Question

solve for x in the equation $x^{2}-8x + 41 = 0$.\
\\(\bigcirc\\) $x = -4\pm\sqrt{37}i$\
\\(\bigcirc\\) $x = -4\pm5i$\
\\(\bigcirc\\) $x = 4\pm\sqrt{37}i$\
\\(\bigcirc\\) $x = 4\pm5i$

Explanation:

Step1: Identify quadratic coefficients

For $ax^2+bx+c=0$, here $a=1$, $b=-8$, $c=41$.

Step2: Use quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Step3: Calculate discriminant

$\Delta = (-8)^2-4(1)(41)=64-164=-100$

Step4: Compute square root of discriminant

$\sqrt{\Delta}=\sqrt{-100}=10i$

Step5: Substitute into formula

$x=\frac{8\pm10i}{2}=4\pm5i$

Answer:

$x = 4 \pm 5i$