Face or base if you are looking for a word
Recall that given the equation of the second degree (or quadratic)
ax ^ 2 + bx + c
Its solutions are:
x = (- b +/- root (b ^ 2-4ac)) / 2a
discriminating:
d = root (b ^ 2-4ac)
If d> 0, then the two roots are real (the radicand of the formula is positive).
If d = 0, then the root of the formula is 0 and, therefore, there is only one solution that is real and of multiplicity 2 (it is a double root).
If d <0, then the two roots are complex and, in addition, one is the conjugate of the other. That is, if one solution is x1 = a + bi, then the other solution is x2 = a-bi (we are assuming that a, b, c are real).
One solution:
A cut point with the x axis
Two solutions:
Two cutting points with the x axis.
Complex solutions:
Does not cut to the x axis
Answer:
2x = 120°, x = 60°, y = 60°
Step-by-step explanation:
2x + x = 180° (eternal angles in a parrallel lines add up to 180°)
3x = 180. divide by 3 in both sides.
x = 60
therefore:
2x = 2(60) = 120°
x = 60°
y = x (crossponding).
if x = 60°, then y = 60°
Done :D
