Question
Find two consecutive natural numbers whose squares have the sum $221.$
✓
Answer
Let the number be $x_1 x+1$
Then $x^2+(x+1)^2=221$
$\Rightarrow x^2 + x^2 + 1 + 2x - 221 = 0$
$\Rightarrow 2x^2 + 2x - 220 = 0$
$\Rightarrow x^2 + x - 110 = 0$
$\Rightarrow x^2 + 11x - 10x - 110 = 0$
$\Rightarrow x(x + 11) - 10(x + 11) = 0$
$\Rightarrow (x = 11) (x - 10) = 0$
$\Rightarrow x = -11 or x = 10$
But $x=-11$ is rejected ...[ $\because$ It cannot been as its is a natural number]
$\therefore x=10$
Hence, required numbers are 10, $10+1$.
i.e., $10$ and $11.$
Then $x^2+(x+1)^2=221$
$\Rightarrow x^2 + x^2 + 1 + 2x - 221 = 0$
$\Rightarrow 2x^2 + 2x - 220 = 0$
$\Rightarrow x^2 + x - 110 = 0$
$\Rightarrow x^2 + 11x - 10x - 110 = 0$
$\Rightarrow x(x + 11) - 10(x + 11) = 0$
$\Rightarrow (x = 11) (x - 10) = 0$
$\Rightarrow x = -11 or x = 10$
But $x=-11$ is rejected ...[ $\because$ It cannot been as its is a natural number]
$\therefore x=10$
Hence, required numbers are 10, $10+1$.
i.e., $10$ and $11.$
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