Can you give me the steps to the answer?
How many real number solutions are there to the equation 0 = –3x² + x – 4?
0
1
2
3
what is the surd?
b^2-4ac
1-(-3*(-4)=1-12=0-11 which means in the quadratic equation
x= (-b +-sqrt(b^2-4ac))/2a
you have two complex solutions, ie
-+sqrt(-11) in the solutions.
zero real solutions.
Thank you but what does 'surd' mean?
http://www.mathsisfun.com/surds.html
You may be more familiar with the term "discriminant."
The discriminant is b^2-4ac, the value inside the square root symbol.
So, in this case, the surd being considered is the square root of the discriminant.
Yes, thanks.
Sure! To find the number of real number solutions to the equation 0 = –3x² + x – 4, we can use the discriminant formula.
1. First, identify the coefficients of each term in the equation: a = -3, b = 1, and c = -4.
2. Then, calculate the discriminant using the formula: D = b² - 4ac. In this case, D = (1)² - 4(-3)(-4) = 1 - 48 = -47.
3. Next, analyze the value of the discriminant:
- If the discriminant is positive (D > 0), then there are two distinct real solutions.
- If the discriminant is zero (D = 0), then there is one real solution.
- If the discriminant is negative (D < 0), then there are no real solutions.
In the given equation, the discriminant is -47, which is negative. Therefore, there are no real number solutions to the equation 0 = –3x² + x – 4.
So, the correct answer is 0.