Does $\sqrt {x^2} = \lvert x\rvert$ Apply for Expressions $x$ with More than One Term?

Question

A proof for the quadratic formula, in my textbook had: $$\left( x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2}$$ For $b^2-4ac > 0,$ $$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$$ So it looks like the operation made to get from the first line to the second was to square both sides.

How come $x+\frac{b}{2a}$ is not $\lvert x+\frac{b}{2a}\rvert$, if $\sqrt {x^2} = \lvert x\rvert$? Does it just only apply when $x \in ℝ$?

Answer 1

Hint: Look at the plus and minus on the right hand side.

$$|x| = y \implies x = \pm y $$

Answer 2

Note that $$\left( x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2} \implies \left| x+\frac{b}{2a} \right| = \sqrt{\frac{b^2-4ac}{4a^2}} \implies x+\frac{b}{2a} = \pm \sqrt{\frac{b^2-4ac}{4a^2}}.$$

Answer 3

We DO have

$|x + \frac b{2a}| = |\sqrt {\frac {b^2 - 4ac}{4a^2}}|$

That means $\pm (x + \frac b{2a}) = \sqrt {\frac {b^2 - 4ac}{4a^2}}$

which means $x + \frac b{2a} = \pm \sqrt{\frac{b^2-4ac}{4a^2}}$

It's just a different (more practical and useful) way of saying the exact same thing.

$|x| = |y| \iff $ one of $x=y; x=-y;-x=y; -x =-y$ is true $\iff x = \pm y$.

Answer 4

It depends which way you are going...

$\sqrt {x^2} = |x|$

We have define the square-root operation to always be the positive root, even though there are two roots.

If we go in the opposite direction.

$x^2 = 4\\ x = \pm 2$

Both roots are valid and we need this plus / minus sign to reflect this.

However you will note in your example:

$\left( x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2}$

For $b^2-4ac > 0$

$x+\frac{b}{2a} = \frac{\pm\sqrt {b^2-4ac}}{2a}$

That line "For $b^2 - 4ac > 0$" is doing the work of an absolute value.

We might say

When $b^2-4ac = |b^2-4ac|$ we can take the square root...

and $|x + \frac {b}{2a}| = |\frac {\sqrt {b^2 - 4ac}} {2a}|$ has the same meaning as $x + \frac {b}{2a} = \pm \frac {\sqrt {b^2 - 4ac}} {2a}$

Answer 5

Your textbook is simply skipping a step that it is assuming the reader will understand. Notice that if you are given $$|a|=b$$ Then you can say that either $$a=b$$ or $$a=-b$$ or, to say both statements in one equation, $$a=\pm b$$ In the proof, you started with $$\bigg(x+\frac{b}{2a}\bigg)^2=\frac{b^2-4ac}{4a^2}$$ and here are the steps that the textbook skipped: $$\sqrt{\bigg(x+\frac{b}{2a}\bigg)^2}=\sqrt\frac{b^2-4ac}{4a^2}$$ $$\bigg|x+\frac{b}{2a}\bigg|=\sqrt\frac{b^2-4ac}{4a^2}$$ $$x+\frac{b}{2a}=\pm\sqrt\frac{b^2-4ac}{4a^2}$$ Do you understand now?