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Basic tricks when taking the square root of equalities and inequalities

schedule Aug 11, 2023
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Theorem.

Taking the square root of both sides in equalities

Consider the following equation:

$$x^2=a$$

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:

$$|x|=a$$

In other words:

$$x^2=a \;\;\;\;\;\;\Longleftrightarrow\;\;\;\;\;\; |x|=\sqrt{a}$$

Intuition. Consider the following equation:

$$$$\label{eq:MogqneHw1YhZtc2nbry} x^2=25$$$$

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

$$x=-5 \;\;\;\;\;\;\text{or}\;\;\;\;\;\; x=5$$

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

$$\sqrt{x^2}=\sqrt{25} \;\;\;\;\;\;\;\Longleftrightarrow\;\;\;\;\;\;\; x=5$$

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

$$\sqrt{x^2}=\sqrt{25} \;\;\;\;\;\;\;\Longleftrightarrow\;\;\;\;\;\;\; |{x}|=5 \;\;\;\;\;\;\;\Longleftrightarrow\;\;\;\;\;\;\; x=-5\;\;\;\;\;\;\text{or}\;\;\;\;\;\;x=5$$

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

Theorem.

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:

$$a^2\lt{b}^2 \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{|b|}$$

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:

$$3^2\lt5^2 \implies |3|\lt|5|$$

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

$$(3)^2\lt(-5)^2\implies |3|\lt|-5|$$

Notice how the absolute value is required in this case.

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

$$(-3)^2\lt(5)^2\implies |-3|\lt|5|$$

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:

$$(-3)^2\lt(-5)^2\implies |-3|\lt|-5|$$

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

Theorem.

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:

$$a^2\lt{b} \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{\sqrt{b}}$$

Proof. Recall from the previous theorem that:

$$a^2\lt{c}^2 \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{|c|}$$

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:

$$a^2\lt{b} \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{|\sqrt{b}|}$$

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:

$$a^2\lt{b} \;\;\;\;\;\implies\;\;\;\;\; |a|\lt{\sqrt{b}}$$

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

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