Radicals: Other Considerations
Purplemath
When working with square roots, we learned how to "rationalize" the denominator of a fraction if it contained a square root.
If the denominator contained a sum or difference containing square roots, we saw that we could use the "difference of squares" formula (in reverse) to get rid of those radicals; in particular, we learned that if we multiplied the fraction, top and bottom, by the conjugate of the denominator, the radicals would cancel out in the denominator.
A Special Case of Rationalizing
If your class has covered the formulas for factoring the sums and differences of cubes, then you might encounter a special case of rationalizing denominators. The reasoning and methodology are similar to the "difference of squares" conjugate process for square roots.
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Simplify:
![2 / (1 + cbrt[4])](radicals/rad112.gif)
I would like to get rid of the cube root, but multiplying by the conjugate won't help much. Yes, the middle term drops out, but that isn't entirely helpful in the case of cubes:
![(1 + cbrt[4])(1 − cbrt[4]) = 1 − (cbrt[4])^2 = 1 − cbrt[16] = 1 − 2 cbrt[2]](radicals/rad113.gif)
But I can "create" a sum of cubes, just as using the conjugate allowed me to create a difference of squares earlier. Using the fact that:
a3 + b3 = (a + b)(a2 − ab + b2)
...and letting a = 1 and b equal the cube root of 4, I get:
![(1 + cbrt[4])(1 − cbrt[4] + (cbrt[4])^2) = 1 + (cbrt[4])^3 = 1 + 4 = 5](radicals/rad114.gif)
If I multiply, top and bottom, by the second factor in the sum-of-cubes formula, then the denominator will simplify with no radicals. So I'll multiply, top and bottom, by this second factor, creating a sum of cubes underneath and thus "clearing the denominator" (and leaving me with just a 5, as shown in the computation above):
![(2/(1 + cbrt[4]))*((1 − cbrt[4] + cbrt[16])/(1 − cbrt[4] + cbrt[16])) = (2 - 2cbrt[4] + 4cbrt[2]) / 5](radicals/rad111.gif)
Naturally, if the sign in the middle of the original denominator had been a "minus", I'd have applied the "difference of cubes" formula to do the rationalization. This sort of "rationalize the denominator" exercise almost never comes up. But if you see this in your homework, then you should expect to see one of these on your next test.
Radicals Expressed With Exponents
Radicals can be expressed as fractional exponents. Whatever is the index of the radical becomes the denominator of the fractional power. For instance:
![sqrt[9] = 2nd-rt[9] = 9^(1/2) = 3](radicals/rad116.gif){kind=small}
The second root became a one-half power. A cube root would be a one-third power, a fourth root would be a one-fourth power, and so forth. This conversion process will matter a lot more once you get to calculus. For now, it allows you to simplify some expressions that you might otherwise not have been able to.
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Express
![cbrt[2] * 4th-rt[2]](radicals/rad115.gif)
as a single radical term.
I can't do anything with the radical product, as it stands. But I can work with fractions (using common denominators), and I can combine exponential terms which have the same base (in this case, a base of 2). So I will convert the radicals to exponential expressions, and then apply exponent rules to combine the factors:
![cbrt[2] 4th-rt[2] = 2^(1/3) 2^(1/4) = 2^(1/3 + 1/4) = 2^(7/12) = 12th-rt[2^7]](radicals/rad117.gif)
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Simplify:
![cbrt[5] / sqrt[5]](radicals/rad120.gif)
I can't work with these radicals; one is a cube root and the other is a square root. But both roots contain a 5, so I can convert the exponential form, and simplify. Then I'll convert back to radical form.
![5^(1/3) / 5^(1/2) = 5^(1/3 − 1/2) = 5^(−1/6) = 1/5^(1/6) = 1/(6th-rt[5]) = (1/(6th-rt[5]))*(6th-rt[5^5]/6th-rt[5^5]) = (6th-rt[5^5]) / 5](radicals/rad121.gif)
Once I'd gotten to where I had a sixth root of five in the denominator, I had to rationalize by multiplying five more copies of the 5, so I could take that 5 out of the radical in the denominator.
Domains of Radical Functions
Usually, we cannot have a negative inside a square root. (The exception is for "imaginary" numbers. If you haven't done the number "i" yet, then you haven't done imaginaries.)
So, for instance, 
is not possible. Do not try to say something like "
", because it's not true, as you can verify by squaring:
We must have a positive value inside the square root. This fact can be important for defining and graphing functions.
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Find the domain of the following function:
The function they've given me has the expression x − 2 inside a square root. The domain of this function is all the allowable x-values that can be plugged into this function.
The fact that I can't have a negative inside the square root requires that the argument of the square root — namely, the x − 2 inside the radical — be zero or greater, so I must have x − 2 ≥ 0. Solving, I get:
domain: x ≥ 2
On the other hand, we CAN have a negative inside a cube root (or any other odd root). For instance:
...because (−2)3 = −8.
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Find the domain of the following:
For 
, there is NO RESTRICTION on the value of x, because x − 2 is welcome to be negative inside a cube root. Then the domain is:
domain: all x
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