Answer:
2 real solutions
Step-by-step explanation:
We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:
b² - 4ac
If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.
Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:
b² - 4ac
(-20)² - 4 * 6 * 1 = 400 - 24 = 376
Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.
<em>~ an aesthetics lover</em>
Answer:F(x)=-x2-3
Step-by-step explanation:We are given a function:
The graph of is also shown in the given question figure.
It is a parabola with vertex at (0,0).
Sign of is positive, that is why the parabola opens up.
General equation of parabola is given as:
Here, in G(x), a = 1
Vertex (h,k) is (0,0).
As seen from the question figure,
The graph of F(x) opens down that is why it will have:
Sign of as negative. i.e.
And vertex is at (0,-3)
Putting the values of a and vertex coordinates,
Hence, the equation of parabola will become:
Answer:
x + (4/ x-2) + (2/ x-1)
Step-by-step explanation:
x + (6x/ x^2 + 2x - x -2)
x + (6x/ (x + 2) X (x - 1))
(6x/ (x + 2) X (x - 1))
(A/ x+2) + (B/ x-1)
(6x/ (x + 2) X (x - 1)) = (A/ x+2) + (B/ x-1)
6x = Ax + Bx - A + 2B
6x = (A+B)x + (-A+2b)
{0 = -A+2B
{6 = A+B
(A,B) = (4, 2)
(4/ x+2) + (2/ x-1)
x + (4/ x-2) + (2/ x-1)
