A business journal investigation of the performance and timing of corporate acquisitions discovered that in a random sample of 2
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Answer:
This problem actually says:
(y + 2), (y + 3) and (2*y^2 - 1) are consecutive terms of an arithmetic progression.
Now, the difference between two consecutive terms in an arithmetic progression is always the same, so we have:
(y + 3) - (y +2) = D
(2*y^2 - 1) - (y + 3) = D
From the first equation we have:
y + 3 - y - 2 = 1 = D
Now we can replace it in the other equation:
2*y^2 - 1 - y - 3 = 1
2*y^2 - y - 5 = 0
Now we need to solve that equation to find the possible values of y.
To solve a quadratic equation of the form:
a*x^2 + b*x + c = 0, we can use the Bhaskara's equation:
in this case, the solutions are:
Then the possible values of y are:
y = (1 + 6.4)/4 = 1.85
y = (1 - 6.4)/5 = -1.35