**Q: Rationalize the denominator:**

**[√(x) +1 ]/ [√(x) – 3 ]**

A: Rationalizing the denominator means to get all of the square-roots out of the denominator!

To rationalize the denominator, we multiple both the top and the bottom by the **conjugate** of the denominator… The conjugate of √(x) – 3 is √(x) + 3 (just change the sign!)

So, multiply the top and bottom by: √(x) + 3.

When you multiply the top and bottom by the same thing, it does not chage the problem. Also, why do we multiply by the conjgate? Because this is what makes the square-roots disappear.. Watch and you will see:

[√(x) +1 ]/ [√(x) – 3 ] = [√(x) +1 ]*[√(x) + 3] / [ [√(x) – 3 ] * [√(x) + 3] ]

Now, FOIL out the top and the bottom… I will do the top first… Remember that √(x)*√(x) = x.

**TOP: **[√(x) +1 ]*[√(x) + 3] = x + 3√(x) + √(x) + 3 = x + 4√(x) + 3

**BOTTOM: **[√(x) – 3 ] * [√(x) + 3] = x + 3√(x) – 3√(x) – 9 = x – 9

Notice how all of the square-roots disappeared in the bottom?

So now we have:

(x + 4√(x) + 3) / (x – 9)

All done.

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thanks that helped alot. not sure if it actually matters but, how does √(x)*√(x) = x? and does that work for all numbers or just x?

Good question. The answer is “yes, it does work for all numbers”.

By definition, the square-root of a number, when multiplied by itself, will give that number back…

So, √(2) * √(2) = √4 = 2

And √10 * √10 = √100 = 10

√8 * √8 = √64 = 8

√x * √x = x

Does that make more sense?