How many real solutions does the following quadratic equation have?: 25x^2 + 60x + 36 = 0

1 answer

To determine the number of real solutions of the quadratic equation \( 25x^2 + 60x + 36 = 0 \), we need to calculate the discriminant, which is given by the formula:

\[ D = b^2 - 4ac \]

where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \).

For the equation \( 25x^2 + 60x + 36 = 0 \):

  • \( a = 25 \)
  • \( b = 60 \)
  • \( c = 36 \)

Now we compute the discriminant \( D \):

\[ D = b^2 - 4ac = 60^2 - 4 \cdot 25 \cdot 36 \]

Calculating \( b^2 \):

\[ 60^2 = 3600 \]

Calculating \( 4ac \):

\[ 4 \cdot 25 \cdot 36 = 100 \cdot 36 = 3600 \]

Now substituting these values back into the discriminant formula:

\[ D = 3600 - 3600 = 0 \]

Since the discriminant \( D = 0 \), this indicates that the quadratic equation has exactly one real solution (a double root).

Thus, the number of real solutions to the equation \( 25x^2 + 60x + 36 = 0 \) is:

\[ \boxed{1} \]