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AP Board Class 10 Maths Chapter 5 Quadratic Equations Ex 5.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Book Answers

AP Board Class 10 Maths Chapter 5 Quadratic Equations Ex 5.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Book Answers
AP Board Class 10 Maths Chapter 5 Quadratic Equations Ex 5.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Book Answers


AP Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks Solutions and answers for students are now available in pdf format. Andhra Pradesh Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Book answers and solutions are one of the most important study materials for any student. The Andhra Pradesh State Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 books are published by the Andhra Pradesh Board Publishers. These Andhra Pradesh Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 textbooks are prepared by a group of expert faculty members. Students can download these AP Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 book solutions pdf online from this page.

Andhra Pradesh Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks Solutions PDF

Andhra Pradesh State Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Books Solutions with Answers are prepared and published by the Andhra Pradesh Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of AP Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Books Answers Solutions, then you are in the right place. Here is a complete hub of Andhra Pradesh State Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Andhra Pradesh Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks. These Andhra Pradesh State Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Andhra Pradesh Board.

Andhra Pradesh State Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Books Solutions

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Class 10th
Subject Maths
Chapters Maths Chapter 5 Quadratic Equations Ex 5.3
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AP Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

10th Class Maths 5th Lesson Quadratic Equations Ex 5.3 Textbook Questions and Answers

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
i) 2x2 + x – 4 = 0
Answer:
Given: 2x2 + x – 4 = 0
⇒ 2x2 + x = 4
⇒ (√2x)2 + x = 4
⇒ (√2x)2 + 2.√2.x.122√ = 4
Now LHS is in the form a2 + 2ab
where b = 122√
Adding b2 = (122√)2 on both sides we get

ii) 4x2 + 4√3x + 3 = 0
Answer:
Given: 4x2 + 4√3x + 3 = 0
⇒ 4x2 + 4√3x = -3
⇒ (2x)2 + 2(2x)√3 = -3
LHS is of the form a2 + 2ab where
where b = √3.
∴ Adding b2 = (√3)2 = 3 on both sides, we get
(2x)2 + 2(2x)(√3) + (√3)2 = -3 + (√3)2
⇒ (2x + √3)2 = -3 + 3 = 0
∴ (2x + √3)2 = 0
⇒ 2x + √3 = 0
⇒ 2x = -√3
⇒ x = −3√2
∴ The roots are −3√2, −3√2.

iii) 5x2 – 7x – 6 = 0
Given quardratic equation = 5x2 – 7x – 6 = 0
∴ 5x2 – 7x – 6
⇒ x2 – 75x = 65, it can be re-written as
x2 – 2.710x = 65 now it is in the form
of a2 – 2ab where a = x, and b = 710
Now adding b2 = (710)2 on both sides, we get

Note: If we take the Q.E. as 5x2 – 7x + 6 = 0, then we get the T.B. answer.

iv) x2 + 5 = -6x
Answer:
The given Q.E. is x2 + 5 = -6x
⇒ x2 + 6x = -5
⇒ (x)2 + 2.(x).3 = -5
Now L.H.S. is of the form a2 + 2ab where b = 3.
Adding b2 = 32 on both sides we get
x2 + 2(x)(3) + 32 = -5 + 32
(x + 3)2 = -5 + 9 = 4
∴ x + 3 = 74 = ± 2
⇒ x = +2 – 3 or – 2 – 3
= -1 or -5 are the roots of the given Q.E.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula,
i) 2x2 + x – 4 = 0
Answer:
Comparing this Q.E. with ax2 + bx + c = 0
a = 2; b = 1; c = -4
x = −𝑏±𝑏2−4𝑎𝑐√2𝑎

ii) 4x2 + 4√3x + 3 = 0
Answer:
Given: 4x2 + 4√3x + 3 = 0
Here a = 4; b = 4√3 ; c = 3

iii) 5x2 – 7x – 6 = 0
Answer:
Given: 5x2 – 7x – 6 = 0
Here a = 5; b = -7 and c = -6

iv) x2 + 5 = -6x
Answer:
Given: x2 + 5 = -6x
⇒ x2 + 6x + 5 = 0
Here a = 1; b = 6; c = 5

Question 3.
Find the roots of the following equations:
i) x – 1𝑥 = 3, x ≠ 0
Answer:
Given: x – 1𝑥 = 3
⇒ x2 + 6x + 5 = 0
⇒ 𝑥2−1𝑥 = 3
⇒ x2 – 1 = 3x
⇒ x2 – 3x – 1 = 0
Here a = 1; b = -3; c = -1

ii) 1𝑥+4 – 1𝑥−7 = 1130, x ≠ -4, 7
Answer:
Given: 1𝑥+4 – 1𝑥−7 = 1130

⇒ x2 – 3x – 28 = -30
⇒ x2 – 3x – 28 + 30 = 0
⇒ x2 – 3x + 2 = 0
⇒ x2 – 2x – x + 2 = 0
⇒ x(x – 2) – 1(x – 2) = 0
⇒ (x – 2) (x – 1) = 0
⇒ x – 2 = 0 (or) x – 1 = 0
⇒ x = 2 or x = 1
⇒ x = 2 or 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 13. Find his present age.
Answer:
Let the present age of Rehman be x years.
3 years ago Rehman’s age = x – 3 and its reciprocal is 1𝑥−3
Rehman’s age 5 years from now = x + 5 and its reciprocal is 1𝑥+5
The sum of the reciprocals


⇒ x2 + 2x – 15 = 3(2x + 2)
⇒ x2 + 2x – 15 = 6x + 6
⇒ x2 + 2x – 15 – 6x – 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 x can’t be negative, x = 7
i.e., Present age of Rehman = 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.
Answer:
Sum of the marks in Mathematics and English = 30
Let Moulika’s marks in Mathematics be x Then her marks in English = 30 – x
If she got 2 more marks in Mathematics, then her marks would be x + 2.
If she got 3 marks less in English then her marks would be 30 – x – 3 = 27 – x
By problem (x + 2) (27 – x) = 210
⇒ x(27 – x) + 2(27 – x) = 210
⇒ 27x – x2 + 54 – 2x = 210
⇒ -x2 + 25x + 54 = 210
⇒ x2 – 25x – 54 + 210 = 0
⇒ x2 – 25x + 156 = 0
⇒ x2 – 12x – 13x + 156 = 0
⇒ x(x – 12) – 13(x 12) = 0
⇒ (x – 12) (x – 13) = 0
⇒ x – 12 = 0 or x – 13 = 0
⇒ x = 12 or x = 13
If x = 12, then marks in Mathematics = 12 English = 30 – 12 = 18
If x = 13, then marks in Mathematics = 13 English = 30 – 13 = 17

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.
Answer:
Let the shorter side of the rectangular field = x m.
Then its longer side = x + 30 m.

The diagonal of a rectangle is also the hypotenuse of the lower triangle Here the diagonal = x + 60
∴ By Pythagoras Theorem
(side)2 + (side)2 = (hypotenuse)2
⇒ (x + 30)2 + x2 = (x + 60)2
⇒ x2 + 60x + 900 + x2 = x2 + 120x + 3600
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30 (x – 90) = 0
⇒ (x – 90) (x + 30) = 0
⇒ x – 90 = 0 (or) x + 30 = 0
⇒ x = +90 (or) x = -30 But ‘x’ can’t be negative.
∴ x = 90 m
i.e., the shorter side x = 90 m Longer side x + 30 = 90 + 30 = 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.
Answer:
Let the large number be x.
8 times larger number = Square of the srnall number = 8x
Square of the larger number = x2
By problem, x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
⇒ x2– 18x + 10x – 180 = 0
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x + 10)(x – 18) = 0
⇒ x + 10 = 0 (or) x – 18 = 0
⇒ x = -10 (or) x = 18
If x = 18, then larger number =18;
(small number)2 = 8 × (+18) = 144
∴ Small number = √144 = 12
The numbers are 18, 12
Note: Discard x = -10.

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.
Answer:
The distance travelled = 360 km.
Let the speed of the train = x kmph.
Time taken to complete a journey =  distance  speed 

⇒ x2 + 5x = 1800
⇒ x2 + 5x – 1800 = 0
⇒ 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 x can’t be negative.
∴ The speed of the train = 40 kmph.

Question 9.
Two water taps together can fill a tank in 938 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 to fill the tank by smaller tap = x (hours)
So the part filled by smaller tap in
1 hour = 1𝑥 × 758 = 758𝑥 ……. (1)
Again then the time taken to fill the tank by larger tap = (x – 10) hours
∴ the part of tank that can be filled by larger tap alone in one hour of time = 1𝑥−10
∴ In 758 hours the part filled by larger tap = 758(1𝑥−10)
∴ By both taps together

⇒ 150x – 750 = 8x2 – 80x
⇒ 8x2 – 80x – 150x + 750 = 0
⇒ 8x2 – 230x + 750 = 0
⇒ 4x2 – 115x + 375 = 0
⇒ 4x2 – 100x – 15x + 375 = 0
⇒ 4x(x – 25) – 15(x – 25) = 0
∴ (4x – 15) (x – 25) = 0 15
⇒ 4x = 15, x = 154 or x = 25
x = 25 hours.
then time taken to fill by larger tap = x – 10 = 25 – 10 = 15 hours
(x cannot be 154 since we have considered ‘x’ as time taken by smaller tap, which is to be higher one)

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 at 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.
Answer:
Let the speed of the passenger train = x kmph.
Then speed of the express train = x + 11 kmph.
Distance travelled = 132 km
We know that time =  distance  speed 

⇒ x2 + 11x = 13 × 11
⇒ x2 + 11x – 1452 = 0
⇒ x2 + 44x – 33x – 1452 = 0
⇒ x(x + 44) – 33 (x + 44) = 0
⇒ (x + 44) (x – 33) = 0
⇒ x + 44 = 0 (or) x – 33 = 0
⇒ x = -44 (or) x = 33
But x can’t be negative.
∴ Speed of the passenger train = x = 33 kmph.
Speed of the express train = x + 11 = 44 kmph.

Question 11.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24m, find the sides of the two squares.
(OR)
If the sum of the areas of two squares is 468 m2 and the difference of their perimeters is 24m, then find the measurements of their sides.
Answer:
Let the side of first square = x m say Then perimeter of the first square = 4x [∵ P = 4 . side]
By problem, perimeter of the second square = 4x + 24 (or) 4x – 24
∴ Side of the second square =

Now sum of the areas of the two squares is given as 468 m2
x2 + (x + 6)2 = 468
⇒ x2 + x2 + 12x + 36 = 468
⇒ 2x2 + 12x + 36 – 468 = 0
⇒ 2x2 + 12x – 432 = 0
⇒ x2 + 6x – 216 = 0
⇒ x2 + 18x – 12x – 216 = 0
⇒ x(x + 18)- 12(x + 18) = 0
⇒ (x + 18) (x – 12) = 0
⇒ x + 18 = 0 (or) x – 12 = 0
⇒ x = -18 (or) 12
But x can’t be negative.
∴ x = 12
i.e., side of the first square = 12
∴ Perimeter = 4 × 12 = 48
∴ Perimeter of the second square = 48 + 24 = 72
∴ Side of the second square = 724 = 18 m.
(or)
x2 + (x – 6)2 = 468
⇒ x2 + x2 – 12x + 36 = 468
⇒ 2x2 – 12x – 432 – 0
⇒ x2 – 6x – 216 = 0
⇒ x2 – 18x + 12x – 216 = 0
⇒ x(x-18) + 12(x-18) = 0
⇒ (x – 18) (x + 12) = 0
⇒ x – 18 = 0 (or) x + 12 = 0
⇒ x = 18 (or) – 12
But x can’t be negative.
∴ x = 18
i.e., side of the first square = 18 m
∴ Perimeter = 4 × 18 = 72
Perimeter of the second square = 72 – 24 = 48
∴ Side of the second square = 484 = 12 m.
i.e., In any way, the sides of the squares are 12m, 18m.

Question 12.
If a polygon of ‘n’ sides has 12n(n – 3) diagonals. How many sides will a polygon having 65 diagonals? Is there a polygon with 50 diagonals?
Answer:
Given: Number of diagonals of a polygon with n-sides = 𝑛(𝑛−3)2
No. of diagonals of a given polygon = 65
i.e., 𝑛(𝑛−3)2 = 65
where n is number of sides of the polygon
⇒ n2 – 3n = 2 × 65
⇒ n2 – 3n – 130 = 0
⇒ n2 – 13n + 10n – 130 = 0
⇒ n(n – 13) + 10(n – 13) = 0
⇒ (n – 13) (n + 10) = O
⇒ n – 13 = 0 (or) n + 10 = 0
⇒ n = 13 (or) n = -10
But n can’t be negative.
∴ n = 13 (i.e.) number of sides = 13.
Also to check 50 as the number of diagonals of a polygon
∴ 𝑛(𝑛−3)2 = 50
⇒ n2 – 3n = 100
⇒ n2 – 3n – 100 = 0
There is no real value of n for which the above equation is satisfied.
∴ There can’t be a polygon with 50 diagonals.


AP Board Textbook Solutions PDF for Class 10th Maths


Andhra Pradesh Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 exam preparation. The AP Board STD 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 5 Quadratic Equations Ex 5.3 Books State Board syllabus with maximum efficiency.

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