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) 3x^{2} – 5x + 2 = 0

(ii) 5x^{2} – 6x – 2 = 0

(iii) 4x^{2} + 3x + 5 = 0

(iv) 4x^{2} + 4√3x + 3 = 0

(v) 2x^{2} + x – 4 = 0

(vi) 2x^{2} + x + 4 = 0

(vii) 4x^{2} + 4bx – (a^{2} – b^{2}) = 0

Solution

Question 2.

Find the roots of the following (RBSESolutions.com) equations (if exists), using Shridhar Acharya quadratic formula.

(i) 2x^{2} – 2√2x + 1 = 0

(ii) 9x^{2} + 7x – 2 = 0

(iii) x + = 3, x ≠ 0

(iv) √2x^{2} + 7x + 5√2 = 0

(v) x^{2} + 4x + 5 = 0

(vi) , 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

⇒ x^{2} + x^{2} + 4x + 4 = 290

⇒ 2x^{2} + 4x + 4 = 290

⇒ 2x^{2} + 4x – 286 = 0

⇒ x^{2} + 2x – 143 = 0

⇒ x^{2} + 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, x^{2} – y^{2} = 45 and y^{2} = 4x

from both equations x^{2} – 4x = 45

⇒ x^{2} – 4x – 45 = 0

⇒ x^{2} – 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,

2x^{2} = (16 – x)^{2} + 164

⇒ 2x^{2} – (16 – x)^{2} – 164 = 0

⇒ 2x^{2} – [256 – 32x + x^{2}] – 164 = 0

⇒ 2x^{2} – 256 + 32x – x^{2} – 164 = 0

⇒ x^{2} + 32x – 420 = 0

⇒ x^{2} + 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

We hope the given RBSE Solutions for Class 10 Maths Chapter 3 Polynomials Ex 3.4 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 3 Polynomials Exercise 3.4, drop a comment below and we will get back to you at the earliest.

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