How do you solve x^2-10x+41=0?

May 11, 2017

Original question: Solve ${x}^{2} - 10 x + 41 = 0$

Since the equation cannot be easily factored, we must use the quadratic equation:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ where $a , b , c$ are from the standard form of a quadratic, $f \left(x\right) = a {x}^{2} + b x + c$

In this question, $a = 1 , b = - 10 , c = 41$, which we can substitute into the quadratic equation and simplify:
$x = \frac{- \left(- 10\right) \pm \sqrt{{\left(- 10\right)}^{2} - 4 \cdot 1 \cdot 41}}{2 \cdot 1}$
$x = \frac{10 \pm \sqrt{100 - 164}}{2}$
$x = \frac{10 \pm \sqrt{- 64}}{2}$

Since we cannot square root a negative number, we use $i$ to denote the imaginary unit, $i = \sqrt{- 1}$, and continue simplifying:
$x = \frac{10 \pm 8 i}{2}$
$x = 5 \pm 4 i$

Therefore, our solutions are:
$x = 5 + 4 i , 5 - 4 i$