Off the top of my head i can tell that B) 3 and A) -1 are two possible roots, you can plug it into a calculator and situate put if there is more
Answer:
16
Step-by-step explanation:
Let's represent this with an equation, where x is the first number. The next number must be x+2, because it has to be even, not x+1, and the rest continue the same way:
x+(x+2)+(x+4)+(x+6)+(x+8) = 100
Solve for x.
5x+20 = 100
5x = 80
<u>x = 16</u>
So, x = 13, x = √3 and x =7i.
now, recall that for an EVEN radical, there are two possible roots, namely is say √3 is say hmmm some value "a", that means that a*a = √3, however, -a*-a is also √3, therefore, ±√3 are two valid values, and therefore -√3 is another one.
now.... keep in mind that, complex solutions or roots, never come all by their lonesome, their sister is always with them, the conjugate, so, for 7i or namely 0 + 7i, her sister is always around, 0 - 7i, which is the other root.
