Question

How many real solutions does the following quadratic equation have?

25x2+60x+36=0


(1 point)
solution(s)
Remaining Attempts : 3

Answer 1

To determine the number of real solutions for the quadratic equation \( 25x^2 + 60x + 36 = 0 \), we can use the discriminant method. The discriminant \( D \) of a quadratic equation of the form \( ax^2 + bx + c = 0 \) is given by:

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

In this case, the coefficients are:
- \( a = 25 \)
- \( b = 60 \)
- \( c = 36 \)

Now, let's calculate the discriminant:

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

Calculating \( 60^2 \):

\[
60^2 = 3600
\]

Calculating \( 4 \cdot 25 \cdot 36 \):

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

Now, substitute these values back into the discriminant:

\[
D = 3600 - 3600 = 0
\]

The discriminant \( D \) is \( 0 \).

According to the properties of the discriminant:
- If \( D > 0 \), there are 2 distinct real solutions.
- If \( D = 0 \), there is exactly 1 real solution (a repeated root).
- If \( D < 0 \), there are no real solutions.

Since \( D = 0 \), the quadratic equation \( 25x^2 + 60x + 36 = 0 \) has **exactly 1 real solution**.