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Three consecutive odd integers are such that the square of the third integer is 33 less than the sum of the squares of the first two. One solution is −5​, −3​, and−1. Find three other consecutive odd integers that also satisfy the given conditions.

Respuesta :

LammettHash
If [tex]x[/tex] is the first integer, then

[tex](x+4)^2=x^2+(x+2)^2-33\iff x^2-4x-45=(x+5)(x-9)=0[/tex]

The other possibility is then [tex]x=9[/tex], so the other two integers would be [tex]x+2=11[/tex] and [tex]x+4=13[/tex].

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