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The property $\sqrt{\text{ab}}$ = $\sqrt{a}$ $\sqrt{b}$ can be used to simplify radicals. The key is to break the number inside the root into two factors, one of which is a perfect square .
$\sqrt{\text{75}}$ | |
$=$ $\sqrt{25\u20223}$ | because 25•3 is 75, and 25 is a perfect square |
$=$ $\sqrt{25}$ $\sqrt{3}$ | because $\sqrt{\text{ab}}$ $=$ $\sqrt{a}$ $\sqrt{b}$ |
$=5$ $\sqrt{3}$ | because $\sqrt{25}$ =5 |
So we conclude that $\sqrt{\text{75}}$ =5 $\sqrt{3}$ . You can confirm this on your calculator (both are approximately 8.66).
We rewrote 75 as $25\u20223$ because 25 is a perfect square. We could, of course, also rewrite 75 as $5\u202215$ , but—although correct—that would not help us simplify, because neither number is a perfect square.
$\sqrt{180}$ | |
$=$ $\sqrt{\mathrm{9\u202220}}$ | because $9\u202220$ is 180, and 9 is a perfect square |
$=$ $\sqrt{9}$ $\sqrt{20}$ | because $\sqrt{\text{ab}}$ $=$ $\sqrt{a}$ $\sqrt{b}$ |
$=3$ $\sqrt{20}$ | So far, so good. But wait! We’re not done! |
$=3$ $\sqrt{\mathrm{4\u20225}}$ | There’s another perfect square to pull out! |
$=3$ $\sqrt{4}$ $\sqrt{5}$ | |
$=3\left(2\right)$ $\sqrt{5}$ | |
$=6$ $\sqrt{5}$ | Now we’re done. |
The moral of this second example is that after you simplify, you should always look to see if you can simplify again .
A secondary moral is, try to pull out the biggest perfect square you can. We could have jumped straight to the answer if we had begun by rewriting 180 as $36\u20225$ .
This sort of simplification can sometimes allow you to combine radical terms, as in this example:
$\sqrt{\text{75}}$ $\u2013$ $\sqrt{\text{12}}$ | |
$=5$ $\sqrt{3}$ $\mathrm{\u20132}$ $\sqrt{3}$ | We found earlier that $\sqrt{\text{75}}$ $=5$ $\sqrt{3}$ . Use the same method to confirm that $\sqrt{\text{12}}$ $=2$ $\sqrt{3}$ . |
$=3$ $\sqrt{3}$ | 5 of anything minus 2 of that same thing is 3 of it, right? |
That last step may take a bit of thought. It can only be used when the radical is the same. Hence, $\sqrt{2}$ + $\sqrt{3}$ cannot be simplified at all. We were able to simplify $\sqrt{\text{75}}$ – $\sqrt{\text{12}}$ only by making the radical in both cases the same .
So why does $5$ $\sqrt{3}$ $\mathrm{\u20132}$ $\sqrt{3}$ $=3$ $\sqrt{3}$ ? It may be simplest to think about verbally: 5 of these things, minus 2 of the same things, is 3 of them. But you can look at it more formally as a factoring problem, if you see a common factor of $\sqrt{3}$ .
$5$ $\sqrt{3}$ $\mathrm{\u20132}$ $\sqrt{3}$ $=$ $\sqrt{3}$ $(5\u20132)=$ $\sqrt{3}$ $\left(3\right)$ .
Of course, the process is exactly the same if variable are involved instead of just numbers!
${x}^{\frac{3}{2}}+{x}^{\frac{5}{2}}$ | |
$={x}^{3}+{x}^{5}$ | Remember the definition of fractional exponents! |
$=\sqrt{{x}^{2}*x}+\sqrt{{x}^{4}*x}$ | As always, we simplify radicals by factoring them inside the root... |
$\sqrt{{x}^{2}}*\sqrt{x}+\sqrt{{x}^{4}}*\sqrt{x}$ | and then breaking them up... |
$=x\sqrt{x}+{x}^{2}\sqrt{x}$ | and then taking square roots outside! |
$=({x}^{2}+x)\sqrt{x}$ | Now that the radical is the same, we can combine. |
It is always possible to express a fraction with no square roots in the denominator.
Is it always desirable? Some texts are religious about this point: “You should never have a square root in the denominator.” I have absolutely no idea why. To me, $\frac{1}{\sqrt{2}}$ looks simpler than $\frac{\sqrt{2}}{2}$ ; I see no overwhelming reason for forbidding the first or preferring the second.
However, there are times when it is useful to remove the radicals from the denominator: for instance, when adding fractions. The trick for doing this is based on the basic rule of fractions: if you multiply the top and bottom of a fraction by the same number, the fraction is unchanged. This rule enables us to say, for instance, that $\frac{2}{3}$ is exactly the same number as $\frac{2\cdot 3}{3\cdot 3}$ = $\frac{6}{9}$ .
In a case like $\frac{1}{\sqrt{2}}$ , therefore, you can multiply the top and bottom by $\sqrt{2}$ .
$\frac{1}{\sqrt{2}}$ = $\frac{1*2}{\sqrt{2}*\sqrt{2}}$ = $\frac{\sqrt{2}}{2}$
What about a more complicated case, such as $\frac{\sqrt{\text{12}}}{1+\sqrt{3}}$ ? You might think we could simplify this by multiplying the top and bottom by $(1+$ $\sqrt{3}$ ), but that doesn’t work: the bottom turns into ${(1+3)}^{2}$ $=1+2$ $\sqrt{3}$ $+3$ , which is at least as ugly as what we had before.
The correct trick for getting rid of $(1+$ $\sqrt{3}$ ) is to multiply it by $(1\u2013$ $\sqrt{3}$ ). These two expressions, identical except for the replacement of $\mathrm{a+}$ by $\mathrm{a-}$ , are known as conjugates . What happens when we multiply them? We don’t need to use FOIL if we remember that
$(x+y)(x-y)={x}^{2}-{y}^{2}$
Using this formula, we see that
$(1+\sqrt{3})(1-\sqrt{3})={1}^{2}-{\left(\sqrt{3}\right)}^{2}=1-3=-2$
So the square root does indeed go away. We can use this to simplify the original expression as follows.
$\frac{\sqrt{\text{12}}}{1+\sqrt{3}}=\frac{\sqrt{12}(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{\sqrt{12}-\sqrt{36}}{1-3}=\frac{2\sqrt{3}-6}{-2}=-\sqrt{3}+3$
As always, you may want to check this on your calculator. Both the original and the simplified expression are approximately 1.268.
Of course, the process is the same when variables are involved.
$\frac{1}{x-\sqrt{x}}$ = $\frac{1\left(x+\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}$ = $\frac{x+\sqrt{x}}{{x}^{2}-x}$
Once again, we multiplied the top and the bottom by the conjugate of the denominator : that is, we replaced $\mathrm{a-}$ with $\mathrm{a+}$ . The formula $(x+a)(x-a)={x}^{2}-{a}^{2}$ enabled us to quickly multiply the terms on the bottom, and eliminated the square roots in the denominator.
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