Question

How many real solutions does the following quadratic equation have?

4x2+x+3=0

Answer 1

dude that cant be right

Answer 2

To determine the number of real solutions for the quadratic equation \( 4x^2 + x + 3 = 0 \), we can use the discriminant. The discriminant \( D \) of a quadratic equation in the form \( ax^2 + bx + c = 0 \) is given by:

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

For the given equation, the coefficients are:

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

Now, we can substitute these values into the discriminant formula:

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

The discriminant \( D \) is negative (\( -47 \)).

In general, the number of real solutions to a quadratic equation is determined by the value of the discriminant:

- If \( D > 0 \), there are two distinct real solutions.
- If \( D = 0 \), there is exactly one real solution (a repeated root).
- If \( D < 0 \), there are no real solutions (the solutions are complex).

Since \( D = -47 < 0 \), the quadratic equation \( 4x^2 + x + 3 = 0 \) has no real solutions.

Thus, the final answer is that the equation has **0 real solutions**.