The squares of two numbers form a geometric progression whose common ratio is 4. The two numbers also form an arithmetic progression and their sum is 3. Find the two numbers of they exist.

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You should come up with the notation for the numbers: x and y.

The problem tells that the squares of the numbers are in geometric progression and the common ratio is 4.

`y^2 = 4*x^2 =gt y = +-2x`

The numbers x and y are terms of arithmetic progression such that:

`x+ y = 3`

You should use the relation `y = +-2x` such that:

`x + 2x = 3 =gt 3x = 3 =gt x = 1 =gt y = 2`

`x - 2x = 3 =gt -x = 3 =gt x = -3 =gt y = 6`

**Hence, evaluating the numbers that could accomplish the given conditions yields x = 1 , y = 2 and x = -3, y = 6.**

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