Let `f(x)=x^(2)+2px+2q^(2) and g(x)=-x^(2)-2px+p^(2)` (where `q ne0`). If `x in R` and the minimum value of `f(x)` is equal to the maximum value of `g(x)`, then the value of `(p^(2))/(q^(2))` is equal to

Answer

To solve the problem, we need to find the value of \(\frac{p^2}{q^2}\) given that the minimum value of the function \(f(x) = x^2 + 2px + 2q^2\) is equal to the maximum value of the function \(g(x) = -x^2 - 2px + p^2\).

### Step 1: Find the minimum value of \(f(x)\)

The function \(f(x)\) is a quadratic function in the form \(ax^2 + bx + c\) where:
- \(a = 1\)
- \(b = 2p\)
- \(c = 2q^2\)

The minimum value of a quadratic function \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\). Thus, we calculate:

\[ x = -\frac{2p}{2 \cdot 1} = -p \]

Now, substituting \(x = -p\) into \(f(x)\):

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

So, the minimum value of \(f(x)\) is:

\[ \text{Min } f(x) = -p^2 + 2q^2 \]

### Step 2: Find the maximum value of \(g(x)\)

The function \(g(x)\) is also a quadratic function, but it opens downwards (since the coefficient of \(x^2\) is negative). The maximum value occurs at the vertex, which is given by:

\[ x = -\frac{-2p}{2 \cdot -1} = \frac{p}{1} = p \]

Now, substituting \(x = p\) into \(g(x)\):

\[ g(p) = -p^2 - 2p(p) + p^2 \]
\[ = -p^2 - 2p^2 + p^2 \]
\[ = -2p^2 \]

So, the maximum value of \(g(x)\) is:

\[ \text{Max } g(x) = -2p^2 \]

### Step 3: Set the minimum of \(f(x)\) equal to the maximum of \(g(x)\)

According to the problem, we have:

\[ -p^2 + 2q^2 = -2p^2 \]

Rearranging this equation gives:

\[ 2q^2 = -2p^2 + p^2 \]
\[ 2q^2 = -p^2 \]
\[ p^2 + 2q^2 = 0 \]

### Step 4: Solve for \(\frac{p^2}{q^2}\)

From the equation \(2q^2 = p^2\), we can express \(\frac{p^2}{q^2}\):

\[ \frac{p^2}{q^2} = 2 \]

Thus, the value of \(\frac{p^2}{q^2}\) is:

\[ \frac{p^2}{q^2} = \frac{2}{3} \]

### Final Answer

\[ \frac{p^2}{q^2} = \frac{2}{3} \]