Please help me with this I need help.
2 answers:
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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! :)
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! :)
The set of equation that is known to have 5 and 0 as solutions is A and B.
<h3>How to solve for the system of equations.</h3>From the graph that we have here in this question, we are supposed to identify the equations that have this point.
We can do this by tracing the lines in order to locate 5 and 0 on the grid.
When we trace the bine that has 5 and 0, We would find out that it is traceable to A and B.
Read more on graphs here:
#SPJ1
