How many real number solutions are there to the equation 0=-3x^2+x-4
A) 0*****
B) 1
C) 2
D) 3
Help on how to solve this.
Thank You!
To find the number of real number solutions to the equation 0 = -3x^2 + x - 4, we can use the discriminant. The discriminant tells us how many solutions there are based on the value of the quadratic equation.
The discriminant formula for a quadratic equation ax^2 + bx + c = 0 is given by: Δ = b^2 - 4ac
If the discriminant (Δ) is positive, there are two real solutions. If Δ is equal to zero, there is one real solution. And if Δ is negative, there are no real solutions.
In the given equation, a = -3, b = 1, and c = -4.
Let's calculate the discriminant: Δ = (1)^2 - 4(-3)(-4)
= 1 - 48
= -47
Since the discriminant is negative (Δ = -47), there are no real solutions to the equation.
So the correct answer is A) 0.
recall the discriminant of a quadratic: b^2-4ac
Here, that is 1^2-4(-3)(-4) = 1+48 = 49
since it is positive, there are 2 real roots.