RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.4

RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.4 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.4.

Board	RBSE
Te×tbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 3
Chapter Name	Polynomials
E×ercise	Exercise 3.4
Number of Questions Solved	5
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Ex 3.4

Question 1.
Solve the following equations by (RBSESolutions.com) the method of completing square.
(i) 3x² – 5x + 2 = 0
(ii) 5x² – 6x – 2 = 0
(iii) 4x² + 3x + 5 = 0
(iv) 4x² + 4√3x + 3 = 0
(v) 2x² + x – 4 = 0
(vi) 2x² + x + 4 = 0
(vii) 4x² + 4bx – (a² – b²) = 0
Solution

Question 2.
Find the roots of the following (RBSESolutions.com) equations (if exists), using Shridhar Acharya quadratic formula.
(i) 2x² – 2√2x + 1 = 0
(ii) 9x² + 7x – 2 = 0
(iii) x + \(\frac { 1 }{ x }\) = 3, x ≠ 0
(iv) √2x² + 7x + 5√2 = 0
(v) x² + 4x + 5 = 0
(vi) \(\frac { 1 }{ x } -\frac { 1 }{ x-2 } =3\) , x ≠ 0, 2
Solution

Question 3.
Find two consecutive odd (RBSESolutions.com) positive integers, sum of whose square is 290.
Solution
Let x and x + 2 are two consecutive odd positive integers.
According to question,
(x)² + (x + 2)² = 290
⇒ x² + x² + 4x + 4 = 290
⇒ 2x² + 4x + 4 = 290
⇒ 2x² + 4x – 286 = 0
⇒ x² + 2x – 143 = 0
⇒ x² + 13x – 11x – 143 = 0
⇒ x(x + 13) – 11(x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x – 11 = 0
⇒ x = 11
Consecutive odd numbers x = 11
x + 2 = 11 + 2 = 13
Hence, two consecutive odd positive integers are 11, 13

Question 4.
The difference of square of two (RBSESolutions.com) numbers is 45 and square of the smaller number is 4 times the larger number. Find two numbers.
Solution
Let x and y are two number where x > y then
according to question, x² – y² = 45 and y² = 4x
from both equations x² – 4x = 45
⇒ x² – 4x – 45 = 0
⇒ x² – 9x + 5x – 45 = 0
⇒ x(x – 9) + 5 (x – 9) = 0
⇒ (x – 9) (x + 5) = 0
⇒ x – 9 = 0
⇒ x = 9
Larger number (x) = 9
square of smaller number = 4 × 9 = 36
Smaller no. = ±√36 = ±6
Hence, numbers are 9, 6 or 9, -6

Question 5.
Divide 16 into two part such that twice (RBSESolutions.com)the square of larger part is more, then 164 from the square of the smaller part.
Solution
Let larger part be x
Smaller part = 16 – x
According to question,
2x² = (16 – x)² + 164
⇒ 2x² – (16 – x)² – 164 = 0
⇒ 2x² – [256 – 32x + x²] – 164 = 0
⇒ 2x² – 256 + 32x – x² – 164 = 0
⇒ x² + 32x – 420 = 0
⇒ x² + 42x – 10x – 420 = 0
⇒ x(x + 42) – 10(x + 42) = 0
⇒ (x + 42) (x – 10) = 0
⇒ x = -42 or x = 10
⇒ x = 10 [x = -42 not possible]
Larger number (x) = 10
Smaller number (16- x) = 16 – 10 = 6
Hence two numbers are 10, 6

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