A natural number is greater than the other by 5. The sum of their squares is 73, then those numbers are

3, 8

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2, 7

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4, 9

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5, 10

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Solution

The correct option is A $3, 8$
Let the two natural numbers be $a$ and $b$
$a^{2} + b^{2} = 73$ .......(1)
$a - b = 5$
$a = b + 5$ ..............(2)
$(b + 5)^{2} + b^{2} = 73$ .........From (1) and (2)
$b^{2} + 10 b + 25 + b^{2} = 73$
$2 b^{2} + 10 b - 48 = 0$
$b^{2} + 5 b - 24 = 0$
$(b + 3) (b - 8) = 0$
$∴ b = 8$
Hence, $a = 3$
Thus, the two numbers are $3$ and $8$

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A natural number is greater than the other by 5. The sum of their squares is 73, then those numbers are

The correct option is A 3, 8Let the two natural numbers be a and ba2+b2=73.......(1)a−b=5 a=b+5..............(2)(b+5)2+b2=73.........From (1) and (2)b2+10b+25+b2=732b2+10b−48=0b2+5b−24=0(b+3)(b−8)=0∴b=8Hence, a=3Thus, the two numbers are 3 and 8