9514 1404 393
Answer:
obtuse
Step-by-step explanation:
The law of cosines tells you ...
b² = a² +c² -2ac·cos(B)
Substituting for a²+c² using the given equation, we have ...
b² = b²·cos(B)² -2ac·cos(B)
We can subtract b² to get a quadratic in standard form for cos(B).
b²·cos(B)² -2ac·cos(B) -b² = 0
Solving this using the quadratic formula gives ...
The fraction ac/b² is always positive, so the term on the right (the square root) is always greater than 1. The value of cos(B) cannot be greater than 1, so the only viable value for cos(B) is ...
The value of the radical is necessarily greater than ac/b², so cos(B) is necessarily negative. When cos(B) < 0, B > 90°. The triangle is obtuse.
Answer:
Step-by-step explanation:
Steps?
A graph shows zeros to be ±3. Factoring those out leaves the quadratic
(x-2)² +1
which has complex roots 2±i.
The function has roots -3, 3, 2-i, 2+i.
I'm not sure which numbers you are referring to, but 3 has the same absolute value as -3, and 7 has the same absolute value as -7, just for an example.