How do you evaluate #(8x - 2) ^ { 2} + 8= - 1#?

1 Answer
Aug 13, 2017

Answer:

See below.

Explanation:

#(8x - 2) ^ { 2} + 8= - 1#

#(8x - 2) ^ { 2} = - 9#

If imaginary numbers are not accepted (assuming all real numbers), then there are no real solutions to this equation, as the square of a real number is always greater than or equal to zero (since #-9# is less than #0#, we cannot solve the equation).

If we accept imaginary numbers, then we continue as follows:

#(8x - 2) ^ { 2} = - 9#

#8x-2=pmsqrt(-9)=pm3i#

#8x=pm3i+2#

#x=(pm3i+2)/8#