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# Basic tricks when taking the square root of equalities and inequalities

*schedule*Aug 11, 2023

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# Taking the square root of both sides in equalities

Consider the following equation:

Where $x\in\mathbb{R}$ is some variable to solve for and $a\in\mathbb{R}$ is a non-negative number. The solution of this equation is given by:

In other words:

Intuition. Consider the following equation:

Without doing any algebraic manipulation, what would the solution be? Clearly, there are two possible solutions:

If we naively take the square root of \eqref{eq:MogqneHw1YhZtc2nbry} without any thought, we only get positive solution:

To also get the second solution, we must take the absolute value of the left side:

This is the reason why we must never forget to include the absolute value when taking the square root of squared variables!

# Taking the square root of both sides of a squared less than b squared

If $a\in\mathbb{R}$ and $b\in\mathbb{R}$, then:

Intuition. In order to prove this, we can show that the result holds for all 4 possible cases:

$a$ and $b$ are both positive.

$a$ is positive and $b$ is negative.

$a$ is negative and $b$ is positive.

$a$ and $b$ are both negative.

Instead of sticking with abstract variables $a$ and $b$, let's make things more concrete by using actual numbers. For the case when $a$ and $b$ are both positive, consider the following inequality:

For the case when $a$ is positive and $b$ is negative:

Notice how the absolute value is required in this case.

For the case when $a$ is negative and $b$ is positive:

Here, the absolute value is not required since $-3\lt5$ is still true, but we can make a stronger claim by including the absolute $\vert{-3}\vert\lt5$. This point is important because many mathematical proofs use this stronger claim.

Finally, for the case when $a$ and $b$ are both negative:

We see that in all 4 cases, our theorem holds true!

# Taking the square root of both sides of a squared less than b

If $a\in\mathbb{R}$ is any number and $b\in\mathbb{R}$ is a non-negative number, then:

Proof. Recall from the previous theorem that:

Where a is any number and c is a non-negative number. Let's define a new variable $b=c^2$. Since $b$ is the square of $c$, we have that $b$ must also be a non-negative number. We substitute $b=c^2$ above to get:

Again, because $b$ is a non-negative number, $\sqrt{b}$ must also be non-negative. This allows us to get rid of the absolute value:

Remember, $b$ is a non-negative number so we've managed to prove the theorem!