6x^2 - 2x + 1 is a quadratic formula from the form ax^2 + bx + c. This form of equation represents a parabola.
Finding 6x^2 - 2x + 1 = 0, means that you need to find the zeroes of the equation.
Δ = b^2 - 4ac
If Δ>0, the equation admits 2 zeroes and 6x^2 - 2x + 1 = 0 exists for 2 values of x.
If Δ<0, the equation doesn't admit any zero, and 6x^2 - 2x + 1 = 0 doesn't exist since the parabola doesn't intersect with the axe X'X
If Δ=0, the equation admits 1 zero, which means that the peak of the parabola is touching the axe X'X.
In 6x^2 - 2x + 1, a=6, b=-2, and c =1.
Δ= b^2 - 4ac
Δ=(-2)^2 - 4(6)(1)
Δ= 4 - 24
Δ= -20
Δ<0 so the parabola doesn't intersect with the Axe X'X, which means there's no solution for 6x^2 - 2x + 1 = 0.
I've added a picture of the parabola represented by this equation under the answer.
Hope this Helps! :)
Answer:
tanA/2.
Step-by-step explanation:
some trigonometric properties:
* tanA = sinA/cosA
* secA = 1/cosA
* cscA = 1/sinA
* cotA = cosA/sinA
1/sinA - cotA
1/sinA - cosA/sinA
1 - cosA / sinA
.... 1 - cosR / sinR = tan(R/2)
1 - cosA / sinA
= tanA/2.
Answer:
A) (-9/7,0), (3/8,0)
Step-by-step explanation:
Zeros of a quadratic function:
x for which y = 0.
In this question:

It's 0 if one of the factors is 0. 6 is never 0, now about the other to:



So (-9/7, 0) is a zero of the quadratic function.
The other is:



(3/8,0) is the other zero.
Thus, the correct answer is given by option A.