Having issues understanding fraction division when applying quadratic formula

Question

I'm trying to apply the quadratic formula, and having trouble understanding how:

$$\frac{-3 ± 3\sqrt{41} }{-18}$$

evaluates to

$$\frac{-1 ±\sqrt{41}}{-6}$$

and not

$$\frac{1}{6}±\frac{\sqrt{41}}{6}$$

$\frac{-3}{-18}$ should evaluate to $\frac{1}{6}$, so where does the $\frac{-1}{6}$ in the answer come from?

Thanks.

Answer 1

All three expressions are equivalent.

The second expression is obtained from the first by cancelling the common factor of $3$ from the numerator and denominator: $$\dfrac{-3 \pm 3\sqrt{41}}{-18} = \dfrac{3(-1 \pm 1 \sqrt{41})}{3 \times (-6)} = \dfrac{-1 \pm \sqrt{41}}{-6}$$

The third expression is obtained from the second by cancelling the common factor of $-1$ from the numerator and denominator, and then using the distributivity rule for division: $$\dfrac{-1 \pm \sqrt{41}}{-6} = \dfrac{(-1)(1 \mp \sqrt{41})}{(-1) \times 6} = \dfrac{1 \mp \sqrt{41}}{6} = \dfrac{1}{6} \mp \dfrac{\sqrt{41}}{6}$$

(My $\mp$ sign means the same thing as the $\pm$ sign; I'm just emphasising that the sign of the $\mp$ sign is the opposite of that of the $\pm$ sign in the chain of equations.)

Answer 2

Note that

$$\frac{-3 ± 3\sqrt{41} }{-18}=\frac{-3 }{-18}\pm\frac{3\sqrt{41} }{-18}=\frac16 \mp\frac{\sqrt{41} }{6}$$

Answer 3

Both versions are correct. $$ \frac{-3 ± 3\sqrt{41} }{-18} = \frac{3\cdot(-1 ± \sqrt{41})}{3\cdot(-6)} = \frac{-1 ± \sqrt{41}}{-6} = \frac{-1}{-6} \pm \frac{\sqrt{41}}{-6} = \frac{1}{6} \mp \frac{\sqrt{41}}{6}. $$

Answer 4

Both are equivalent.

$$\frac{-3\pm3\sqrt{41}}{-18} = \frac{3(-1 ±\sqrt{41})}{3(-6)} = \frac{-1 \pm\sqrt{41}}{-6} = \frac{-1}{-6} \pm \frac{\sqrt{41}}{-6} = \frac{1}{6}\mp\frac{\sqrt{41}}{6}$$

Since your solutions involve adding and subtracting $\frac{\sqrt{41}}{6}$, using $\mp$ or $\pm$ makes no difference. (The use of $\mp$ simply shows that the sign of $\frac{\sqrt{41}}{6}$ is opposite the sign of $\sqrt{41}$ in the original fraction.)