When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$?

Question

The solution to the quadratics is given by

$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and $\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}$ if $x$ and $y$ are negative, but given that we have $a^2$, is this justified?

Is $\sqrt{x/y^2} = \sqrt{x}/y$? Is $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ always true?

Answer 1

The two possible values of $\sqrt{4a^2}$ for complex $a$ differ only by a sign (one is $2a$, the other is $-2a$), so the set of solutions given by the formula $$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$$ is the same as that given by the formula $$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}.$$

(The "plus" in one formula may or may not correspond to "plus" in the other one, but this doesn't matter.)

Answer 2

When $y> 0$, you have

$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{y}.$$

When $y<0$, you have

$$\sqrt{\frac x{y^2}}=\frac{\sqrt{x}}{\sqrt{y^2}}=\frac{\sqrt{x}}{-y}.$$

Answer 3

Since, in general, for all $\;z\in\Bbb R\;,\;\;\sqrt{z^2}=|z|\;$, we have that

$$\sqrt{y^2}=y\iff y\ge 0$$

and thus

$$\sqrt\frac x{y^2}=\frac{\sqrt x}y\iff y>0\;\;\text{(because clearly it can't be in this case}\;\;y=0)$$

Answer 4

$$\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$$

Is "the solution" to $ax^2 + bx + c$, provided that $a,b,c$ are real numbers and $a \ne 0$.

$4a^2 \ge 0$ for all real numbers $a$. Since we are excluding $a = 0$, then $4a^2 > 0$. We can then say

$$\sqrt{4a^2} = \begin{cases} 2a, & \text{if $a \ge 0$} \\ -2a, & \text{if $a < 0$} \end{cases}$$

The real problem has to do with $b^2 - 4ac$ since it can be positive, negative, or $0$. But, in all three cases, we will have

$$\sqrt{\dfrac{b^2-4ac}{4a^2}}= \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$

so we end up with

\begin{align} \dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}} &= \dfrac{b}{2a}\pm \left( \pm \dfrac{\sqrt{b^2-4ac}}{2a} \right) \\ &= \dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a} \\ \end{align}

By the way, for any real number $x$, $\sqrt{x^2} = |x|$.

Answer 5

That's a good question.

But for the sake of the quadratic equation we don't need $\sqrt{x/y^2} = \sqrt{x}/y$. We just need $\sqrt{x/y^2} = \pm \sqrt{x}/y$.

If $y^2 = c \ne 0$ then $y = \{x_1,x_2\}$ where $x_1 = -x_2$. This is true for all $c$, real or complex. Arbitrarily declare one of them, say $x_1$, to be the $\sqrt{c}$ then we can always have $\sqrt{y^2} = \pm y$. Always.

(If $y^2 = 0 \implies y = 0$ and $0 = \pm 0$.)

If $y^2 = c = r*e^{i\theta} \in \mathbb C$ then $y = \{\sqrt{r}e^{i(\theta/2},\sqrt{r}e^{i\theta/2 + \pi}\}$ = $\pm \sqrt{r} e^{i\theta/2}$

So $\sqrt{x/y^2} = \pm \sqrt{x}/y$. So equation is good even for complex numbers.