The set of values of \(p\) for which the roots of the equation \(3x^2+2x+p(p–1) = 0\) are of opposite sign is
(0, 1)
To determine the set of values of \(p\) for which the roots of the equation \(3x^2+2x+p(p–1) = 0\) are of opposite sign, we need to understand the properties of quadratic equations.
For a standard quadratic equation of the form \(ax^2 + bx + c = 0\), if its roots are of opposite signs, it implies two conditions:
Therefore, the primary condition for roots to be of opposite signs is that their product must be less than zero.
Let's compare the given equation \(3x^2+2x+p(p–1) = 0\) with the standard form \(ax^2 + bx + c = 0\).
The product of the roots of a quadratic equation is given by the formula \(\frac{c}{a}\).
For the given equation, the product of roots (\(\alpha\beta\)) is:
\(\alpha\beta = \frac{p(p–1)}{3}\)
As established, for the roots to be of opposite signs, their product must be negative:
\(\alpha\beta < 0\)
So, we must have:
\(\frac{p(p–1)}{3} < 0\)
To solve the inequality \(\frac{p(p–1)}{3} < 0\), we can multiply both sides by 3 (a positive number, so the inequality sign does not change):
\(p(p–1) < 0\)
To find the values of \(p\) that satisfy this inequality, we can identify the critical points where the expression \(p(p–1)\) equals zero. These points are \(p=0\) and \(p=1\).
We can use a sign analysis method by checking the sign of \(p(p–1)\) in the intervals defined by these critical points:
| Interval | Sign of \(p\) | Sign of \((p-1)\) | Sign of \(p(p-1)\) |
|---|---|---|---|
| \(p < 0\) | Negative (\(-\)) | Negative (\(-\)) | Positive (\(+\)) |
| \(0 < p < 1\) | Positive (\(+\)) | Negative (\(-\)) | Negative (\(-\)) |
| \(p > 1\) | Positive (\(+\)) | Positive (\(+\)) | Positive (\(+\)) |
From the table, we can see that \(p(p–1) < 0\) when \(p\) is in the interval between 0 and 1 (exclusive of 0 and 1).
Therefore, the set of values of \(p\) for which the roots are of opposite sign is \((0, 1)\).
The set of values for \(p\) that ensures the roots of the quadratic equation \(3x^2+2x+p(p–1) = 0\) are of opposite sign is \((0, 1)\).
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