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The given problem presents the quartic equation:

\[
x^4 - px^2 + qx - r = 0
\]
where \(p\), \(q\), and \(r\) are real constants, and it is stated that the equation has two pairs of equal roots. We are tasked with showing that:

\[
p^2 + 4r = 0
\]
and determining the value of \(q\).

### Step 1: Factorization using two pairs of equal roots

Since the equation has two pairs of equal roots, we can express the quartic equation in a factored form as:

\[
(x^2 + \alpha)^2 = 0
\]
or equivalently:

\[
(x^2 - a)^2 = 0
\]
where \(a\) is some real number. Expanding this gives:

\[
(x^2 - a)(x^2 - a) = x^4 - 2ax^2 + a^2 = 0
\]

Comparing this with the general form of the quartic equation:

\[
x^4 - px^2 + qx - r = 0
\]

### Step 2: Compare coefficients

From the expanded form \(x^4 - 2ax^2 + a^2\), we can directly compare it with the given quartic equation:

1. The coefficient of \(x^4\) is 1 in both cases.
2. The coefficient of \(x^2\) is \(-p\), and from the factorized form, it is \(-2a\). Therefore:
   \[
   p = 2a
   \]
3. The constant term \(r = a^2\).

### Step 3: Deriving the equation \(p^2 + 4r = 0\)

From the above, we know:
\[
p = 2a \quad \text{and} \quad r = a^2
\]
Now, substitute these values into the equation \(p^2 + 4r = 0\):
\[
(2a)^2 + 4a^2 = 0
\]
\[
4a^2 + 4a^2 = 0
\]
\[
8a^2 = 0
\]
This implies:
\[
a = 0
\]

### Step 4: Value of \(q\)

Since \(a = 0\), the equation simplifies to \(x^4 = 0\), meaning there is no linear term (no \(qx\) term). Therefore, the value of \(q = 0\).

### Final Result:

- \(p^2 + 4r = 0\) holds, and
- \(q = 0\).

Would you like further details on any step or additional examples? 

Here are 5 related questions to explore further:

1. How can the roots of a quartic equation be derived in general?
2. What other types of factorizations can be applied to quartic equations with real coefficients?
3. How do you determine the nature of roots for a quartic equation?
4. Can the discriminant of a quartic equation provide insight into the equality of roots?
5. How does the factor theorem apply to polynomials of higher degrees?

**Tip:** When dealing with quartic equations, identifying symmetric structures such as equal roots can simplify factorization and reduce the equation to simpler terms.

The quartic equation x^4 - px^2 + qx - r = 0, where p, q, and r are real constants, has two pairs of equal roots. Show that p^2 + 4r = 0 and state the value of q.

Explore the solution to the quartic equation x^4 - px^2 + qx - r = 0, showing that p^2 + 4r = 0 and determining the value of q. Learn about factorizations and properties of equations with equal roots. Suitable for students in Grades 11-12.

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