Question

The sum of the squares of two consecutive positive integers is 841. If larger number is x, then the algebraic representation will b

Answer 1

Step-by-step explanation:

The sum of the squares of two consecutive positive integers is 841.

Assume that the numbers are (x) and (x + 1) respectively.

As said in question, sum and square of those numbers is 841. i.e.

→ (x)² + (x + 1)² = 841

Used identity: (a + b)² = a² + b² + 2ab

→ x² + x² + 1 + 2x = 841

→ 2x² + 2x = 841 - 1

→ 2x² + 2x = 840

Take 2 as common,

→ 2(x² + x ) = 2(420)

→ x² + x = 420

→ x² + x - 420 = 0

Hence, the algebraic representation of the situation is x² + x - 420 = 0.

Answer 2

Answer:

The final answer is $x^2 + x -420 =0$

Step-by-step explanation:

Given,

We have the sum of the squares of two consecutive numbers to be equal to 841. We need to find the algebraic representation of the numbers if the larger number is x.

If the larger number is x, Then the lesser number will be equal to x-1 since they are consecutive numbers. Next we find our equation by putting them all together as said in the question.

$x^2$ $+$ $(x-1)^2$ $=$ $841$

We open the brackets and,

$x^2 + x^2 +2x + 1 =841$

$2x^2 +2x =840$

$x^2 + x -420 =0$

This is our final equation.

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