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Question

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

The correct answer 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.

Quadratic Equation Roots Condition

For a standard quadratic equation of the form \(ax^2 + bx + c = 0\), if its roots are of opposite signs, it implies two conditions:

  1. The product of the roots must be negative.
  2. The discriminant (\(\Delta = b^2 - 4ac\)) must be positive, ensuring real roots. However, if the product of the roots (\(\frac{c}{a}\)) is negative, the discriminant (\(b^2 - 4ac\)) will always be positive because \(ac\) will be negative, making \(-4ac\) positive. So, if the product of roots is negative, real roots are guaranteed.

Therefore, the primary condition for roots to be of opposite signs is that their product must be less than zero.

Identifying Coefficients

Let's compare the given equation \(3x^2+2x+p(p–1) = 0\) with the standard form \(ax^2 + bx + c = 0\).

  • The coefficient of \(x^2\) is \(a = 3\).
  • The coefficient of \(x\) is \(b = 2\).
  • The constant term is \(c = p(p–1)\).

Product of Roots Calculation

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:

\(\frac{p(p–1)}{3} < 0\)

Solving the Inequality for p

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 \((0, 1)\).

Conclusion for p Values

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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