Answer:
[tex]a=\pm i\sqrt{26}}[/tex]
Step-by-step explanation:
[tex]6^2+a^2=10\\\\36+a^2=10\qquad\text{subtract 36 from both sides}\\\\a^2=-26<0\\\\\bold{NO\ REAL\ SOLUTIONS}\\\\\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\a^2=-26\to a=\pm\sqrt{-26}\\\\a=\pm\sqrt{(-1)(26)}\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\a=\pm\sqrt{-1}\cdot\sqrt{26}\\\\a=\pm i\sqrt{26}[/tex]
