Question
Find the two natural numbers which differ by $5$ and the sum of whose squares is $97$.
✓
Answer
Let the two numbers be $x$ and $x + 5$.
From the given information,
$x^2 + (x + 5)^2 = 97$
$2x^2 + 10x + 25 – 97 = 0$
$2x^2 + 10x – 72 = 0$
$x^2 + 5x – 36 = 0$
$(x + 9) (x – 4) = 0$
$x = -9$ or $4$
Since, $-9$ is not a natural number. So, $x = 4$.
Thus, the numbers are $4$ and $9$.
From the given information,
$x^2 + (x + 5)^2 = 97$
$2x^2 + 10x + 25 – 97 = 0$
$2x^2 + 10x – 72 = 0$
$x^2 + 5x – 36 = 0$
$(x + 9) (x – 4) = 0$
$x = -9$ or $4$
Since, $-9$ is not a natural number. So, $x = 4$.
Thus, the numbers are $4$ and $9$.
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