Gujarat Board Solutions Class 10 Maths Chapter 4 Quadratic Equations

 Gujarat Board Solutions Class 10 Maths Chapter 4 Quadratic Equations

Gujarat Board Textbook Solutions Class 10 Maths Chapter 4 Quadratic Equations

Ex 4.1

Question 1.
Check whether the following are quadratic equations:

Solution:

∴It is of the form ax2 + bx + c = 0, where
a = 1, b = 0, c = 7.
Hence, the given equation is a quadratic equation.

(ii) RHS = ( – 2)(3 – x) = – 6 + 2x
x2 – 2x = – 6 + 2x
x2 – 2x + 6 – 2x = 0
x2 – 4x + 6 = 0
It is ofthe form ax2 + bx + c = 0, where
a = 1, b = – 4, c = 6.
Hence, the given equation is a quadratic equation.

(iii) LHS = (x – 2)(x + 1)
– x2 – x – 2
RHS = (x – 1) (x + 3)
= x2 + 2x – 3
x2 – x – 2 = x2 + 2x – 3
– x – 2 – 2x + 3 = 0
-3x + 1 = 0
It is not of the form ax2 + bx + c = 0
Hence, the given equation is not a quadratic equation
(iv) LHS = (x – 3) (2x + 1)
= 2x2 – 6x + x – 3
= 2x2 – 5x – 3
RHS = x(x + 5) = x2 + 5x
2x2 – 5x – 3 = x2 + 5x
= 2x2 – x2 – 5x – 5x – 3 = 0
x2 – 10x – 3 = 0
It is of the form ax2 + bx + c = 0, where
a= 1, b=10, c = -3
Hence, the given equation is a quadratic equation

It is of the form ax2 + bx + c = 0
where a = 1, b = -11, c = 8
Hence, the given equation is a quadratic equation.

(vi) RHS = (x – 2)2 = x2 – 4x + 4
x2 + 3x + 1 =x2 – 4x +4
3x + 4x + 1 – 4 = 0
7x – 3 = 0
It is not of the form ax2 +bx + c = 0
Hence the given equation is not a quadratic equation.

It is not of the form ax2 +bx + c = 0
Hence the given equation is not a quadratic equation.

It is not of the form ax2 +bx + c = 0 where a = 2, b = -13, c = 9.
Hence the given equation is not a quadratic equation.

Question 2.
Represent the following problems situations in the form of quadratic equations:

(i) The area of a rectangle plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. Find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. Find the integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages 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 uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.

Solution:
(i) Let breadth of the rectangle = x m. Then,
length of the rectangle = (2x + 1) m
∴ Area of the rectangle
= (2x + 1)xm2
According to the given condition,
(2x + 1)x = 528
2x2 + x – 528 = 0
which is a quadratic equation in x.
Solving this equation by factorization
method, we get

But breadth cannot be – ve.
∴ x = 16.
Hence, breadth of the rectangle = 16 m and
length of the rectangle = 2 x 16 + 1 = 33 m.

(ii) Let two consecutive positive integers are x and (x+ 1). Then
x (x + 1) = 306
x2 + x – 306 = 0
which is a quadratic equation in x.
Solving this equation by factorization
method, we get

But integers are +ve.
∴ x = 17.
Hence, two consecutive positive integers are 17 and 18.

(iii) Let Rohan’s present age = x years
Then, Rohan’s mother age = (x + 26) years
Rohan’s age after 3 years (x + 3) years
and Rohan’s mother age after 3 years
= (x + 29) years
According to the given condition,
(x + 3) (x + 29) = 360
x2 + 32x + 87 = 360
x2 + 32x – 273 = 0
Which is a quadratic equation in r solving it by factorization method.
x2 + 39x – 7x – 273 = 0
x(x + 39) – 7(x + 39) = 0
(x + 39) (x – 7) = 0
= x + 39.
(or)
x = 7
But age cannot be -ve,
– x = 7
Hence, Rohan’s present age a 7 years.

(iv) Let the speed of the train = x km/h
Distance = 480 km
Time taken by the tram to cover the distance of 48Okm = 480/x h.
Now, according to the given condition,

which is a quadratic equation in x.
Solving it by factorization method,
x2 – 40x + 32x – 1280 = 0
= x (x – 40) + 32(x – 40) = 0
(x – 40) (x + 32) = 0
x – 40 = 0
(or)
x+32 = 0
x = 40
or
x = – 32.
But speed of the train cannot be – ve.
x = 40.
Hence, the speed of the train =40 km/h.

Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorization method:

Solution:

Hence, the roots of the given quadratic equation are 5 and – 2.

x = – 2 or x = 3/2
Hence, the roots of the given quadratic equation are – 2 and 3/2

Hence, the roota of the given quadratic equation are – 5/√2 and – √2.


So, this root is repeated twice.
Hence, both the roots of the given quadratic equation are 1/10.

Question 2.
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. 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 ₹) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was 750. Find out the number of toys produced on that day.
Solution.
(i) Let the number of marbles John had = x
Then, the number of marbles Jivanti had = (45 – x)
According to the given question,

Hence, if the number of marbles John had = 36,
then the number of marbles Jivanti had
= 45 – 36 = 9.
or, if the number of marbles John had = 9,
then the number of marbles Jivanti had = 45 – 9 = 36.

(ii) Let the number of toys produced on said
day = x and, cost of production – (55 – x)
According to the given question,
x (55 – x) = 750
55x – x2 – 750 = 0

Hence, the number of toys produced on said day are 30 or 25.

Question 3.
Find two numbers whose sum is 27, and product is 182.
Solution:
Let the first number be x. So, second number = 27 – x.
According to question,

If x = 13 then other number = 27 – x
= 27 – 13 = 14
If x = 14 then other number = 27 – x
= 27 – 14 = 13
Hence, required two numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose square is 365.
Solution:
Let first positive integer be x.
So, other positive integer = x + 1
According to question,

Here x = – 14 does not exist.
Therefore, required positive integers are
x = 13 and x + 1 = 13 + 1 = 14.

Question 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.
Solution:
Let the base (BC) of the right triangle = x cm.
So, altitude (AB) = (x – 7) cm
and hypotenuse (AC) = 13 cm

According to pythagoras theorem,

Since side of triangle is never negative.
So, x = 12
Now required sides of the triangle are
AB = x – 7
= 12 – 7 = 5 cm
BC = x = 12 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 produces on that day. If the total cost of production on that day was 90. Find the number of articles produced and cost of each articles.
Solution.
Let the number of articles produced be x.
Therefore, cost of production of each article
(in ₹) on that particular day = ₹ (2x + 3)
So, total cost of production that day = ₹ x (2x + 3)
According to question,

Since x = – 15/2 is not possible.
So, x = 6
Hence, number of articles production be x = 6
and cost of production of each article
= ₹ (2x + 3)
= ₹ (2 x 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:

Solution

  

Question 2.
Find the roots of the quadratic equations given in Q.1, above by applying the quadratic formula.
Solution:
(i)
We have,
2x2 – 7x + 3 = 0
Here, a = 2, b = – 7 and c = 3
Now, b2 – 4ac = (7)2 – 4 (2)(3)
= 49 – 24 = 25
Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 – 7x + 3 = 0
has distinct roots i.e., α and β
Now,

Therefore, the required roots are 3 and 1/2
(ii)
We have,
2x2 + x – 4 = 0
Here, a = 2, b = l and c = – 4
Now, b2 – 4ac = (1)2 – 4(2)(-4)
= 1 + 32 = 33
Since, b2 – 4ac > 0
Therefore, the quadratic equation 2x2 + x – 4 = 0 has distinct roots i.e., α and β
Now,

(iv)
We have,
2x2 + x + 4 = 0.
Here, a = 2, b = 1 and c = 4
Now, b2 – 4ac =(1)2 – 4(2)(4)
= 1 – 32 = – 31
Since, b2 – 4ac < 0
Therefore, the quadratic equation 2x2 + x + 4 = 0 has no real roots.

Question 3.
Find the roots of the gollowing equations:

solution:

(ii)
We have

So the roots of the equation are 2 and 1.

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is Find his present age.
Solution:
Let the present age of Rehman be x years.
3 years ago, his age was (x – 3) years and
5 years from now his age will be (x +5) years.
Now, according to the given problem, we have

x2 + 2x – 15 = 3(2x + 2)
x2 + 2x – 15 = 6x +6
x2 + 2x – 6x – 15 – 6 = 0
x2 – 4x – 21 = 0
x2 – 7x + 3x – 21 = 0
x (x – 7) + 3(x – 7) = 0
(x – 7)(x + 3) = 0
x – 7 = 0 or x + 3 = 0
x = 7 or x = – 3
But age cannot be -ve.
x = 7.
Hence, the present age of Rehman = 7 years.

Question 5.
In a class, 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 would have been 210. Find her marks in the two subjects.
Solution:
Let marks got by Shefali in Mathematics = x
Then, marks got by her in English = (30 – x)
Now, according to the given problem, we have
(x+ 2)(30 – x – 3) = 210
(x + 2)(27 – x) = 210
27x – x2 + 54 – 2x – 210 = 0
– x2 + 25 x -156 = 0
x2 – 25 x + 156 = 0
x2 – 13x – 12x + 156 = 0
x(x – 13) – 12(x – 13) = 0
(x – 13)(x – 12) = 0
x – 13 = 0 or x – 12 = 0
x = 13 or x = 12
Hence, if marks got by Shefali in Mathematics 13
then marks got by her in English 30 – 13 = 17
and if marks got by Shefali in Mathematics = 12,
then marks got by her in English = 30 – 12 = 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 the shorter side of the field = x m
Then, longer side of the field = (x + 30) m and
diagonal of the field = (x + 60) m
Now, in right MBC, by Pythagoras theorem,
We have
AB2+ BC2 = AC2
(x + 30)2 + x2 = (x + 60)2
x2 + 60 x +900 + x2 = x2 + 120x + 3600
2x2 + 60x + 900 = x2 + 120x + 3600
2x2 + 60x – 120x + 900 – 3600 = 0
x2 – 60x – 2700 = 0
which is a quadratic equation in X.
We can solve it by factorization method
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x + 30) + (x – 90) = 0
x + 30 = 0 or x – 90 = 0
x = -30 or x = 90
But x = – 30 is not possible because aide cannot be negative.
x = 90
So, breadth of the rectangle = 90 m and length
of the rectangle = 90 + 30 = 120 m.

Question 7.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, would have taken 1 hour less for the sanie journey. Find the speed of the train.
Solution:
Let the speed of the train = x km/h
Distance = 360 km
Time taken to cover 360 km

x2 + 5x = 1800
x2 + 5x – 1800 = 0
which is a quadratic equation in x. It can be solved by factorization method.
x2 + 45x – 40x – 1800 = 0
x (x + 45) – 40(x + 45) = 0
– (x + 45)(x – 40) = 0
x + 45 = 0 or x – 40 = 0
x = – 45 or x = 40
But speed of the train cannot be -ve.
x = 40
Hence, speed of the train = 40 km/h.

Question 8.
Two water taps together can fill a tank in hours. The larger tap 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.
Solution:

 

Question 9.
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.
Solution:

 

Question 10.
Sum of the area of the two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of first square = x m
and the side of second square = y m
Then, the area of first square = x2 m2
and the area of second square = y2 m2
Perimeter of the first square = 4x m
and perimeter of the second square = 4y m
x2 + y2 = 468 ……………. (1)
and 4x – 4y = 24
x – y = 6 ……………(2)
From equation (2), we get
x = 6 + y …………..(3)
Putting this value of x in (1), we get
(6 + y)2 + y2 = 468
36 + 12y + y2 + y2 = 468
2y2 + 12y + 36 – 468 = 0
= 2y 2 + 12y – 432 = 0
y2 + 6y – 216 = 0
which is a quadratic equation in y.
We can solve this equation by quadratic formula.
Here,
a = 1, b = 6, c = – 216
D = b2 – 4ac
= (6)2 – 4 x 1 x ( – 216)
= 36 + 864 = 900

But the side of a square cannot be – ve.
y = 12
Putting this value ofy in (3), we get
x = 6 + 12 = 18
Hence, the side of the first square = 18 m
and the side of the second square = 12 m.

Ex 4.4

Question 1.
Find the discriminant ofthe following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – 4√3x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution:
(i) We have
2x2 – 3x + 5 = 0
Here, a = 2, b = – 3, c = 5
Now, D = b2 – 4ac
( – 3)2 – 4(2)(5)
= 9 – 40 = – 31
Since, D < 0
Therefore, the given quadratic equation has no real roots.

2x2 – 6x + 3 = O
Here, a = 2, b = – 6, c = 3
Now, D = b2 – 4ac
= (- 6)2 – 4(2)(3)
= 36 – 24 = 12
Since, D > 0
Therefore, the given quadratic equation has real and district roots.

Question 2.
Find the value of k for each of the following quadratic equations so that they have equal roots.
(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
Solution:
(i) We have,
2x2 + kx + 3 = 0
Here, a = 2, b = k,c = 3
Now, D = b2 – 4ac
(k)2 – 4(2)(3)
= k2 – 24
For equal roots,
D = 0
k2 – 24 = 0
k2 = 24

(ii) We have,
kx (x – 2) + 6 = 0
Here, a = k, b = – 2k, c = 6
Now, D = b2 – 4ac
D = ( – 2k)2 – 4(k)(6)
D = 4k2 – 24k
For equal roots,
D = 0
4k2 – 24k = 0
4k(k – 6) = 0
4k = 0 or k – 6 = 0
k = 0 or k = 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 the breadth of the rectangular mango grove be x m
So, length = 2x m
According to the question,
Area = 800 m2
2x (x) = 800
2x2 = 800
x2 = 400
x2 – 400 = 0
Here, a = 1, b = 0, c = – 400
so, D = b2 – 4ac
= – 4 x 1 x (- 400)
= 1600 > 0
Thus, designing the rectangular mango grove is possible.
x2 = 400
x =± √400 = ± 20
x = + 20 and – 20
x = – 20 is not possible.
So, x = 20
Hence, breadth of the rectangular mango grove is 20 m and length is 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 the present age of first friend be x years.
So, present age of other be (20 – x),
4 years ago,
Age of first friend = (x – 4) years
and age of second friend = (20 – x – 4) years
= (16 – x) years
According to question,
(x – 4) (16 – x) = 48
= 16x – x2 – 64 + 4x = 48
– x2 + 20x – 112 = 0
x2 – 20x + 112 = 0
Here, a = 1, b = – 20, c = 112
D = b2 – 4ac
= ( – 20)2 – 4(1)(112)
=400 – 448 = – 48
Since, D < 0
So, the given quadratic equation has no real roots and hence the given situation is not possible.