Understanding the Problem: Common Roots of Quadratic Equations

The question asks for a relationship between the coefficients \(p\) and \(q\) if the two given quadratic equations, \(x^2 - px + q = 0\) and \(x^2 + qx - p = 0\), share a common root. If a value of \(x\) satisfies both equations, we can use this fact to find the condition on \(p\) and \(q\).

Finding the Common Root Condition

Let \(\alpha\) be the common root of the two equations. This means that \(\alpha\) must satisfy both equations:

Equation 1: \(\alpha^2 - p\alpha + q = 0\)

Equation 2: \(\alpha^2 + q\alpha - p = 0\)

We now have a system of two equations with the common root \(\alpha\) and the coefficients \(p\) and \(q\). We can eliminate \(\alpha^2\) by subtracting the second equation from the first:

\(( \alpha^2 - p\alpha + q ) - ( \alpha^2 + q\alpha - p ) = 0 - 0\)

\(\alpha^2 - p\alpha + q - \alpha^2 - q\alpha + p = 0\)

Combining like terms:

\((-p - q)\alpha + (q + p) = 0\)

\(-(p+q)\alpha + (p+q) = 0\)

Factor out the term \((p+q)\):

\((p+q)(1 - \alpha) = 0\)

This equation implies that either \((p+q) = 0\) or \((1 - \alpha) = 0\).

Therefore, if the two quadratic equations have a common root \(\alpha\), then either \(p+q = 0\) or \(\alpha = 1\).

Analyzing the Two Cases

We have two possible cases arising from the common root condition:

Case 1: \(p+q = 0\)

If \(p+q = 0\), then \(q = -p\). Substitute \(q = -p\) into the original equations:

• First equation: \(x^2 - px + (-p) = 0 \implies x^2 - px - p = 0\)
• Second equation: \(x^2 + (-p)x - p = 0 \implies x^2 - px - p = 0\)

In this case, the two original equations become identical: \(x^2 - px - p = 0\). Identical equations have all their roots in common. So, if \(p+q = 0\), the equations have common roots (the roots of \(x^2 - px - p = 0\)).

Case 2: \(\alpha = 1\)

If the common root is \(\alpha = 1\), we substitute \(\alpha = 1\) into either of the original equations. Using the first equation:

\(1^2 - p(1) + q = 0\)

\(1 - p + q = 0\)

Rearranging this equation gives the condition on \(p\) and \(q\):

\(p - q - 1 = 0\)

Substituting \(\alpha=1\) into the second equation:

\(1^2 + q(1) - p = 0\)

\(1 + q - p = 0\)

Rearranging gives the same condition:

\(p - q - 1 = 0\)

Conclusion on the Condition

Our analysis shows that if the two equations have a common root, then either \(p+q = 0\) or \(p - q - 1 = 0\). This means the existence of a common root implies that the condition \((p+q = 0) \text{ OR } (p - q - 1 = 0)\) must be true.

Looking at the given options:

The condition \(p - q - 1 = 0\) derived from Case 2 is presented as Option 4. The existence of a common root implies that *at least one* of \(p+q=0\) or \(p-q-1=0\) must hold. Option 4 represents one of these possibilities.

Let's also check if Option 4 being true guarantees a common root. If \(p - q - 1 = 0\), then \(q = p-1\). The equations become \(x^2 - px + (p-1) = 0\) and \(x^2 + (p-1)x - p = 0\). Substituting \(x=1\) into the first equation: \(1^2 - p(1) + (p-1) = 1 - p + p - 1 = 0\). Substituting \(x=1\) into the second equation: \(1^2 + (p-1)(1) - p = 1 + p - 1 - p = 0\). Since \(x=1\) satisfies both equations when \(p - q - 1 = 0\), having \(p - q - 1 = 0\) guarantees that \(x=1\) is a common root.

Thus, the existence of a common root implies \((p+q=0) \lor (p-q-1=0)\), and one of the options listed is \(p-q-1=0\). This condition is sufficient for a common root to exist.

Review Table: Conditions and Implications

Given the options, \(p - q - 1 = 0\) is the condition that is directly derived from the case where the common root is 1, and the existence of a common root implies this case is possible (unless \(p+q=0\)). In the context of multiple-choice questions, Option 4 is the intended correct answer representing a key necessary component of the condition for a common root.

Final Answer Derivation

If the equations \(x^2 - px + q = 0\) and \(x^2 + qx - p = 0\) have a common root \(\alpha\), we found that either \(p+q=0\) or \(\alpha=1\). If \(\alpha=1\), substituting into either equation gives \(1 - p + q = 0\), which is \(p - q - 1 = 0\). This condition is presented as Option 4.

Additional Information: Common Root Formula

For two general quadratic equations \(a_1x^2 + b_1x + c_1 = 0\) and \(a_2x^2 + b_2x + c_2 = 0\), the condition for having a common root is \((a_1c_2 - a_2c_1)^2 = (a_1b_2 - a_2b_1)(b_1c_2 - b_2c_1)\).

For the given equations: \(a_1=1, b_1=-p, c_1=q\) \(a_2=1, b_2=q, c_2=-p\)

\(a_1c_2 - a_2c_1 = (1)(-p) - (1)(q) = -(p+q)\)

\(a_1b_2 - a_2b_1 = (1)(q) - (1)(-p) = p+q\)

\(b_1c_2 - b_2c_1 = (-p)(-p) - (q)(q) = p^2 - q^2 = (p-q)(p+q)\)

The condition becomes:

\((-(p+q))^2 = (p+q)((p-q)(p+q))\)

\((p+q)^2 = (p+q)^2(p-q)\)

\((p+q)^2 - (p+q)^2(p-q) = 0\)

\((p+q)^2 [1 - (p-q)] = 0\)

\((p+q)^2 (1 - p + q) = 0\)

\((p+q)^2 (q - p + 1) = 0\)

This equation is satisfied if and only if \(p+q = 0\) or \(q - p + 1 = 0\). The condition \(q - p + 1 = 0\) is equivalent to \(p - q - 1 = 0\).

So, the existence of a common root is equivalent to \((p+q=0) \lor (p-q-1=0)\). Among the given options, only \(p-q-1=0\) is part of this necessary and sufficient condition.