Question

What are the solution of x^2-2x+17=0

Answer 1

x = 1 - 4i or x = 1 + 4i

Explanation:

`x^2-2x+17=0\qquad\text{subtract 17 from both sides}\\\\x^2-2x=-17\\\\x^2-2(x)(1)=-17\qquad\text{add}\ 1^2\ \text{to both sides}\\\\x^2-2(x)(1)+1^2=-17+1^2\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-1)^2=-17+1\\\\(x-1)^2=-16<0\Rightarrow\boxed{\text{NO REAL SOLUTION}}\ because\ x^2\geq0\\\\\text{In the set of complex numbers}\\\\i=√(-1)\\\\(x-1)^2=-16\iff x-1=\pm√(-16)\\\\x-1=-√((16)(-1))\ \vee\ x-1=√((16)(-1))\\\\x-1=-√(16)\cdot√(-1)\ \vee\ x-1=√(16)\cdot\sqrt{-1)`

`x-1=-4i\ \vee\ x-1=4i\qquad\text{add 1 to both sides}\\\\x=1-4i\ \vee\ x=1+4i`