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.
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.