
Smaller number = y
According to 1st condition,
x2 – y2 = 180 ……………(1)
According to 2nd condition,
y2 = 8x
From (1) and (2), we get
x2 – 8x = 180
Or x2 – 8x – 180 = 0
Compare it with ax2 + bx + c = 0
∴ a = -1, b = -8, c = -180
and b2 – 4ac = (-8)2 – 4 × 1 × (-180)
= 64 + 720 = 784 > 0
Question 8.
A train travels 360 km ¡t 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.
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 to intermediate stations), if the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Question 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.
Solution:
In case of larger square
Let length of each side of square = x m
Area of square = x2 m2
Perimeter of square = 4x m
In case of smaller square:
Let lenth of each side of square = y m
Area of square = y2 m2
Perimeter of square = 4y m
According to 1st condition,
x2 + y2 = 468 …………….(1)
According to 2nd condition,
4x – 4y = 24
Or 4(x – y) = 24
Or x – y = 6
x = 6 + y
From (1) and (2), we get
(6 + y)2 + y2 = 468
Or 36 + y2 + 12y + y2 = 468
Or 2y2 + 12y + 36 – 468 = 0
Or 2y2 + 12y – 432 = 0
Or y2 + 6y – 216 = 0
Compare it with ay2 + by + c = 0
∴ a = 1, b = 6, c = -216
and b2 – 4ac = (6)2 – 4 × 1 × (- 216) = 36 + 864 = 900 > 0
∵ length of square cannot be negative
So, we reject y = – 18
∴ y = 12
From (2), x = 6 + 12 = 18
Hence, sides of two squares are 12 m and 18 m.
Ex 4.4
Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i)2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = o
(iii) 2x2 – 6x + 3 = 0
Solution:
(i) Given quadratic equation is, 2x2 – 3x + 5 = 0
Compare it with ax2 + bx + c = 0
a = 2, b = -3, c = 5
D = b2 – 4ac
= (-3)2 4 × 2 × 5
= 9 – 40 = -31 < 0
Hence, given quadratic equation has no real roots.
Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) Given quadratic equation is : 2x2 + kx + 3 = 0
Compare it with ax2 + bx + c = 0
∴ a = 2, b = k, c = 3
∵ roots of the given quadratic equation are equal.
∴ D = 0
b2 – 4ac = 0
Or(k)2 – 4 × 2 × 3 = 0
Or k2 – 24 = 0
Or k2 = 24
Or k = ±√24
Or k = ±2√6.
(ii) Given quadratic equation is:
kx (x – 2) + 6 = 0
Or k – 2kx + 6 = 0
Compare it with ax2 + bx + c = 0
∴ a = k, b = -2k, c = 6
∵ roots of the given quadratic equation are equal
∴ b2 – 4ac = 0
Or(-2k)2 – 4 × k × 6 = 0
Or 4k2 – 24k = 0
Or 4k[k – 6]= 0
Either 4k = 0 Or k- 6 = 0
k = 0 Or k = 6
∴ k = 0, 6.
Question 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.
Solution:
Let breadth of rectangular grove = x m
and length of rectangular grove = 2x m
Area of rectangular grove = length × breadth
= [x × 2x] m2 = 2 × 2 m2
According to question
2x2 = 800
x2 = 800/2 = 400
x = ± √400
x = ± 20.
∵ length of rectangle cannot be negative.
So, we reject x = -20
∴ x = 20
∴ breadth of rectangular grove = 20 m
and length of rectangular grove = (2 × 20) m = 40 m.
Question 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.
Solution:
Let age of one friend = x years
and age of 2nd friend = (20 – x) years
Four years ago,
Age of 1st friend = (x – 4) years
Age of 2nd friend = (20 – x – 4) years = (16 – x) years
∴ Their product = (x – 4) (16 – x)
= 16x – x2 – 64 + 4x
= – x2 + 20x – 64
According to Question
– x2 + 20x – 64 = 48
Or – x2 + 20x – 64 – 48 = 0
Or – x2 + 20x – 112 = 0
Or x2 – 20x + 112 = 0 …………….(1)
Compare it with ax2 + bx + c = 0
∴ a = 1, b = -20, c = 112
D = b2 – 4ac
= (-20)2 – 4× 1 × 112
= 400 – 448 = -48 < 0
∴ roots are not real
then no real value of x satisfies the quadratic equation (1).
Hence, given situation is not possible.
Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
Solution:
Let length of rectangular park = x m
Breadth of rectangular park = y m
∴ Perimeter of rectangular park = 2 (x + y) m
and area of rectangular park = xy m2
According to 1st condition
2 (x + y) = 80
x + y = 80/2 = 40
y = 40 – x …………(1)
According to 2nd condition,
xy = 400
x (40 – x) = 400 [using (1)]
Or 40x – x2 = 400
Or 40x – x2 – 400 = 0
Or x2 – 40x + 400 = 0
Compare it with ax2 +bx + c = 0
a = 1, b = -40, c = 400
D = b2 – 4ac
= (-40)2 – 4 × 1 × 400
= 1600 – 1600 = 0
When x = 20 then from (1)
y = 40 – 20 = 20
∴ Length and breadth of rectangular park are equal of measure 20 m.
Hence, given rectangular park exist and it is a square.