Solve for real numbers the equation: x 1 - 1 + 2 x 2 - 4 + 3 x 3 - 9 + . . . . . . + n x n - n 2 = 1 2 x 1 + x 2 + . . . . . . + x n

Given, x 1 - 1 + 2 x 2 - 4 + 3 x 3 - 9 + . . . . . . + n x n - n 2 = 1 2 x 1 + x 2 + . . . . . . + x n = x 1 - 1 2 + 2 x 2 - 2 2 + ⋯ + n x n - n 2 = 1 2 x 1 + x 2 + ⋯ + x n ⇒ 2 x 1 - 1 2 + 2 x 2 - 2 2 + ⋯ + n x n - n 2 = x 1 + x 2 + ⋯ + x n ⇒ 2 x 1 - 1 2 + 2 x 2 - 2 2 + ⋯ + n x n - n 2 = x 1 + x 2 + ⋯ + x n - 1 2 + 2 2 + 3 2 + . . . . + n 2 + 1 2 + 2 2 + 3 2 + . . . . + n 2 ⇒ x 1 - 1 2 + x 2 - 2 2 + ⋯ + x n - n 2 - 2 . 1 x 1 - 1 - 2 . 2 x 2 - 2 2 - ⋯ - 2 . n x n - n 2 + 1 2 + 2 2 + 3 2 + . . . . + n 2 = 0 ⇒ x 1 - 1 - 2 x 1 - 1 + 1 + x 2 - 2 2 - 4 x 2 - 2 2 + 2 2 + ⋯ + x n - n 2 - 2 n x n - n 2 + n 2 = 0 .... ⇒ x 1 - 1 - 1 2 + x 2 - 2 2 - 2 2 + . . . . . + ⋯ + x n - n 2 - n 2 = 0 By hypothesis, the sum of these squares must be equal to 0 , hence all squares are equal to 0 . Thus x 1 - 1 = 1 , x 2 - 2 2 = 2 , … , x n - n 2 = n . The unique solution to the equation is x 1 = 2 , x 2 = 8 , … , x n = 2 n 2 .

E M B I B E

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise

NCERT Solutions for Chapter: Polynomials, Exercise 1: Exercise

Author:NCERT

NCERT Mathematics Solutions for Exercise - NCERT Solutions for Chapter: Polynomials, Exercise 1: Exercise

Attempt the practice questions on Chapter 2: Polynomials, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Textbook of Competency Based Questions for Class IX solutions are prepared by Experienced Embibe Experts.

Questions from NCERT Solutions for Chapter: Polynomials, Exercise 1: Exercise with Hints & Solutions

HARD

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 5

The area of a rectangle is $(3 x^{2} + x - 2)$ square units. Its width is $(1 + x)$ units. What is the length of the rectangle?

HARD

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 6

A polynomial is divided by $(x - 1) .$ The quotient obtained is $3 x^{3} - x^{2} - x - 4$ , and the remainder is $- 5 .$ Which polynomial meets these conditions?

HARD

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 7

What is the common factor of $x^{3} - x^{2}$ and $- 22 x^{2} + 142 x - 120 ?$

HARD

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 8

A polynomial is expressed as: $p (x) = x^{3} + x^{2} - x - 1$ . At what values of $x$ is the polynomial $p (x) = 0$

HARD

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 18

If $X = \sqrt[3]{7 + 5 \sqrt{2}} - \frac{1}{\sqrt[3]{7 + 5 \sqrt{2}}}$ , then the value of $x^{3} + 3 x - 14$ is equal to

MEDIUM

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 29

If $x = \sqrt{9 + \sqrt{77}}$ , then the value of $\frac{1}{11} {(x + \frac{2}{x})}^{2}$ is equal to

HARD

9th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class IX>Chapter 2 - Polynomials>Exercise>Q 41

Solve for real numbers the equation:

$\sqrt{x_{1} - 1} + 2 \sqrt{x_{2} - 4} + 3 \sqrt{x_{3} - 9} + . . . . . . + n \sqrt{x_{n} - n^{2}} = \frac{1}{2} (x_{1} + x_{2} + . . . . . . + x_{n})$