Are two obtuse triangles always, sometimes, or never similar
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! :)