how many real number equations does this equation have?
0= -7x^2+6x+3
A. 1 solution*
B. 2 solutions
C. no solutions
D. infintie solutions
should be "how many real number equations..."
Recall the determinant: b^2-4ac
If it is zero, there is a single root. Here, the determinant is 6^2+4(-7)(3) = 120
So, your choice is incorrect.
FYI, regarding the determinant:
positive: 2 roots
zero: 1 root
negative: no real roots
This is because in the quadratic formula the roots are
x = (-b±√(b^2-4ac))/2a
so, if the determinant is positive, you are adding/subtracting a real nonzero value from -b/2a
you can see what happens if it is zero or negative.
It's called the discriminant because it discriminates among the kinds of roots.
b² - 4 a c
is the discriminant not determinant.
thanks for the catch. I also failed to correct his "equations" for "solutions"
Must have been a Freudian slip
(that's when you say one thing, but mean your mother)
To determine how many solutions an equation has, we can use the quadratic formula:
The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / 2a
Let's apply this formula to the given equation, 0 = -7x^2 + 6x + 3.
Comparing this with the standard quadratic form ax^2 + bx + c = 0, we have:
a = -7
b = 6
c = 3
Plugging these values into the quadratic formula, we get:
x = (-6 ± √(6^2 - 4(-7)(3))) / (2*(-7))
x = (-6 ± √(36 + 84)) / (-14)
x = (-6 ± √120) / (-14)
x = (-6 ± √(4 * 30)) / (-14)
x = (-6 ± 2√30) / (-14)
Simplifying further, we have:
x = (3 ± √30) / 7
Since these are two distinct solutions, the equation has 2 solutions.
Therefore, the correct answer is B. 2 solutions.