(p, q) is point such that p and q are integers, `p ge 50` and the equation `px^(2)+qx+1=0` has real roots. Square of the least distance of the point from the origin is

Answer

To solve the problem, we need to find the square of the least distance of the point (p, q) from the origin, given that p and q are integers, \( p \geq 50 \), and the quadratic equation \( px^2 + qx + 1 = 0 \) has real roots.

### Step-by-step Solution:

1. **Understanding the Distance Formula**:
The square of the distance from the origin (0, 0) to the point (p, q) is given by:
\[ D^2 = p^2 + q^2 \]
We need to minimize \( D^2 \).

2. **Condition for Real Roots**:
For the quadratic equation \( px^2 + qx + 1 = 0 \) to have real roots, the discriminant must be non-negative. The discriminant \( \Delta \) is given by:
\[ \Delta = q^2 - 4p \cdot 1 = q^2 - 4p \]
For real roots, we require:
\[ q^2 - 4p \geq 0 \implies q^2 \geq 4p \]

3. **Substituting the Minimum Value of p**:
Since \( p \geq 50 \), we will take the minimum value \( p = 50 \):
\[ q^2 \geq 4 \cdot 50 = 200 \]

4. **Finding Minimum Integer Value of q**:
We need to find the smallest integer \( q \) such that \( q^2 \geq 200 \). The smallest integer \( q \) that satisfies this condition is:
\[ q = \lceil \sqrt{200} \rceil = \lceil 14.14 \rceil = 15 \]
(since \( 15^2 = 225 \) which is greater than 200).

5. **Calculating the Minimum Value of \( D^2 \)**:
Now we can calculate \( D^2 \) using \( p = 50 \) and \( q = 15 \):
\[ D^2 = p^2 + q^2 = 50^2 + 15^2 = 2500 + 225 = 2725 \]

6. **Conclusion**:
The square of the least distance of the point (p, q) from the origin is:
\[ \boxed{2725} \]