Math, asked by ⱮøøɳƇⲅυѕɦεⲅ, 1 month ago

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.


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Answers

Answered by itzkanika85
5

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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Warm regards:Miss Chikchiki

Answered by heroyar
3

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)

:-)

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