Respuesta :

Start by declaring variables.
Since we have two consecutive integers, let your variables be x and x + 1.

Thus, we can say that:
x² + (x + 1)² = 41
x² + x² + 2x + 1 = 41
2x² + 2x - 40 = 0
x² + x - 20 = 0
(x + 5)(x - 4) = 0

So, x = -5 or 4.

Thus, we can have -5 and -4,
or 4 and 5.
shockwave89
≡ We know that:
⇔ Consider the first integer is [tex]a[/tex]
⇔ So, the next integer is [tex](a+1)[/tex]

≡ Solution:
⇒ [tex](a)^{2}+(a+1)^{2}=41[/tex]
⇒ [tex]a^{2}+(a^{2}+2a+1)=41[/tex]
⇒ [tex]2a^{2}+2a-40=0[/tex]
⇒ [tex](2a-8)(a+5)[/tex]
⇒ [tex]a_{1}=\frac{8}{2}=\boxed{4}[/tex]║[tex]a_{2}=-5[/tex]

∴ So, there are 2 options, [tex]\boxed{4}and\boxed{5}[/tex] or [tex]\boxed{-5}and\boxed{-4}[/tex]
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