First get rid of the exponent: 6x + 3 = 243. then subtract 3 from both sides: 6x = 240. lastly divide both side by 6: x = 40. be sure to check your work by plugging it back in.
Do recall that squaring and the *radical sign* cancel each other out... like so:(

)

= a
When you put it that way, it isn't enough :P
(

)

= a
(

)

=?
so you start with
(

)

=

8x+1=25 <-- subtract 1 to both sides
8x=24 <- divide 8 to both sides
x= 3
To find out if it's an extraneous solution ask yourself: It mustn't result in a radical that I like to call... 'illegal'. Plug it into the radicand 8x+1 and make sure you get something that is not a negative number.... so, DO you get a negative number when you plug in x = 3 into the radicand?
(extraneous solution is a invalid solution)
x=3 not extraneous
<span>To be both complex and imaginary, the answer would have to include "i" as well as have the form "a + bi". The first example, 3i - 5, fits. This would look to be the only example that does fit. The negative square root of 14 would be imaginary, but not complex.</span>
Any number raised to the power of 0 is 1.
Therefore, 15^0 = 1
Weird that the only choices are given as integrals with respect to

. (Integration along the

axis would be way easier, but whatever)
The fourth option should do it.