Answer:
-6i
Step-by-step explanation:
Complex roots always come in pairs, and those pairs are made up of a positive and a negative version. If 6i is a root, then its negative value, -6i, is also a root.
If you want to know the reasoning, it's along these lines: to even get a complex/imaginary root, we take the square root of a negative value. When you take the square root of any value, your answer is always "plus or minus" whatever the value is. The same thing holds for complex roots. In this case, the polynomial function likely factored to f(x) = (x+8)(x-1)(x^2+36). To solve that equation, you set every factor equal to zero and solve for the x's.
x + 8 = 0
x = -8
x - 1 = 0
x = 1
x^2 + 36 = 0
x^2 = -36 ... take the square root of both sides to get x alone
x = √-36 ... square root of an imaginary number produces the usual square root and an "i"
x = ±6i
<span>–122 < –3(–2 – 8x) – 8x
-122< 6 + 24x - 8x
-122 < 6 + 16x
-122 - 6 < 16x
-128<16x
-128/16<x
-8<x
x>-8
answer is </span>B. x > –8
Answer:
Step-by-step explanation:
Define two points on the line:
Use the slope formula and the defined points to find the slope:
Substitute the found slope and one of the points into the point-slope form of a linear equation:
Simplify:
