Answer
True
Work Step by Step
We will solve the equation using the property which says that if $x^{2}=a$, then $x=\pm\sqrt a$.
Step 1: $(3x-4)^{2}+6=4$
Step 2: $(3x-4)^{2}=4-6$
Step 3: $(3x-4)^{2}=-2$
Step 4: $3x-4=\pm \sqrt {-2}$
Step 5: $3x-4=\pm \sqrt {-1\times2}$
Step 6: $3x-4=\pm (\sqrt {-1}\times\sqrt {2})$
Step 7: $3x-4=\pm (i\times\sqrt {2})$ [as $i=\sqrt {-1}$]
Step 8: $3x-4=\pm (i\sqrt {2})$
Step 9: $3x=4\pm (i\sqrt {2})$
Step 10: $x=\frac{4+i\sqrt {2}}{3}$ or $x=\frac{4-i\sqrt {2}}{3}$
Therefore, the solution set is {$\frac{4-i\sqrt {2}}{3},\frac{4+i\sqrt {2}}{3}$}. So, the statement is true and the solution set consists of two, non-real complex numbers.
