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प्रश्न
Find two natural numbers which differ by 3 and whose squares have the sum of 117.
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उत्तर १
Let the numbers be x and x-3. Then,
x2 + (x-3)2 = 117,
⇒ x2 + x2 + 9 - 6x =117
⇒ 2 x2 -6x - 108 = 0
⇒ x2 - 3x - 54 = 0
⇒ x2 - 9x + 6x - 54 = 0
⇒ x (x - 9) + 6(x - 9) = 0
⇒ (x-9) (x+6 ) = 0
⇒ x = 9 (As the number have to be natural number)
Hence the numbers are 6 and 9.
उत्तर २
Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)2 = 117
⇒ x2 + x2 + 6x + 9 = 117
⇒ 2x2 + 6x + 9 – 1117 = 0
⇒ 2x2 + 6x – 108 = 0
⇒ x2 + 3x – 54 = 0 ...(Dividing by 2)
⇒ x2 + 9x – 6x – 54 = 0
⇒ x(x + 9) –6(x + 9) 0
⇒ (x + 9)(x – 6) = 0
Either x + 9 = 0,
then x = –9,
but it is not a natural number.
or
x - 6 = 0,
then x = 6
∴ First natural number = 6
and second numberr = 6 + 3 = 9.
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