Quadratic Equations Solution of TS & AP Board Class 10 Mathematics
Check whether the following are quadratic equation:
(x + 1)2 = 2(x - 3)
Answer:(x + 1)2 = 2(x - 3)
LHS = (x + 1)2 = x2 + 1 + 2x
RHS = 2(x – 3) = 2x – 6
∴ x2 + 1 + 2x = 2x - 6
⇒ x2 + 1 + 2x – 2x + 6 = 0
⇒ x2 + 7 = 0 which is of the form ax2 + bx + c = 0
∴ it is a quadratic equation.
Question 2.
Check whether the following are quadratic equation:
x2 - 2x = ( - 2)(3 - x)
Answer:
x2 - 2x = ( - 2)(3 - x)
LHS = x2 – 2x
RHS = ( - 2)(3 - x) = - 6 + 2x
∴ x2 - 2x = - 6 + 2x
Or x2 - 2x + 6 – 2x = 0
⇒ x2 + 6 which is of the form ax2 + bx + c = 0
∴ it is a quadratic equation.
Question 3.
Check whether the following are quadratic equation:
(x - 2)(x + 1) = (x - 1) (x + 3)
Answer:
(x - 2)(x + 1) = (x - 1) (x + 3)
LHS = (x - 2)(x + 1) = x2 + x - 2x - 2 = x2 - x - 2
RHS = (x - 1) (x + 3) = x2 + 3x - x - 3 = x2 + 2x - 3
Equating LHS and RHS
x2 - x - 2 = x2 + 2x - 3
⇒ - 3x + 1 = 0 which is not of the form ax2 + bx + c = 0
Hence it is not a quadratic equation.
Question 4.
Check whether the following are quadratic equation:
(x - 3)(2x + 1) = x(x + 5)
Answer:
(x - 3)(2x + 1) = x(x + 5)
Here LHS = (x - 3)(2x + 1) = 2x2 - 6x + x - 3 = 2x2 - 5x - 3
RHS = x(x + 5) = x2 + 5x
Equating LHS and RHS
2x2 - 5x - 3 = x2 + 5x
x2 - 10x - 3 = 0 which is of the form ax2 + bx + c = 0
∴ it is a quadratic equation.
Question 5.
Check whether the following are quadratic equation:
(2x - 1) (x - 3) = (x + 5)(x - 1)
Answer:
(2x - 1) (x - 3) = (x + 5)(x - 1)
Here LHS = (2x – 1)(x - 3) = 2x2 –x - 6x + 3 = 2x2 - 7x + 3
RHS = (x + 5)(x - 1) = x2 + 5x – x - 5 = x2 + 4x - 5
Equating LHS and RHS
2x2 - 7x + 3 = x2 + 4x - 5
x2 – 11x + 8 = 0
which is of the form ax2 + bx + c = 0
∴ it is a quadratic equation.
Question 6.
Check whether the following are quadratic equation:
x2 + 3x + 1 = (x - 2)2
Answer:
x2 + 3x + 1 = (x - 2)2
Here RHS = (x - 2)2
⇒ x2 + 4 - 4x
Equating LHS and RHS
x2 + 3x + 1 = x2 + 4 - 4x
7x - 3 = 0
Which is not of the form ax2 + bx + c = 0
∴ it is not a quadratic equation.
Question 7.
Check whether the following are quadratic equation:
(x + 2)3 = 2x(x2 - 1)
Answer:
(x + 2)3 = 2x(x2 - 1)
LHS = (x + 2)3 = x3 + 23 + 3 × x × 2(x + 2)
(∵(a + b)3 = a3 + b3 + 3ab(a + b))
= x3 + 8 + 6x2 + 12x
RHS = 2x3 - 2x
Equating LHS and RHS
x3 + 8 + 6x2 + 12x = 2x3 - 2x
- x3 + 6x2 + 14x + 8 = 0
which is not of the form ax2 + bx + c = 0
∴ it is not a quadratic equation.
Question 8.
Check whether the following are quadratic equation:
x3 - 4x2 - x + 1 = (x - 2)3
Answer:
x3 - 4x2 - x + 1 = (x - 2)3
RHS = (x - 2)3 = x3 - 23 - 3 × x × 2(x - 2)
(∵(a + b)3 = a3 + b3 + 3ab(a + b))
= x3 - 8 - 6x2 + 12x
Equating LHS and RHS
x3 - 4x2 –x + 1 = x3 - 8 - 6x2 + 12x
2x2 - 13x + 9 = 0
which is of the form ax2 + bx + c = 0
∴ it is a quadratic equation.
Question 9.
Represent the following situations in the form of quadratic equation:
The area of a rectangular plot is 528m2. The length of the plot is one meter more than twice its breadth. We need to find the length and breadth of the plot.
Answer:
Let length = L and breath = B
Area of rectangle (A) = L × B
According to the given conditions
L = 2B + 1
∴ A = (2B + 1) B
⇒ 528 = 2b2 + b
⇒ 2b2 + b - 528 = 0
Which is a quadratic equation of the form ax2 + bx + c = 0
Question 10.
Represent the following situations in the form of quadratic equation:
The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let us suppose the two consecutive numbers to be x and x + 1
According to the given condition x(x + 1) = 306
x2 + x - 306 = 0, which is the required quadratic equation of the form ax2 + bx + c = 0
Question 11.
Represent the following situations in the form of quadratic equation:
Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Answer:
Let age of Rohan = x
⇒ Rohan’s mother age = x + 26
After 3 years, Rohan’s age = x + 3
and mother’s age = (x + 29) years
According to the given condition
(x + 3) (x + 29) = 360
x2 + 32x - 273 = 0, which is quadratic equation of the form ax2 + bx + c = 0
x2 + 39x - 7x -273 = 0
x(x + 39) -7 (x+ 39) = 0
(x-7)(x+39)= 0
x = 7 or x = -39
but age cannot be negative.
hence, x = 7
Question 12.
Represent the following situations in the form of quadratic equation:
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 the speed of the train be x
Time taken to cover 480 km =
If the speed had been 8 km/h less than the train would take 3 hours to cover the same distance
Then speed will be x - 8 and time will be =
Speed =
⇒
⇒ 480x - 3x2 - 3840 + 24x = 480x
⇒ - 3(x2 - 8x + 1280) = 0
⇒ (x2 - 8x + 1280) = 0
which is the required quadratic equation of the form ax2 + bx + c = 0
Answer:
⇒ x2 - 5x + 2x - 10 = 0
⇒ x(x - 5) + 2(x - 5) = 0
⇒ (x + 2) (x - 5) = 0
The roots of the equation are those value for which (x + 2) (x - 5) = 0
⇒ x = - 2 or 5
- 2 and 5 are the roots of the given equation.
Question 2.
Find the roots of the following quadratic equation by factorization:
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
The roots of the equation are those value for which (x + 2) (2x - 3) = 0
⇒
Answer:
x(x + 4) = 12
⇒ x2 + 4x - 12 = 0
⇒ x2 + 6x - 2x - 12 = 0
⇒ x( x + 6) - 2( x + 6) = 0
⇒ (x - 2) (x + 6) = 0
The roots of the equation are those value for which (x - 2) (x + 6) = 0
⇒ x = 2 or - 6
2 and - 6 are the roots of the given equation.
Question 7.
Find the roots of the following quadratic equation by factorization:
3x2 - 5x + 2 = 0
Answer:
3x2 - 5x + 2 = 0
⇒ 3x2 – 2x - 3x + 2 = 0
⇒ x(3x - 2) - 1(3x - 2) = 0
⇒ (x - 1) (3x - 2) = 0
The roots of the equation are those value for which (x - 1) (3x - 2) = 0
⇒
Hence
are the roots of the given equation.
Question 8.
Find the roots of the following quadratic equation by factorization:
Answer:
⇒ x2 - 2x - 3 = 0
⇒ x2 – 3x + 1x - 3 = 0
⇒ x(x - 3) + 1(x - 3) = 0
⇒ (x + 1) (x - 3) = 0
The roots of the equation are those value for which (x + 1) (x - 3) = 0
⇒ x = - 1 or 3
Hence - 1 and 3 are the roots of the given equation.
Question 9.
Find the roots of the following quadratic equation by factorization:
3(x - 4)2 - 5(x - 4) = 12
Answer:
3(x - 4)2 - 5(x - 4) = 12
⇒ 3( x2 + 16 - 8x) - 5x + 20 - 12 = 0
⇒ 3x2 + 48 – 24x - 5x + 20 - 12 = 0
⇒ 3x2 – 29x + 56 = 0
⇒ 3x2 – 21x - 8x + 56 = 0
⇒ 3x(x - 7) - 8( x - 7) = 0
⇒ (3x - 8) (x - 7) = 0
The roots of the equation are those value for which (3x - 8) (x - 7) = 0
⇒
are the roots of the given equation.
Question 10.
Find two numbers whose sum is 27 and product is 182.
Answer:
let the two numbers be x and y
According to the given condition
x + y = 27 ⇒ y = 27 - x
And x × y = 182
⇒ x (27 –x) = 182
⇒ 27x – x2 - 182 = 0
⇒ x2 – 27x + 182 = 0
⇒ x2 – 13x – 14x + 182 = 0
⇒ x (x - 13) - 14 (x - 13) = 0
⇒ (x - 14) (x - 13) = 0
The roots of the equation are those value for which(x - 14) (x - 13) = 0
⇒ x = 14 or 13
Hence two numbers are 14 and 13
Question 11.
Find two consecutive positive integers, sum of whose square is 613.
Answer:
Let the two consecutive positive integers be x and x + 1
According to the given condition
x2 + (x + 1)2 = 613
⇒ x2 + x2 + 1 + 2x = 613
⇒ 2x2 + 2x – 612 = 0
⇒ x2 + x - 306 = 0
⇒ x2 + 18x – 17x - 306 = 0
⇒ x(x + 18) - 17(x + 18) = 0
⇒ (x + 18)(x - 17) = 0
The roots of the equation are those value for which(x + 18)(x - 17) = 0
⇒ x = 17 or - 18
Hence the two positive integers are 17 and 18 or - 17 and - 18
Question 12.
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:
Consider a right angle ΔABC
Let the base of the Δ be x
⇒ Altitude = x - 7
In the right-angled triangle
According to the Pythagoras theorem
(Hypotenuse)2 = (base)2 + (altitude)2
⇒ 132 = x2 + (x - 7)2
⇒ 169 – x2 – x2 - 49 + 14x = 0
⇒ 2x2 - 14x – 120 = 0
⇒ x2 – 7x - 60 = 0
⇒ x2 – 12x + 5x - 60 = 0
⇒ x( x - 12) + 5( x – 12) = 0
⇒ (x + 5) (x - 12) = 0
⇒ x = - 5 or 12
But base cannot be negative, so x = 12 cm
⇒ altitude = 12 - 7 = 5 cm
Hence 5 cm and 12 cm are the two sides of the given triangle.
Question 13.
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 Rs.90 , find the number of articles produced and the cost of each articles.
Answer:
Let x be the number of articles produced on that day
Thus cost of production of each article = 2x + 3
Total cost = Rs 90
According to the given condition
x(2x + 3) = 90
⇒ 2x2 + 3x - 90 = 0
⇒ 2x2 - 12x + 15x – 90 = 0
⇒ 2x(x - 6) + 15(x - 6) = 0
⇒ (2x + 15) (x - 6) = 0
⇒
Since the number of articles produced cannot be negative ∴ x = 6
And the cost of each article = 2 × 6 + 3 = Rs 15
Number of articles = 6
Cost of each article = 15
Question 14.
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Answer:
Let the length and the breath of the rectangle be l and b respectively.
According to the given condition
l × b = 40 m2 ( area = length × breath)…………..1
and
Perimeter = 2(l + b)
2(l + b) = 28
⇒ l + b = 14
⇒ b = 14 –l
Putting value of b in 1
l( 14 - l) = 40
⇒ l2 – 14 l + 40 = 0
⇒ l2 – 10l – 4l + 40 = 0
⇒ l( l - 10) – 4(l - 10) = 0
⇒ (l - 4) ( l - 10) = 0
⇒ l = 4 or 10 m
Hence length is 4 m
And breath is 10 m.
Question 15.
The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm, the find its base and altitude.
Answer:
let the base of the triangle be x cm and altitude be y cm
Given base = 4 + y
Area of the triangle = 1/2 × base × altitude
48 = 1/2 × (4 + y) × y
⇒ 96 = 4y + y2
⇒ y2 + 4y – 96 = 0
⇒ y2 + 12y - 8 y – 96 = 0
⇒ y( y + 12) - 8 ( y + 12) = 0
⇒ (y - 8) ( y + 12) = 0
⇒ y = 8 or - 12
Since altitude cannot be negative ∴ y = 8
⇒ base = 4 + 8 = 12 cm
Hence, Base = 12 cm; Altitude = 8 cm
Question 16.
Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train If after two hours they are 50 km. apart find the average speed of each train.
Answer:
Let x km /hour be the speed of the first train
Then the speed of the second train is (x - 5) km/hour
Given that they are 50 km apart after 2 hours
As per the given conditions
The first train travels from O to A and
The 2nd train travels from O to B
Also given that AB = 50 Km
Since the OAB is a right angled triangle
∴ By Pythagoras theorem
OA2 + OB2 = AB2………….1
Now distance = speed × time
⇒ OA = x × 2
Also OB = 2(x - 5)
Putting value of OB and OA in 1
⇒ 8 x2 – 40x + 100 = 2500
⇒ x2 – 5x – 300 = 0
⇒ x(x - 20) + 15(x - 20) = 0
⇒ (x + 15) (x - 20) = 0
⇒ x = - 15 or 20
Since speed cannot be negative ∴ the speed of the 1st train = 20 km/hour
And speed of 2nd train is 15 km/hour.
Question 17.
In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was D 1600. How many boys are there in the class?
Answer:
Let the number of girls be x and number of boys be y
Given that the total number of students = 60
⇒ x + y = 60 ⇒ y = 60 –x
Total money collected = 1600
According to the given condition
1600 = xy + yx
⇒ 1600 = 2 xy
⇒ xy = 800
⇒ x( 60 –x) = 800
⇒ x2 - 60x + 800 = 0
⇒ x2 – 20x – 40x + 800 = 0
⇒ x ( x – 20) – 40 (x – 20) = 0
⇒ (x - 40) (x - 20) = 0
⇒ x = 40 or 20
Hence if the number of girls is 40, then the number of boys is 20
And if the number of girls is 20, then the number of boys is 40.
Question 18.
A motor boat heads upstream a distance of 24 km on a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained a constant speed, what was its speed?
Answer:
let the speed of the stream be x km/hour
∴ The speed of the boat upstream = (x - 3) km/hr and the speed of the boat downstream = (x + 3)km/hr
Similarly,
According to the given condition
⇒ 24(x + 3) + 24 (x - 3) = 6 (x + 3) (x - 3)
⇒ 48x = 6x2 – 54
⇒ x2 - 8x - 9 = 0
⇒ x2 - 9x + x – 9 = 0
⇒ x( x - 9) + (x - 9) = 0
⇒ (x + 1) (x - 9) = 0
⇒ x = - 1 or 9
Since the speed of the stream cannot be negative so x = 9 km/hr
Hence the speed of the stream is 9 km/hr.
Exercise 5.3
Find the roots of the following quadratic equation if they exist.
2x2 + x - 4 = 0
Answer:
2x2 + x - 4 = 0
The equation is of the form ax2 + bx + c = 0
Since a≠0 ,so it exists
We use the quadratic formula to evaluate the roots
The two roots are
Question 2.
Find the roots of the following quadratic equation if they exist.
4x2 + 4√3 + 3 = 0
Answer:
4x2 + 4√3 + 3 = 0
The equation is of the form ax2 + bx + c = 0
Since a≠0 ,so it exists
We use the quadratic formula to evaluate the roots
The two roots are
Question 3.
Find the roots of the following quadratic equation if they exist.
5x2 - 7x - 6 = 0
Answer:
5x2 - 7x - 6 = 0
The equation is of the form ax2 + bx + c = 0
Since a≠0 ,so it exists
We use the method of factorization to solve it
⇒ 5x2 - 10x + 3x - 6 = 0
⇒ 5x(x - 2) + 3(x - 2) = 0
⇒ (5x + 3)(x - 2) = 0
⇒ x
Question 4.
Find the roots of the following quadratic equation if they exist.
x2 - 6x + 5 = 0
Answer:
x2 - 6x + 5 = 0
The equation is of the form ax2 + bx + c = 0
Since a≠0 ,so it exists
We use the method of factorization to solve it
⇒ x2 - 5x - x + 5 = 0
⇒ x(x - 5) - 1(x - 5) = 0
⇒ (x - 1)(x - 5) = 0
⇒ x = 1,5
Question 5.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
2x2 + x - 4 = 0
Answer:
We use the quadratic formula to evaluate the roots
The two roots are
Question 6.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
4x2 + 4√3 + 3 = 0
Answer:
We use the quadratic formula to evaluate the roots
The two roots are
Question 7.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
5x2 - 7x - 6 = 0
Answer:
We use the method of factorization to solve it
⇒ 5x2 - 10x + 3x - 6 = 0
⇒ 5x(x - 2) + 3(x - 2) = 0
⇒ (5x + 3)(x - 2) = 0
⇒ x
Question 8.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
x2 - 6x + 5 = 0
Answer:
We use the method of factorization to solve it
⇒ x2 - 5x - x + 5 = 0
⇒ x(x - 5) - 1(x - 5) = 0
⇒ (x - 1)(x - 5) = 0
⇒ x = 1,5
Answer:
⇒ 11(x2 - 3x - 28) = - 330
⇒ x2 - 3x - 28 = - 30
⇒ x2 - 3x + 2 = 0
We use the method of factorization to solve
⇒ x2 - 2x - x + 2 = 0
⇒ x(x - 2) - 1(x - 2) = 0
⇒ (x - 1)(x - 2) = 0
⇒ x = 1, 2
Question 11.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is
Find his present age.
Answer:
Let Present age of Rehman be x
Three Years ago his age was = x - 3
Five years from now his age will be = x + 5
According to the problem:
⇒ 6(x + 1) = (x - 3)(x + 5)
⇒ 6x + 6 = x2 + 2x - 15
⇒ x2 - 4x - 21 = 0
⇒ x2 - 7x + 3x - 21 = 0
⇒ x(x - 7) + 3(x - 7) = 0
⇒ (x + 3)(x - 7) = 0
⇒ x = - 3 ,7
Since age cannot be negative so ,
Present age is 7 years
Question 12.
In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 marks more in Mathematics and 3 marks less in English, the product of her marks have been 210. Find her marks in the two subjects.
Answer:
Maths = 12, English = 18 (or)
Maths = 13, English = 17
Let marks in Mathematics be x , marks in English be 30 - x
When marks in Mathematics is increased by 2 it becomes x + 2
When marks in English is decreased by 3 it becomes 27 - x
According to the problem the product of the two marks is 210
(x + 2)(27 - x) = 210
⇒ - x2 + 25x + 54 = 210
⇒ x2 - 25x + 156 = 0
Performing factorization we get:
⇒ x2 - 13x - 12x + 156 = 0
⇒ x(x - 13) - 12(x - 13) = 0
⇒ (x - 12)(x - 13) = 0
⇒ x = 12,13
There will be two answers
Mathematics = 12 ,English = (30 - 12) = 18
Mathematics = 13, English = (30 - 13) = 17
Question 13.
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:
120 m; 90 m
Let the shorter side be x
Longer side = x + 30
Diagonal = x + 60
Applying Pythagoras theorem
Diagonal2 = Longer side2 + shorter side2
⇒ (x + 60)2 = (x + 30)2 + x2
⇒ x2 + 120x + 3600 = x2 + 60x + 900 + x2
⇒ x2 - 60x - 2700 = 0
Performing factorization we get:
⇒ x2 - 90x + 30x - 2700 = 0
⇒ x(x - 90) + 30(x - 90) = 0
⇒ (x + 30)(x - 90) = 0
⇒ x = - 30, 90
Neglecting the negative term
Shorter Side = 90 m
Longer side = (90 + 30) = 120 m
Question 14.
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:
18, 12; - 18, - 12
Let the larger number be x and smaller number be y
x2 – y2 = 180 …Equation (i)
Since the square of the smaller number is 8 times the larger number, so
⇒ x2 – 8x - 180 = 0
Performing factorization we get:
⇒ x2 - 18x + 10x - 180 = 0
⇒ x(x - 18) + 10(x - 18) = 0
⇒ (x + 10)(x - 18) = 0
x = - 10,18
Rearranging Equation (i)
y2 = x2 - 180
Putting x = - 10 we get
y2 = - 80
Since the square is negative so the solution is neglected
Putting x = 18 we get
y2 = 144
⇒ y =
12
Answer: 18,12 ;18, - 12
Question 15.
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 be x
Distance = 360 km
Time taken for a speed of x km/h
When speed is (x + 5) km/h time taken
According to the problem:
⇒ 360(x + 5) = 360x + x(x + 5)
⇒ x2 + 5x - 1800 = 0
Performing factorization we get:
x2 + 45x - 40x - 1800 = 0
⇒ x(x + 45) - 40(x + 45)
⇒ (x - 40)(x + 45) = 0
x = 40, - 45
Since Speed cannot be negative, so
Speed = 40 km/h
Question 16.
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 time taken by the tap of larger diameter be t and smaller diameter be t + 10
Part of tank filled in 1 hour by the tap of larger diameter
Part of tank filled in 1 hour by the tap of smaller diameter
Part of tank filled in 1 hour when both the taps were working together
So we can say:
⇒ 150t + 750 = 8(t2 + 10t)
⇒ 8t2 - 70t - 750 = 0
⇒ 4t2 - 35t - 375 = 0
Performing factorization we get:
⇒ 4t2 - 60t + 25t - 375 = 0
⇒ 4t(t - 15) + 25(t - 15) = 0
⇒ (4t + 25)(t - 15) = 0
t
Since time is always a positive quantity, so
Time taken by the tap of larger diameter = 15 h
Time taken by the tap of smaller diameter = (15 + 10) = 25 h
Question 17.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (with out 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 speed of passenger train be x and express train be x + 11
Distance travelled = 132 km
Time difference between the two trains = 1 hours
⇒ 132x + 1452 - 132x = x(x + 11)
⇒ x2 + 11x - 1452 = 0
Performing factorization we get:
⇒ x2 + 44x - 33x - 1452 = 0
⇒ x(x + 44) - 33(x + 44) = 0
⇒ (x - 33)(x + 44) = 0
⇒ x = 33, - 44
Since speed cannot be negative so
Speed of Passenger train = 33 kmph
Speed of Express Train = (33 + 11) = 44 kmph
Question 18.
Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of the two squares.
Answer:
Let the side of one square be x, another square be (x + 6)
Sum of Areas = 468 m2
x2 + (x + 6)2 = 468
2x2 + 12x - 432 = 0
⇒ x2 + 6x - 216 = 0
Performing factorization we get:
⇒ x2 + 18x - 12x - 216 = 0
⇒ x(x + 18) - 12(x + 18) = 0
⇒ (x - 12)(x + 18) = 0
⇒ x = 12, - 18
Neglecting the negative value , we get
Side of squares = 12 m ,18 m
Question 19.
A ball is thrown vertically upward from the top of a building 96m tall with an initial velocity 80m/second. The distances of the ball from the ground after t seconds is s = 96 + 80t - 4.9t2. After how may seconds does the ball strike the ground.
Answer:
s = 96 + 80t - 4.9t2
When the ball reaches the ground, s = 0
4.9t2 - 80t - 96 = 0
Since time cannot be negative so negative root is neglected
⇒ Time taken = 17.44 s
Question 20.
If a polygon of ‘n’ sides has
diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Answer:
No. of diagonals =
n(n - 3) = 2 × 65
⇒ n2 - 3n = 130
⇒ n2 - 3n - 130 = 0
Performing factorization we get:
⇒ n2 - 13n + 10n - 130 = 0
⇒ n(n - 13) + 10(n - 13) = 0
⇒ (n + 10)(n - 13) = 0
n = 13, - 10
Since no. of sides cannot be negative so
No. of Sides = 13
When No. of Diagonals is 50
n(n - 3) = 50 × 2
⇒ n2 - 3n - 150 = 0
Discriminant = (9 - 4 × 1 × ( - 150)) = 609
Since 609 is not a perfect square so n can never be a whole number.
Hence 50 diagonals are not possible
Exercise 5.4
Find the nature of the roots of the following quadratic equation. If real roots exist, find them:
2x2 - 3x + 5 = 0
Answer:
2x2 - 3x + 5 = 0
Discriminant = ( - 3)2 - 4 × 2 × 5 = - 31
Since The Discriminant is negative so the roots are imaginary
Question 2.
Find the nature of the roots of the following quadratic equation. If real roots exist, find them:
3x2 - 4√3 x + 4 = 0
Answer:
3x2 - 4√3 x + 4 = 0
Discriminant = ( - 4√3)2 - 4 × 3 × 4 = 0
Since The Discriminant is zero so the roots are equal
The roots are
Question 3.
Find the nature of the roots of the following quadratic equation. If real roots exist, find them:
2x2 - 6x + 3 = 0
Answer:
2x2 - 6x + 3 = 0
Discriminant(D) = ( - 6)2 - 4 × 2 × 3 = 12
Since The Discriminant is positive so the roots are real and distinct
The roots are evaluated by using the formula
The roots are
Question 4.
Find the value of k for each of the following quadratic equation, so that they have two equal roots.
2x2 + kx + 3 = 0
Answer:
2x2 + kx + 3 = 0
If the quadratic Equation has equal root then their Discriminant = 0
k2 - 4 × 2 × 3 = 0
⇒ k2 = 24
⇒ k = ±2√6
Question 5.
Find the value of k for each of the following quadratic equation, so that they have two equal roots.
kx(x - 2) + 6 = 0
Answer:
kx(x - 2) + 6 = 0
⇒ kx2 - 2kx + 6 = 0
If the quadratic Equation has equal root then their Discriminant = 0
4k2 - 4 × k × 6 = 0
⇒ k2 - 6k = 0
⇒ k(k - 6) = 0
k = 0,6
If k = 0 then the equation is no longer Quadratic
Answer : k = 6
Question 6.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800m2? If so, find its length and breadth.
Answer:
Let Breadth be x and length be 2x
Area = (Length × Breadth) = 800m2
2x2 = 800
⇒ x2 = 400
Discriminant = 4 × 1 × 400 = 1600
Hence it is possible
⇒ x = ±20
Neglecting the negative term we have,
Breadth = 20m
Length = 40m
Question 7.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present ages.
Answer:
Let age of one friend be x ,other 20 - x
Four years ago there age was (x - 4) ,(16 - x)
Product of age four years ago = 48
So we can say,
(x - 4)(16 - x) = 48
⇒ x2 - 20x + 112 = 0
Discriminant = (- 20)2 - 4 × 1 × 112 = - 48
Since the discriminant is negative so the roots of the equation are imaginary.
Answer: The above situation is not possible
Question 8.
Is it possible to design a rectangular park of perimeter 80m. and area 400m2? If so, find its length and breadth.
Answer:
Let Length of rectangular park be x and breadth be (40 - x)
Area = 400 m2
⇒ x(40 - x) = 400
⇒ x2 - 40x + 400 = 0
Discriminant = ( - 40)2 - 4 × 1 × 400
⇒ Discriminant = 1600 - 1600 = 0
Since Discriminant is 0 so it is possible
x2 - 40x + 400 = 0
⇒ (x - 20)2 = 0
⇒ x = 20 m
Answer: Length = 20m, Breadth = 20m
