Question

How many real solutions does the following quadratic equation have?
4x^2+x+3=0
two real solutions
one real solution
no real solution
three real solutions

Answer 1

To determine how many real solutions the quadratic equation \( 4x^2 + x + 3 = 0 \) has, we can use the discriminant, which is given by the formula:

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

For the quadratic equation \( ax^2 + bx + c = 0 \), the coefficients are:
- \( a = 4 \)
- \( b = 1 \)
- \( c = 3 \)

Now, we can calculate the discriminant:

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

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

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

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