Question

using the quadratic formula to solve $x^2 + 20 = 2x$, what are the values of $x$?
○ $1\pm\sqrt{21}i$
○ $-1\pm\sqrt{19}i$
○ $1\pm2\sqrt{19}i$
○ $1\pm\sqrt{19}i$

Explanation:

Step1: Rewrite in standard form

Rewrite \(x^2 + 20 = 2x\) as \(x^2 - 2x + 20 = 0\). Here, \(a = 1\), \(b = -2\), \(c = 20\).

Step2: Apply quadratic formula

Quadratic formula: \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Substitute values: \(x=\frac{-(-2)\pm\sqrt{(-2)^2 - 4(1)(20)}}{2(1)}=\frac{2\pm\sqrt{4 - 80}}{2}=\frac{2\pm\sqrt{-76}}{2}\).

Step3: Simplify the square root

Simplify \(\sqrt{-76}=\sqrt{4\times(-19)} = 2\sqrt{19}i\). Then \(x=\frac{2\pm2\sqrt{19}i}{2}=1\pm\sqrt{19}i\).

Answer:

\(1\pm\sqrt{19}i\) (corresponding to the option "1±√19i")