BSEB Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 Textbook Solutions PDF: Download Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Book Answers |
Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Textbooks Solutions PDF
Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Textbooks. These Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.Bihar Board Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Books Solutions
Board | BSEB |
Materials | Textbook Solutions/Guide |
Format | DOC/PDF |
Class | 10th |
Subject | Maths Chapter 4 Quadratic Equations Ex 4.4 |
Chapters | All |
Provider | Hsslive |
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BSEB Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Textbooks Solutions with Answer PDF Download
Find below the list of all BSEB Class 10th Maths Chapter 4 Quadratic Equations Ex 4.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:BSEB Bihar Board Class 10th Maths Solutions Chapter 4 Quadratic Equations 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 – 43‾√x + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution:
(i) The given equation is 2x2 – 3x + 5 = 0.
Here, a = 2, b = – 3 and c = 5.
∴ D = b2 – 4ac = (- 3)2 – 4 × 2 × 5 = 9 – 40 = – 31 < 0
So, the given equation has no real roots.
(ii) The given’equation is 3x2 – 43‾√x + 4 = 0
Here, a = 3, b = – 43‾√ and c = 4.
∴ D = b2 – 4ac = (- 43‾√)2 – 4 × 2 × 5 = 9 – 40 = – 31 < 0
So, the given equation has real equal roots, given by
(iii) The given equation is 2x2 – 6x + 3 = 0.
Here, a = 2, b = – 6 and c = 3
∴ D = b2 – 4ac = (- 6)2 – 4 × 2 × 3 = 36 – 24 = 12 > 0
So, the given equation has real roots, given by
∴ The roots of the given equation are 3+3√2 and 3−3√2.
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) The given equation is 2x2 + kx + 3 = 0.
Here, a = 2, b = k and c = 3
∴ D = b2 – 4ac = k2 – 4 × 2 × 3 = k2 – 24
The given equation will have real and equal roots, if
D = 0 i.e; k2 – 2kx + 6 = 0
Here, a = k, b = – 2k and c = 6
∴ D = b2 – 4ac = (- 2k)2 – 4 × k × 6 = 4k2 – 24k
The given equation will have real and equal roots, if D = 0.
So, 4k2 – 24k = 0 or 4k(k – 6) = 0
i.e., 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 2x be the length and x be the breadth of a rectangular mango grove.
Area = (2x)(x) = 800. [Given]
So, x2 = 400 or x = 20 [Since side cannot be negative]
The value of x is real so to design of grove is possible
Its length = 40 m and breadth = 20 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 of the friends = x years. Then, age of the other friend = 20 – x.
Four years ago,
Age of one of the friends = (x – 4) years
and age of the other friend = (20 – x – 4) years
= (16 – x) years
According to condition:
(x – 4)(16 -x) = 48
or 16x – x2 – 64 + 4x = 48
or x2 – 20x + 112 = 0
Here, a = 1, b = – 20 and c = 112.
∴ D = b2 – 4ac = (- 20)2 – 4 × 1 × 22
= 400 – 448 = – 48 < 0
So, the given equation has no real roots.
Thus, the 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 be x metres and breadth be y metres.
∴ Perimeter = 80 m
or 2(x + y) = 80 or x + y = 40 ……………….. (1)
Also, Area = 400 m2
So, xy = 400
or x(40 – x) = 400 [Using (1)]
or 40x – x2 = 400 or x2 – 40x + 400 = 0
Here, a = 1, b = – 40 and c = 400.
∴ D = b2 – 4ac = (- 40)2 – 4 × 1 × 400
= 1600 – 1600 = 0
So, the given equation has equal real roots. To design a rectangular park is possible.
∴ Its length and breadth is given by
x2 – 40x + 400 = 0
or (x – 20)2 = 0 i.e., x – 20 = 0 or x = 20
∴ Length = 20 m, Breadth = (40 – 20) m = 20 m.
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