Respuesta :
Answer:
-4 and -3.
Step-by-step explanation:
Let the numbers be x and x+1.
From the given information:
x^2 + (x + 1)^2 + x = 21
x^2 + x^2 + 2x + 1 + x - 21 = 0
2x^2 + 3x - 20 = 0
(2x - 5 )(x + 4 ) = 0
x = 5/2 or -4 We ignore 5/2 as its not an interger
So the required integers are -4 and -3.
Two consecutive numbers are x and x+1.
Clearly, x is the lesser, and x+1 is the largest.
The sum of their squares is
[tex]x^2+(x+1)^2 = x^2+x^2+2x+1 = 2x^2+2x+1[/tex]
If we add this sum and the lesser we have
[tex]2x^2+2x+1+x=2x^2+3x+1[/tex]
We want this quantity to be 21, so we have
[tex]2x^2+3x+1=21 \iff 2x^2+3x-20=0[/tex]
The solutions of this equation are
[tex]x=-4,\quad x=\dfrac{5}{2}[/tex]
Since we want integers, the numbers are -4 and -3.
