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,
$x^2+(x-3)^2=117$
$\Rightarrow x^2+x^2-6 x+9=117$
$\Rightarrow 2 x^2-6 x+9-117=0$
$\Rightarrow 2 x^2-6 x-108=0$
$\left.\Rightarrow x^2-3 x-54=0 \text { (Dividing by } 2\right)$
$\Rightarrow x^2-9 x+6 x-54=0$
$\Rightarrow x(x-9)+6(x-9)=0$
$\Rightarrow(x-9)(x+6)=0$
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,
$x^2+(x-3)^2=117$
$\Rightarrow x^2+x^2-6 x+9=117$
$\Rightarrow 2 x^2-6 x+9-117=0$
$\Rightarrow 2 x^2-6 x-108=0$
$\left.\Rightarrow x^2-3 x-54=0 \text { (Dividing by } 2\right)$
$\Rightarrow x^2-9 x+6 x-54=0$
$\Rightarrow x(x-9)+6(x-9)=0$
$\Rightarrow(x-9)(x+6)=0$
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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