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Andhra Pradesh SSC Class 10 Solutions For Maths – Quadratic Equations (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 5 Quadratic Equations
Exercise 5.1:
Question 1:
Check whether the following are quadratic equations:
AP SSC 10th Class Textbook Solutions
Solution :
- LHS=(x+1)2=x2+2x+1
Therefore, (x+1)2= 2(x-3) can be written as
x2+2x+1=2x-6
i.e., x2+7=0
It is in the form of ax2+bx+c=0
Therefore, the given equation is quadratic equation. - RHS= (-2)(3-x)= -6+2x
Therefore, x2-2x= (-2)(3-x)
x2-2x= -6+2x
x2-4x+6=0
It is in the form of ax2+bx+c=0
Therefore, the given equation is quadratic equation. - Here, LHS=(x-2)(x+1)=x2+x-2x-2=x2-x-2 RHS=(x-1)(x+3)=x2+3x-x-3=x2+2x-3
So, x2-x-2= x2+2x-3
i.e., -3x+1=0
It is not in the form of ax2+bx+c=0
Therefore, the given equation is not a quadratic equation. - LHS=(x-3)(2x+1)=2x2+x-6x-3=2x2-5x-3
RHS=x(x+5)=x2+5x
So, 2x2-5x-3= x2+5x
i.e., x2-10x-3=0
It is in the form of ax2+bx+c=0
Therefore, the given equation is quadratic equation. - LHS=(2x-1)(x-3)=2x2-6x-x+3=2x2-7x+3
RHS=(x+5)(x-1)=x2-x+5x-5= x2+4x-5
So, 2x2-7x+3= x2+4x-5
x2-11x+8=0
It is in the form of ax2+bx+c=0
Therefore, the given equation is quadratic equation. - LHS=x2+3x+1
RHS=(x-2)2=x2-4x+4
So, x2+3x+1=x2-4x+4
7x-3=0
It is not in the form of ax2+bx+c=0
Therefore, the given equation is not a quadratic equation. - LHS=(x+2)3=x3+3x2+12x+8
RHS=2x(x2-1)=2x3-2x
So, x3+3x2+12x+8=2x3-2x
-x3+3x2+14x+8=0
It is not in the form of ax2+bx+c=0
Therefore, the given equation is not a quadratic equation. - RHS=(x-2)3=x3-3x2+12x-8
Therefore, x3-4x2-x+1=x3-3x2+12x-8
-x2-13x+9=0
It is in the form of ax2+bx+c=0
Therefore, the given equation is quadratic equation.
Question 2:
Represent the following situations in the form of quadratic equations:
- The area of a rectangular plot is 528 m2. The length of the plot is one metre more than twice its bread We need to find the length and breadth of the plot.
- The product of two consecutive positive integers is 306. We need to find the integers.
- 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.
- 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 :
- Let length=l and breadth=b
Area of rectangle= length x breadth
Given, The length of the plot (in meters) is one more than twice its breadth.
So, l=2b+1
Area=(2b+1) (b)
528=2b2+b
2b2+b-528=0
The quadratic equation is 2b2+b-528=0 - Assume the two consecutive integers are x and x+1.
given, the product of two consecutive integer=306
x(x+1)=306
x2+x=306
x2+ x – 306 = 0
The required quadratic equation is x2+ x – 306=0 - Let age of Rohan = x
Then, Rohan’s mother’s age = x+26
After 3 year,
Age of Rohan = (x+3)years
Rohan’s mother age = (x+29)years
Product of their ages = 360years
(x+3)(x+29) = 360
x2 + 29x+3x + 87 = 360
x2 + 32x – 273 = 0
The required equation is x2 + 32x – 273 = 0 - Let the speed of the train be x.
Time taken to cover 480km =
If the speed had been 8km/h less then the train would take 3 hours to cover the same distance.
Exercise 5.2:
Question 1:
Find the roots of the following quadratic equations by factorisation:
Question 1(i):
x2 – 3x – 10 = 0
Solution :
We have x2 – 3x – 10 = 0
x2 – 3x – 10 = x2 – 5x + 2x – 10
= x(x + 2) -5(x + 2)
= (x + 2)(x – 5)
The roots of x2 – 3x – 10 = 0 are the values of x for which (x + 2)(x – 5) = 0
Therefore, x + 2 = 0 or x – 5 = 0
i.e., x = -2 or x = 5
Therefore, the roots of x2 – 3 – 10 = 0 are -2 and 5.
Question 1(ii):
2x2 + x – 6 = 0
Solution :
We have 2x2 + x – 6 = 0
2x2 + x – 6 = 2x2 + 4x – 3x – 6
= 2x(x + 2) -3(x + 2)
= (x + 2)(2x – 3)
The roots of 2x2 + x – 6 = 0 are the values of x for which (x + 2)(2x – 3) = 0
Therefore, x + 2 = 0 or 2x – 3 = 0
Question 1(iii):
\(\sqrt { 2 } { x }^{ 2 }7x+5\sqrt { 2 } =0\)
Solution :
Question 1(iv):
\(2{ x }^{ 2 }-x+\frac { 1 }{ 8 } =0\)
Solution :
Question 1(v):
100x2 – 20x + 1 = 0
Solution :
We have, 100x2 – 20x + 1 = 0
100x2 – 20x + 1 = 100x2 – 10x – 10x + 1
= 10x(10x – 1) – 1(10x – 1)
= (10x – 1)(10x – 1)
The roots of 100x2 – 20x + 1 = 0 are the values of x for which (10x – 1)(10x – 1) = 0.
Therefore, 10x – 1 = 0
Question 1(vi):
x(x + 4) = 12
Solution :
We have, x(x + 4) = 12
x2 + 4x – 12 = x2 + 6x – 2x – 12
= x(x + 6) – 2(x + 6)
= (x + 6)(x – 2)
The roots of x2 + 4x – 12 = 0 are the values of x for which (x + 6)(x – 2) = 0.
Therefore, x + 6 = 0 or x – 2 = 0
i.e., x = -6 or x = 2.
Therefore, the roots of x2 + 4x – 12 = 0 are -6 and 2.
Question 1(vii):
3x2 – 5x + 2 = 0
Solution :
We have, 3x2 – 5x + 2 = 0
3x2 – 5x + 2 = 3x2 – 2x – 3x + 2
= x(3x – 2) – 1(3x – 2)
= (3x – 2)(x – 1)
The roots of 3x2 – 5x + 2 = 0 are the values of x for which (3x – 2)(x – 1) = 0
Therefore, 3x-2=0 or x-1=0
Question 1(viii):
\(x-\frac { 3 }{ x } =2\)
Solution :
x2 – 3 = 2x ⇒ x2 – 2x – 3 = 0 = x2 – 3x + x – 3
= x(x – 3) + 1(x – 3)
= (x – 3)(x + 1)
The roots of x2 – 2x – 3 = 0 are the values of x for which (x – 3)(x + 1) = 0
Therefore, x – 3 = 0 or x + 1 = 0
i.e., x = 3 or x = -1
Therefore, the roots of x2 – 2x – 3 = 0 are 3 and -1.
Question 1(ix):
3(x – 4)2 – 5(x – 4) = 12
Solution :
We have, 3(x – 4)2 – 5(x – 4) = 12
3(x2 – 8x + 16)- 5x + 20 = 12
⇒3x2 – 24x + 48 – 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 3x2 – 29x + 56 = 0 are the values of x for which (3x – 8) (x – 7) = 0.
Therefore, 3x – 8 = 0 or x – 7 = 0
Question 2:
Find two numbers whose sum is 27 and product is 182.
Solution :
Let the two numbers be x and y.
Given,
x + y = 27 (1)
and
x × y = 182 (2)
From equation (1)
y = 27 – x
substitute value of y in equation (2)
x – 13 = 0 or x – 14 = 0
x = 13 or 14.
The two numbers are 13 and 14.
Question 3:
Find two consecutive positive integers, sum of whose squares is 613.
Solution :
Let two positive integer be ‘x’ and second integer be x + 1.
According to question,
So, x2 + (x + 1)2 = 613
⇒ x2 + x2 + 2x = 1 = 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
Therefore, the roots of x2 + (x + 1)2 = 613 are -18 and 17.
Thus the integers are 17 and 18 or -17 and -18.
Question 4:
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 Base = x
Altitude = (base – 7) = x – 7
Hypotenuse2 = altitude2 + base2
132 = (x – 7)2 + x2
169 = x2 – 14x + 49 + x2
169 = 2x2 – 14x + 49
2x2 – 14x + 49 – 169 = 0
2x2 – 14x – 120 = 0
x2 – 7x – 60 = 0 (dividing by 2)
x2 – 12x + 5x – 60 = 0
x(x – 12) + 5(x – 12) = 0
(x – 12)(x + 5) = 0
So, x = 12 or x = -5
Base cannot be negative and hence base = 12 cm.
Since base = 12 cm, we have,
altitude = base – 7 cm = 5 cm.
The altitude = 5 cm and base = 12 cm.
Question 5:
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 article.
Solution :
Question 6:
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters.
Solution :
Let length and breadth of the rectangle be l and b respectively. Then,
Area of rectangle = l x b = 40 (1)
and
Perimeter = 2(l + b) = 28 (2)
2(l + b) = 28
(l + b) = 14
b = 14 – l
Substituting the value of b in equation(1)
l x(14- l)=40
Hence, length = 10 m and breadth = 4 m.
Question 7:
The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm then find its base and altitude.
Solution :
Let the base be x and the altitude be y.
Given,
Base = 4 + altitude
= 4 + y
Hence, base = 12 cm and altitude = 8 cm.
Question 8:
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.
Solution :
Question 9:
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?
Solution :
Let number of girls be x and boys be y
Total number of student,
60 = x + y (1)
Total money collected,
1600 = xy + yx
⇒1600 = 2xy
⇒xy = 800 (2)
From (1)equation,
y = 60 – x
Substituting the value of y in equation (2)
800 = x(60 – x)
⇒800 = 60x – x2
⇒x2 – 60x + 800 = 0
⇒x2 – 20x – 40x + 800 = 0
⇒x(x – 20) – 40(x – 20) = 0
⇒ (x – 20)(x – 40) = 0
⇒x – 20 = 0 or x – 40 = 0
⇒x = 20 or x = 40
If the number of girls is 20, then the number of boys is 40.
And if the number of girls is 40, then the number of boys is 20.
Question 10:
A motor boat heads upstream a distance of 24km 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?
Solution :
Let the speed of the stream be x km/hr.
Therefore, the speed of the boat upstream = (x – 3) km/hr and the speed of boat downstream = (x + 3) km/hr
Since x is speed of the stream, it cannot be negative. So, we ignore the value x = -1.
Therefore, the speed of the stream is 9 km/hr.
Exercise 5.3:
Question 1:
Find the roots of the following quadratic equations, if they exist.
Question 1(i):
2x2 + x – 4 = 0
Solution :
Question 1(ii):
\(4{ x }^{ 2 }+4\sqrt { 3 } x+3=0\)
Solution :
Question 1(iii):
5x2 – 7x – 6 = 0
Solution :
Question 1(iv):
x2 + 5 = -6x
Solution :
x2 + 5 = -6x
x2 + 6x = -5
x2 + 6x + 32 = -5 + 32 (Adding in 32 on both sides)
(x + 3)2 = -5 + 9
(x + 3)2 = 4
(x + 3) = ±2
x = -3 + 2 or x = -3 – 2
x = -1 or x = -5.
Question 2:
Find the roots of the quadratic equations given in Q. 1 above by applying the quadratic formula.
Solution :
Question 3:
Find the roots of the following equations:
Solution :
Multiplying throughout by x, we get x2 – 1 = 3x
i.e., x2 – 3x – 1 = 0, which is a quadratic equation.
Here, a = 1, b = -3, c = -1
So, b2 – 4ac = 9 + 4 = 13 > 0
As x ≠ -4, 7,
Multiplying the equation by (x + 4)(x – 7), we get
30((x – 7) – (x + 4)) = 11(x + 4)(x – 7)
-30 = (x2 – 7x + 4x – 28)
-30 = (x2 – 3x – 28)
-30 = x2 – 3x – 28
x2 – 3x – 28 + 30 = 0
x2 – 3x + 2 = 0
(x – 1)(x – 2) = 0
x = 1 and 2 are roots of the given equation
Question 4:
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now 1/3. Find his present age.
Solution :
x=-3 and x=7
Since age cannot be negative,
Therefore discarding x = -3,we get,
x = 7
Hence, Rehman’s present age is 7 years.
Question 5:
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 would have been 210. Find her marks in the two subjects.
Solution :
Let the marks in Mathematics be x and in English be y
As given in question,
x + y = 30, ….(1)
If she got 2 marks more in mathematics and 3 marks less in English
Then, x + 2 and y – 3 and, product of her marks,
x y = 210.
(x + 2)(y – 3) = 210 ….(2)
From equation (1)
x = 30 – y.
Substituting the value of x in equation (2)
(y – 3)(32 – y) = 210
32y – y2 – 96 + 3y= 210
-y2 + 35y – 96 = 210
-y2 + 35y – 306 = 0
y2 – 35y+ 306 = 0,
This is a quadratic equation in y.
i.e., y = 18 or 17
Hence, x = 12 or 13
Thus, Moulika got 12 in Mathematics and 18 in English or She got 13 in Mathematics and 17 in English.
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 rectangle be ‘x’, longer side be ‘y’ and diagonal be ‘z’.
According to given condition,
z = x + 60——–(1)
y = x + 30——–(2)
Since the angles of a rectangle are right angles,
Therefore, applying Pythagoras theorem
z2 = x2 + y2 (Since one part of a rectangle is the triangle with diagonal as the hypotenuse)
Substituting (1) and (2) in the above equations (x + 60)2 = x2 + (x + 30)2
x2 + 2(x)(60) + (60)2 = x2 + x2 + 2(x)(30) + (30)2
x2 + 120x + 3600 = 2x2 + 60x + 900
x2 – 60x – 2700 = 0
(x – 90)(x + 30) = 0
x = 90 or -30
x cannot be negative, hence, x = 90 m and y = 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.
Solution :
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.
Solution :
i.e., x = 45 or -40
But speed can’t be negative so the speed is 45. Hence the speed of the train = 45 – 5 = 40 km/hr.
Question 9:
Two water taps together can fill a tank in \(9\frac { 3 }{ 8 }\) 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.
Solution :
Question 10:
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangaluru (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 :
Let speed = x and time = y
For the express train,
Speed= x + 11 and time = y – 1
For passenger train, xy = 132
i.e., x = 44 or x = -33
But speed cannot be negative, so the speed is 44.
The speed of the express train is 44 km/hr
The speed of the passenger train is 33 km/hr.
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 :
Let the side of the 1st square (sq1) be ‘x’ and side of the 2nd square (sq2) be ‘y’.
Since, Perimeter(square) = 4 × (side)
Therefore,
P(sq1) = 4 × x and P(sq2) = 4 × y
From the above condition,
P(sq1) – P(sq2) = 24 (Assuming ‘x’ is greater than ‘y’)
(4x) – (4y) = 24
4(x – y) = 24
(x – y) = 24/4
x – y = 6 ————-(1)
Also, Area(square) = (side)2
Therefore,
A(sq1) = x2and A(sq2) = y2
From the above condition,
x2 + y2 = 468 ——-(2)
But, x2 + y2 = (x – y)2 + 2xy——(3)
468 = (6)2 + 2(xy)
468 = 36 + 2(xy)
432 = 2(xy)
2(xy) = 432
xy = 432/2
xy = 216—————(4)
x = 216/y————–(5)
Substituting (5) in (1)
(216/y) – y = 6
(216 – y2)/y = 6
216 – y2 = 6y
y2 + 6y – 216 = 0
Therefore,
y = 12 or y = -18
But, since ‘y’ is the length of the square it cannot be
negative.
Hence, discarding y = -18
Therefore, y = 12——–(6)
Substituting (6) in (5),we get,
x = 216/12
x = 18
Side of square1 = 18 m
Side of square2 = 12 m
Question 12:
A ball is thrown vertically upward from the top of a building 96 m tall with an initial velocity 80 m/second. The distance ‘s’ 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.
Solution :
Given,
16t2 – 80t – 96 = 0
t2 – 5t – 6 = 0
(t – 6)(t + 1) = 0
Hence, t = 6 or t = -1
But, time cannot be negative so, t= 6seconds.
Question 13:
If a polygon of ‘n’ sides has 1/2 n (n-3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Solution :
But the side should be an integer, so a polygon cannot have 50 diagonals.
Exercise 5.4:
Question 1:
Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
- \(2{ x }^{ 2 }-3x+5=0\)
- \(3{ x }^{ 2 }-4\sqrt { 3 } x+4=0\)
- \(2{ x }^{ 2 }-6x+3=0\)
Solution :
- The given equation is in the form of ax2+bx+c=0, where a=2, b=-3, c=5.
Therefore, the discriminate
b2-4ac=9-40=-31 < 0
So, the given equation has no real roots. - The given equation is the form of ax2 + bx + c = 0, where
- The given equation is in the form of ax2 + bx + c = 0, where
a = 2, b = -6, c = 3.
Therefore, the discriminate
b2 – 4ac = 36 – 24 = 12
.
Question 2:
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
- 2x2 + kx + 3 = 0
- kx (x – 2) + 6 = 00
Solution :
- The given equation is in the form of ax2 + bx + c = 0, where
a = 2, b = k, c = 3.
Given they are equal roots. So,
b2 – 4ac = 0
K2 – 24 = 0
k=±√24=±2√6 - Kx(x-2)+6=0
Kx2-2kx+6=0
The given equation is in the form of ax2+bx+c=0, where
a=k, b= -2k, c=6.
Given they are equal roots. So,
b2-4ac=0
4k2-24k=0
4K(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 m ? If so, find its length and breadth.
Solution :
Yes it is possible, let length=x and breadth =y
x=2y
Area=length × breadth
800 =2y×(y)
800=2y2
y2=400
y=20
So, x=40.
Ans: length=40m and breadth=20m.
Question 4:
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.
Solution :
No. The 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 :
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 5 Quadratic Equations are helpful to complete your math homework.
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