Question
Find two natural numbers which differ by $3$ and whose squares have the sum $117$.
✓
Answer
Let first number $= x$
Then second number $= x - 3$
According to the condition,
$x^2 + (x - 3)^2 = 117$
$\Rightarrow x^2 + x^2 - 6x + 9 = 117$
$\Rightarrow 2x^2 - 6x + 9 - 117 = 0$
$\Rightarrow 2x^2 - 6x - 108 = 0$
$\Rightarrow x^2 - 3x - 54 = 0 (Dividing by 2)$
$\Rightarrow x^2 - 9x + 6x - 54 = 0$
$\Rightarrow x(x - 9) + 6(x - 9) = 0$
$\Rightarrow (x - 9)(x + 6) = 0$
Either x - 9 = 0, then x = 9
Or $x + 6 = 0$, then $x = -6$ which is not a natural number,
First natural number $= 9$
and second number $= 9 - 3 = 6$
Then second number $= x - 3$
According to the condition,
$x^2 + (x - 3)^2 = 117$
$\Rightarrow x^2 + x^2 - 6x + 9 = 117$
$\Rightarrow 2x^2 - 6x + 9 - 117 = 0$
$\Rightarrow 2x^2 - 6x - 108 = 0$
$\Rightarrow x^2 - 3x - 54 = 0 (Dividing by 2)$
$\Rightarrow x^2 - 9x + 6x - 54 = 0$
$\Rightarrow x(x - 9) + 6(x - 9) = 0$
$\Rightarrow (x - 9)(x + 6) = 0$
Either x - 9 = 0, then x = 9
Or $x + 6 = 0$, then $x = -6$ which is not a natural number,
First natural number $= 9$
and second number $= 9 - 3 = 6$
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