Understanding the Quadratic Equation and its Roots

The given problem involves a quadratic equation of the form \(x^2 + px + q = 0\). We are told that the roots of this equation are \(p\) and \(q\), and that both \(p\) and \(q\) are non-zero.

For any quadratic equation in the standard form \(ax^2 + bx + c = 0\), there are well-known relationships between the coefficients (\(a\), \(b\), \(c\)) and the roots (\(\alpha\), \(\beta\)). These relationships are:

• Sum of roots: \(\alpha + \beta = -\frac{b}{a}\)
• Product of roots: \(\alpha \beta = \frac{c}{a}\)

Applying Root Properties to Solve for p and q

In our specific equation, \(x^2 + px + q = 0\), we have:

• \(a = 1\)
• \(b = p\)
• \(c = q\)

The roots are given as \(\alpha = p\) and \(\beta = q\).

Using the Sum of Roots Property

The sum of the roots is \(p + q\). According to the property, this sum must equal \(-\frac{b}{a}\). Substituting the values from our equation:

\(p + q = -\frac{p}{1}\)

\(p + q = -p\)

Now, let's rearrange this equation to relate \(q\) and \(p\):

\(q = -p - p\)

\(q = -2p\) (Equation 1)

Using the Product of Roots Property

The product of the roots is \(p \times q\). According to the property, this product must equal \(\frac{c}{a}\). Substituting the values from our equation:

\(p \times q = \frac{q}{1}\)

\(pq = q\) (Equation 2)

Finding the Possible Values of q

We have two equations based on the properties of roots:

1. \(q = -2p\)
2. \(pq = q\)

We are also given that \(p\) and \(q\) are non-zero roots, which means \(p \neq 0\) and \(q \neq 0\).

Let's analyze Equation 2: \(pq = q\)

Since we know \(q \neq 0\), we can divide both sides of the equation by \(q\):

\(\frac{pq}{q} = \frac{q}{q}\)

\(p = 1\)

Now we have found the value of \(p\). We can substitute this value into Equation 1 to find the value of \(q\):

\(q = -2p\)

\(q = -2(1)\)

\(q = -2\)

Verifying the Non-Zero Condition

We found \(p = 1\) and \(q = -2\). Both of these values are non-zero, which satisfies the condition given in the problem.

Thus, the only pair of non-zero roots \(p, q\) that satisfies the equation \(x^2 + px + q = 0\) is \(p=1\) and \(q=-2\).

Conclusion on the Number of Possible Values for q

From our analysis, we found that \(q\) must be equal to \(-2\). There is only one such value that \(q\) can take while satisfying all the conditions of the problem.

Therefore, there is only one possible value for \(q\).

Revision Table: Key Concepts Review

Additional Information: Understanding Quadratic Equations

A quadratic equation can have real or complex roots. The nature of the roots is determined by the discriminant, \(\Delta = b^2 - 4ac\).

• If \(\Delta > 0\), there are two distinct real roots.
• If \(\Delta = 0\), there is exactly one real root (a repeated root).
• If \(\Delta < 0\), there are two distinct complex roots (conjugate pairs).

In this problem, after finding \(p=1\) and \(q=-2\), the equation becomes \(x^2 + 1x - 2 = 0\), or \(x^2 + x - 2 = 0\). The roots of this specific equation are the values of \(x\) that satisfy it. We can factor this equation:

\(x^2 + 2x - x - 2 = 0\)

\(x(x+2) - 1(x+2) = 0\)

\((x-1)(x+2) = 0\)

The roots are \(x=1\) and \(x=-2\). According to the problem statement, these roots are \(p\) and \(q\).

• If we assign \(p=1\) and \(q=-2\), this matches our derived values for \(p\) and \(q\). Both are non-zero.
• If we assign \(p=-2\) and \(q=1\), then substituting into the original equation \(x^2 + px + q = 0\) would give \(x^2 - 2x + 1 = 0\). The roots of this equation are found by factoring: \((x-1)^2 = 0\), which gives a single root \(x=1\). However, the problem states the roots are \(p\) and \(q\), implying two roots (possibly equal, but here non-zero). If the roots were \(p=-2\) and \(q=1\), the roots of \(x^2 - 2x + 1 = 0\) must be \(-2\) and \(1\). This is not true, as the only root is \(1\). Thus, this case is not possible.

This confirms that the only consistent assignment for the roots \(p\) and \(q\) satisfying the conditions is \(p=1\) and \(q=-2\).