Q. For the following quadratic equation, determine the nature of the roots. The equation is \(x^2 + 10x + 25 = 0\).

Problem

Determine the nature of the roots of the quadratic equation:

\[x^{2} + 10x + 25 = 0\]

1. Identify the coefficients.For a quadratic in the form \(ax^{2} + bx + c = 0\), the coefficients are:

   \(a = 1,\; b = 10,\; c = 25\).

2. Compute the discriminant.The discriminant is defined by \(\Delta = b^{2} – 4ac\). Compute it using the coefficients:

   \[\Delta = 10^{2} – 4 \cdot 1 \cdot 25 = 100 – 100 = 0.\]

3. Interpret the discriminant.The sign of \(\Delta\) determines the nature of the roots:
   • If \(\Delta > 0\), two distinct real roots.
   • If \(\Delta = 0\), one real repeated root (a double root).
   • If \(\Delta < 0\), two complex conjugate roots.

   Here \(\Delta = 0\), so the quadratic has one real repeated root (a double root).

4. Find the root explicitly (optional verification).Using the quadratic formula:

   \[x = \dfrac{-b \pm \sqrt{\Delta}}{2a} = \dfrac{-10 \pm \sqrt{0}}{2 \cdot 1} = \dfrac{-10}{2} = -5.\]

   Alternatively, factor the quadratic: \((x + 5)^{2} = x^{2} + 10x + 25\), so \((x + 5)^{2} = 0\) gives \(x = -5\) with multiplicity 2.

Conclusion: The equation has one real repeated root (a double root): \(x = -5\).