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Question
Let us find two natural numbers which differ by 3 and whose squares have the sum 117.
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Solution
Let the numbers be x and x - 3
By the given hypothesis,
๐ฅ2 + (๐ฅ - 3)2 = 117
⇒ ๐ฅ2 + ๐ฅ2 + 9 - 6๐ฅ - 117 = 0
⇒ 2๐ฅ2 - 6๐ฅ - 108 = 0
⇒ ๐ฅ2 - 3๐ฅ - 54 = 0
⇒ ๐ฅ(๐ฅ - 9) + 6(๐ฅ - 9) = 0
⇒ (x - 9) (x + 6) = 0
⇒ x = 9 or x = -6
Considering positive value of x
x = 9, x - 3 = 9 - 3 = 6
∴ The two numbers be 9 and 6.
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