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cc.2.1.hs.f.7 solve for x: $x^2 - 4x + 5 = 0$ $\\bigcirc\\ 2 \\pm i$ $\…

Question

cc.2.1.hs.f.7
solve for x:
$x^2 - 4x + 5 = 0$
$\bigcirc\\ 2 \pm i$
$\bigcirc\\ 2 \pm 2i$
$\bigcirc\\ 4 \pm i$
$\bigcirc\\ 2 \pm 3i$

Explanation:

Step1: Identify coefficients

For quadratic equation \(ax^2 + bx + c = 0\), here \(a = 1\), \(b = -4\), \(c = 5\).

Step2: Use quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\)

First, calculate discriminant \(D = b^2 - 4ac = (-4)^2 - 4\times1\times5 = 16 - 20 = -4\).

Step3: Simplify square root of discriminant

\(\sqrt{D}=\sqrt{-4}=2i\) (since \(\sqrt{-1}=i\)).

Step4: Substitute into quadratic formula

\(x=\frac{-(-4)\pm2i}{2\times1}=\frac{4\pm2i}{2}=2\pm i\).

Answer:

\(2\pm i\) (the option: \(2\pm i\))