Question

how many real solutions does the following quadratic equation have 4x^2 + x + 3 = 0

Answer 1

To determine how many real solutions a quadratic equation has, we can use the discriminant. For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant \( D \) is given by:

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

For the equation \( 4x^2 + x + 3 = 0 \), we have:

- \( a = 4 \)
- \( b = 1 \)
- \( c = 3 \)

Now, we can calculate the discriminant:

\[
D = (1)^2 - 4(4)(3) = 1 - 48 = -47
\]

Since the discriminant \( D \) is negative (\(-47\)), it indicates that the quadratic equation has no real solutions. Thus, the equation \( 4x^2 + x + 3 = 0 \) has **0 real solutions**.