If p and q are the roots of `x^(2)-4x+1=0` , choose from the options below, the correct equation whose roots are given by `(p+q)^(2)` and `(p-q)^(2)` .

Options

A. `x^(2)-12x+168=0`
B. `x^(2)-10x+90=0`
C. `x^(2)-28x+192=0`
D. `x^(2)-36x+320=0`

Answer

To solve the problem, we need to find the roots of the quadratic equation \(x^2 - 4x + 1 = 0\) and then determine the new equation whose roots are \((p+q)^2\) and \((p-q)^2\), where \(p\) and \(q\) are the roots of the original equation.

### Step 1: Find the roots \(p\) and \(q\) of the equation \(x^2 - 4x + 1 = 0\)

We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -4\), and \(c = 1\).

Calculating the discriminant:
\[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 1 = 16 - 4 = 12 \]

Now, substituting the values into the quadratic formula:
\[ x = \frac{4 \pm \sqrt{12}}{2} \]
\[ x = \frac{4 \pm 2\sqrt{3}}{2} \]
\[ x = 2 \pm \sqrt{3} \]

Thus, the roots are:
\[ p = 2 + \sqrt{3}, \quad q = 2 - \sqrt{3} \]

### Step 2: Calculate \(p + q\) and \(p - q\)

Now, we find:
\[ p + q = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4 \]
\[ p - q = (2 + \sqrt{3}) - (2 - \sqrt{3}) = 2\sqrt{3} \]

### Step 3: Calculate \((p + q)^2\) and \((p - q)^2\)

Next, we calculate the squares:
\[ (p + q)^2 = 4^2 = 16 \]
\[ (p - q)^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12 \]

### Step 4: Form the new quadratic equation with roots \(16\) and \(12\)

The new roots are \(16\) and \(12\). We can use the fact that if \(r_1\) and \(r_2\) are the roots of a quadratic equation, the equation can be formed as:
\[ x^2 - (r_1 + r_2)x + r_1r_2 = 0 \]

Calculating the sum and product of the roots:
\[ r_1 + r_2 = 16 + 12 = 28 \]
\[ r_1 \cdot r_2 = 16 \cdot 12 = 192 \]

Thus, the quadratic equation is:
\[ x^2 - 28x + 192 = 0 \]

### Final Answer

The equation whose roots are \((p + q)^2\) and \((p - q)^2\) is:
\[ \boxed{x^2 - 28x + 192 = 0} \]