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# radical in denomintor

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Intern
Joined: 01 Jan 2020
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23 Jan 2020, 08:53
quick question, how do you simplify the expression 3/ sqrt (x-1) ?
Since the entire expression x-1 is under sqrt sign, multiplying both top and bottom by sqrt x+1 wont work right?
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24 Jan 2020, 03:37
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quick question, how do you simplify the expression 3/ sqrt (x-1) ?
Since the entire expression x-1 is under sqrt sign, multiplying both top and bottom by sqrt x+1 wont work right?

If the entire denominator of a fraction is under a square root, then you can just multiply the numerator and denominator by that root. So in your case, you'd do this:

$$\left( \frac{3}{\sqrt{x - 1}} \right) \left( \frac{ \sqrt{x-1} }{ \sqrt{x-1} } \right) = \frac{3 \sqrt{x-1} }{x - 1}$$

That might not look any simpler than what you started with, but it's required in math to get roots out of denominators, so you need to do it.

It's only when you're adding a root to something else (or adding two roots, or subtracting) that you need to do something similar to what you suggested. So if you had this fraction:

$$\frac{1}{1 + \sqrt{x}}$$

then, because we're adding a root to something else, we'd need to use the difference of squares pattern (the "conjugate" if you've seen that word) to get the root out of the denominator. So here we'd multiply by 1 - √x in both the numerator and denominator:

$$\frac{1}{1 + \sqrt{x}} = \left( \frac{1}{1 + \sqrt{x}} \right) \left( \frac{1 - \sqrt{x}}{1 - \sqrt{x}} \right) = \frac{1 - \sqrt{x}}{1 - x}$$
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Joined: 01 Jan 2020
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24 Jan 2020, 05:06
Right, that makes sense now. Thank you!!
Re: radical in denomintor   [#permalink] 24 Jan 2020, 05:06
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# radical in denomintor

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