Question #9e76e

2 Answers
Jul 30, 2015

Answer:

In precalculus and calculus, typically, if #m/n# is a completely reduced positive rational number, then #b^{m/n}=root(n){b^{m}}=(root(n){b})^{m}# is defined as a unique real number for #b\geq 0# when #n# is even and for all real values of #b# when #n# is odd. Other conventions can hold when you allow complex number roots. If #m/n# is negative, take reciprocals.

Explanation:

For example, #64^{3/2}=(root(2)64)^{3}=8^3=512#, but usually #(-64)^{3/2}# is left undefined in precalculus and calculus (unless you are in the unit about complex numbers).

On the other hand, #64^{5/3}=(root(3)64)^{5}=4^{5}=1024# and #(-64)^{5/3}=(root(3)(-64))^{5}=(-4)^{5}=-1024#.

To simplify radicals with variable radicands typically involves using properties of exponents after you've converted them to exponential form.

For instance:

#root(3){x^{7}}*root(5)(x^{9})=x^{7/3}*x^{9/5}=x^{7/3+9/5}=x^{(35+27)/15}=x^{62/15}=root(15){x^{62}}#

This is a useful skill in, for example, calculus, where rewriting radicals as powers helps you to "differentiate" or "integrate" the resulting function, which are very important things to be able to do.

Jul 30, 2015

Answer:

The root of x is defined for all real numbers if the root n is odd, and for positive numbers if the root n is even.

Explanation:

The root of x is defined for all real numbers if the root n is odd, and for positive numbers if the root n is even.

#sqrt(64)=8#
#root(3)(8)=2#

#sqrt(-64)# is not defined under real numbers. It's actually a complex number #+-8i#.
However #root(3)(-8)=-2# because odd roots of negative numbers are defined.

As for your second question, simplifying radicals with variable radicands, it would be best if you had an example. Generally you factor out perfect squares (if the root is 2), perfect cubes (if the root is 3), etc., take the nth root of what you can, and leave the rest as your second factor.

Remember #root(n)(ab)=root(n)(a)##root(n)(b)#

A couple of examples:

#sqrt(8x^4y^5)#=#sqrt(4x^4y^4)*sqrt(2y)# = #2x^2y^2sqrt(2y)#

#root(3)(54x^5y^3)=root(3)(27x^3y^3)root(3)(2x^2)# = #3xyroot(3)(2x^2)#
(You have to realize that 27 is a perfect cube factor of 54. Any variable expression whose exponent is a multiple of 3 is a perfect cube, multiple of 4 is a perfect fourth power, etc.)