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

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

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Answer 1 · 2017-05-11T22:25:32

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

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

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

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

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

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

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