MHB Three Consecutive Odd Integers

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The discussion focuses on finding three consecutive odd integers where the square of the first plus the square of the third equals 170. Participants confirm the setup and suggest solving for x, with a recommendation to express x as 2n + 1 to ensure it remains odd. A correction is made regarding the equation, clarifying that it should be x^2 + (x + 4)^2 = 170. There is also a mention of a sign switch in the factorization, leading to a revised equation. The conversation emphasizes careful attention to detail in solving the problem.
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Find three consecutive odd integers such that the square of the first plus the square of the third is 170.

See picture for the set up.

View attachment 7356

Is the set up correct?
 

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Re: Three Executive Odd Integers

RTCNTC said:
Find three consecutive odd integers such that the square of the first plus the square of the third is 170.

See picture for the set up.

Is the set up correct?
That's good. Just solve for x and make sure your answer is odd. Another addition to this would be to set x = 2n + 1, which forces x to be odd, but you don't actually have to take this step.

-Dan

Edit: Whoops! Your equation should only contain two numbers: [math]x^2 + (x + 4)^2 = 170[/math]. Sorry about that.
 
Re: Three Executive Odd Integers

topsquark said:
That's good. Just solve for x and make sure your answer is odd. Another addition to this would be to set x = 2n + 1, which forces x to be odd, but you don't actually have to take this step.

-Dan

Edit: Whoops! Your equation should only contain two numbers: [math]x^2 + (x + 4)^2 = 170[/math]. Sorry about that.

Ok. Thanks.
 

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Re: Three Executive Odd Integers

RTCNTC said:
See picture reply.

Correct?
You switched signs. It's [math](x - 7)(x + 11) = x^2 + (-7 + 11)x - 77 = x^2 + 4x - 77[/math].

-Dan
 
Re: Three Executive Odd Integers

I rushed through the problem. Thanks for the correction.
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