Question
Find three consecutive odd integers, the sum of whose squares is $83$.
✓
Answer
Let the three numbers be $x, x + 2, x + 4$
According to statements,
$(x)^2 + (x + 2)^2 + (x + 4)^2 = 83$
$\Rightarrow x^2 + x^2 + 4x + 4 + x^2 + 8x + 16 = 83$
$\Rightarrow 3x^2 + 12x + 20 = 83$
$\Rightarrow 3x^2 + 12x + 20 - 83 = 0$
$\Rightarrow 3x^2 + 12x - 63 = 0$
$\Rightarrow x^2 + 4x - 21 = 0$
$\Rightarrow x^2 + 7x - 3x - 21 = 0$
$\Rightarrow x(x + 7) -3(x + 7) = 0$
$\Rightarrow (x - 3)(x + 7) = 0$
Either $x - 3 = 0$,
then $x = 3$
or
$x + 7 = 0,$
then $x = -7$
∴ Number will be $3, 3 + 2, 3 + 4 = 3, 5, 7$
or
Numbers will be $-7, -7 + 2, -7 + 4 = -7, -5. -3$.
According to statements,
$(x)^2 + (x + 2)^2 + (x + 4)^2 = 83$
$\Rightarrow x^2 + x^2 + 4x + 4 + x^2 + 8x + 16 = 83$
$\Rightarrow 3x^2 + 12x + 20 = 83$
$\Rightarrow 3x^2 + 12x + 20 - 83 = 0$
$\Rightarrow 3x^2 + 12x - 63 = 0$
$\Rightarrow x^2 + 4x - 21 = 0$
$\Rightarrow x^2 + 7x - 3x - 21 = 0$
$\Rightarrow x(x + 7) -3(x + 7) = 0$
$\Rightarrow (x - 3)(x + 7) = 0$
Either $x - 3 = 0$,
then $x = 3$
or
$x + 7 = 0,$
then $x = -7$
∴ Number will be $3, 3 + 2, 3 + 4 = 3, 5, 7$
or
Numbers will be $-7, -7 + 2, -7 + 4 = -7, -5. -3$.
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