Question

Heya ❤❤❤❤

If the sum of squares of two consecutive odd positive integers is 290 . What are the numbers ?

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Answer 1

$\huge\mathfrak{Ello\:Dear}$❤

Let the first odd integer be x.

So,the second consecutive odd integer will be x+2.

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According to the question.

${ (x)}^{2} + {(x + 2)}^{2} = 290 \\ {x}^{2} + {x}^{2} + 4x + 4 = 290 \\ 2 {x}^{2} + 4x + 4 = 290 \\ 2 {x}^{2} + 4x = 286 \\ 2 {x}^{2} + 4x - 286 = 0 \\ 2( {x}^{2} + 2x - 143) = 0 \\ {x}^{2} + 2x - 143 = 0 \\ {x}^{2} + 13x - 11x - 143 = 0 \\ x(x + 13) - 11(x + 13) = 0 \\ (x + 13)(x - 11) = 0\\ x - 11 = 0 \: or \: x + 13 = 0 \\ x = 11 \: or \: x = - 13$

We always take positive value of x.

Therefore, x = 11 and x + 2 = 13

The two consecutive positive odd integers whose square's sum is 290 are 11 and 13.

Hope it helps uhh✌

$\bold{@PhilophobiaXD}$

Answer 2

Here's the Solution :

Let the two positive odd consecutive integers be x and x+2

According to question ,

${x}^{2} + ( {x + 2})^{2} = 290 \\ \\ = > {x}^{2} + {x}^{2} +4x + 4 = 290 \\ \\ = > 2 {x}^{2} + 4x + 4 - 290 = 0 \\ \\ = > 2{ x }^{2} + 4x - 286 = 0 \\ \\ = > 2( {x}^{2} + 2x - 143) = 0 \\ \\ = > {x}^{2} + 2x - 143 = 0 \\ \\ = > {x}^{2} + 13x - 11x - 143 = 0 \\ \\ = > x(x + 13) - 11(x + 13) = 0 \\ \\ = > (x - 11)(x + 13) = 0 \\ \\ now \\ x - 11 = 0 \\ = > x = 11 \\ \\ and \\ x + 13 = 0 \\ = > x = - 13$

Since we are given to find odd positive integers so we shall avoid -13 here .

Therefore , required odd positive integers are 11 and 11+2=13 .