A positive real number is 4 less than another. If the sum of the squares of the two numbers is 72, then find the numbers.
☛ Nᴇᴇᴅ Qᴜᴀʟɪᴛʏ Aɴsᴡᴇʀ
Answers
Answer:
Possible pairs with sum as 72 and with at least 1 perfect square element is :
(1,71), (4, 68), (9, 63), (16, 56), (25, 47), (36, 36), (49, 23), (64, 8)
There is only one pair (36, 36) or (6², 6²) where both elements are perfect squares. But, it doesn't satisfy the condition that one number is 4 less than other number.
Such integer numbers do not exist.
Non integer solution :
a² + (a - 4)² = 72
a² + a² - 8a + 16 = 72
a² - 4a - 28 = 0
a = {4 ± √(16 + 112)}/2
a = ½(4 ± 8√2) = 2 ± 4√2.
Other number is (a - 4)
2 + 4√2 - 4 = 2(2√2 - 1) or 2 - 4√2 - 4 = -2(2√2 + 1)
But since a - 4 is stated to be positive real number, it should be 2(2√2 - 1)
Step-by-step explanation:
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Answer:
Possible pairs with sum as 72 and with at least 1 perfect square element is :
(1,71), (4, 68), (9, 63), (16, 56), (25, 47), (36, 36), (49, 23), (64, 8)
There is only one pair (36, 36) or (6², 6²) where both elements are perfect squares. But, it doesn't satisfy the condition that one number is 4 less than other number.
Such integer numbers do not exist.
Non integer solution :
a² + (a - 4)² = 72
a² + a² - 8a + 16 = 72
a² - 4a - 28 = 0
a = {4 ± √(16 + 112)}/2
a = ½(4 ± 8√2) = 2 ± 4√2.
Other number is (a - 4)
2 + 4√2 - 4 = 2(2√2 - 1) or 2 - 4√2 - 4 = -2(2√2 + 1)
But since a - 4 is stated to be positive real number, it should be 2(2√2 - 1)
:-)