Simplifying Algebraic Expressions 2

<p><span style="font-size:22px"><a href="https://www.prepswift.com/quizzes/quiz/prepswift-simplifying-algebraic-expressions-2" target="_blank">Simplifying Algebraic Expressions 2 Exercise</a></span></p><p>Sometimes we can simplify complex roots like this:</p> <p>$\sqrt{y^4-10y^2+25}$</p> <p>We can use Identity 3 to recognize this is the same as $\sqrt{(y^2-5)^2}$ which is just $|y^2-5|$. We can only cancel the square root if we put the absolute value symbols ($|$ )around the result.</p> <p>Some complex fractions can be simplified too, but they take a bit of thinking:</p> <p>$\frac{2}{a^2-25}-\frac{1}{a^2+5a}$&nbsp;</p> <p>We can start by factoring out the denominators using our identities:</p> <p>$\frac{2}{(a+5)(a-5)}-\frac{1}{a(a+5)}$</p> <p>To get a common denominator (and cancel out what they don&#39;t share in common) we need to multiply the left by $\frac{a}{a}$ and the right by $\frac{a-5}{a-5}$. This will give both fractions a denominator of $a(a-5)(a+5)$:</p> <p>$\frac{2a}{a(a-5)(a+5)}-\frac{a-5}{a(a-5)(a+5)} = \frac{2a-a+5}{a(a-5)(a+5)} = \frac{a+5}{a(a-5)(a+5)}$</p> <p>Finally, we can cancel out the $a+5$ to get:</p> <p>$\frac{1}{a(a-5)}$</p> <p>Note that $a$ cannot equal $5$, $-5$, or $0$!</p> <blockquote> <p>Why does $-5$ not work?</p> <p>And indeed, it does when we look at the final expression&nbsp;$\frac{1}{a(a-5)}$. However, look at the&nbsp;<em>original</em>&nbsp;expression instead. Does $-5$ work in that case? No.</p> <p>And why? Because when you cancel out the $a + 5$, you&#39;re implicitly assuming that $a \neq -5$. Otherwise the domain of the expression will change - that is, values that could not work (such as $a = -5$) suddenly become allowed. And that&#39;s not OK - we simplify to make things easier for us, but the expression must functionally be&nbsp;<u>the same thing</u> (unless you know what you&#39;re doing - but that&#39;s not the subject of this discussion).&nbsp;<br /> <br /> Another simple example is to consider the equation $x^2 = x$. A common mistake is to divide by $x$ on both sides, making the result $x = 1$. But notice that $x = 0$ works in the original equation! The reason is similar - when dividing by $x$, you&#39;re assuming that $x \neq 0$ when that need not be the case.&nbsp;<br /> <br /> (We&#39;ll cover the concept of a domain in a later video/entry)</p> </blockquote>

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