How many real solutions does the following quadratic equation have

4x^2 + x + 3 = 0

1 answer

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

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

In our equation, the coefficients are:

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

Now, let's calculate the discriminant:

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

Since the discriminant \(D\) is negative (\(D < 0\)), this means that the quadratic equation has no real solutions; instead, it has two complex solutions.

Thus, the answer is that the equation \(4x^2 + x + 3 = 0\) has 0 real solutions.