Question

Solve for x in the equation x squared minus 8 x + 41 = 0.
x = negative 4 plus-or-minus StartRoot 37 EndRoot i
x = –4 ± 5i
x = 4 plus-or-minus StartRoot 37 EndRoot i
x = 4 ± 5i

Answer 1

x = 4 ± 5i

Option (4) is correct.

Step-by-step explanation:

The given equation is

$x^2-8x+41=0$

By Sridharacharya's rule

$x=\frac{-(-8)\pm \sqrt{(-8)^2-4\times 1\times 41}}{2\times 1}$

$\implies x=\frac{8\pm\sqrt{64-164}}{2}$

$\implies x=\frac{8\pm\sqrt{-100}}{2}$

$\implies x=\frac{8\pm10\sqrt{-1}}{2}$

$\implies x=\frac{8\pm 10i}{2}$     (∵ $\sqrt{-1}=i$)

$\implies x=4\pm5i$

Hope this answer is helpful.

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