How many real solutions does the following quadratic equation have?
4x^2 + x + 3 = 0
A. two real solutions
B. no real solutions
C. three real solutions
D. one real solution
Using the quadratic formula, we can find the discriminant of the quadratic equation. The discriminant is calculated as $b^2 - 4ac$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation. In this case, $a = 4$, $b = 1$, and $c = 3$, so the discriminant is $1^2 - 4(4)(3) = 1 - 48 = -47$. Since the discriminant is negative, the quadratic equation has no real solutions. Therefore, the correct answer is $\boxed{\text{B. no real solutions}}$.
To find the number of real solutions for a quadratic equation, we can use the discriminant. The discriminant is the part of the quadratic formula under the square root sign, which is b^2 - 4ac.
For the given equation, 4x^2 + x + 3 = 0, we can observe that:
a = 4
b = 1
c = 3
Now, let's calculate the discriminant:
Discriminant = b^2 - 4ac
Discriminant = (1)^2 - 4(4)(3)
Discriminant = 1 - 48
Discriminant = -47
Since the discriminant is negative (-47), there are no real solutions for the quadratic equation.
Therefore, the answer is B. no real solutions.
To determine the number of real solutions of a quadratic equation, we can use the discriminant formula. The discriminant is calculated as b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
In this case, for the equation 4x^2 + x + 3 = 0, we have a = 4, b = 1, and c = 3. Now let's plug these values into the discriminant formula:
Discriminant = b^2 - 4ac
Discriminant = (1)^2 - 4(4)(3)
Discriminant = 1 - 48
Discriminant = -47
Since the discriminant is negative (-47), there are no real solutions to the quadratic equation 4x^2 + x + 3 = 0.
Therefore, the answer is B. no real solutions.