Simplification of Radical Expressions
Key Questions
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Generally, you don't want to have radical at the denominators. So, let's say that we want to simplify the expression
#\frac{\sqrt{a}}{\sqrt{b}}# , where#a# and#b# can be any expression you want. Since, of course,#\frac{\sqrt{b}}{\sqrt{b}}=1# , we can multiply it without changing the value of our expression, so we have#\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{a}}{\sqrt{b}} \cdot \frac{\sqrt{b}}{\sqrt{b}}# . The advantage is that now we observe that#\sqrt{b} \cdot \sqrt{b}=b# , and so our expression becomes#\frac\{\sqrt{ab}}{b}# , and we got rid of the radical at the denominator. -
Expression with a a square root , cubed root or other fractional exponents in the expression
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This is easy! If you want to multiply this are the rules: First coefficients are multiplied with each other and the sub-radical amounts each other, placing the latter product under the radical sign common and the result is simplified.
Let's go:
#2sqrt5# times# 3sqrt10# #2sqrt5 × 3sqrt10 = 2 × 3sqrt(5×10)=6sqrt50# #= 6sqrt(2·5^2)# # = 30sqrt2# Now if you want to divide, then the coefficients are divided among themselves and sub-radical amounts each other, placing the latter quotient under the radical common and the result is simplified.
#2root3 (81x^7)# by#3root3( 3x^2)# #(2root3 (81x^7)) /(3root3 (3x^2)) = 2/3root3 ((81x^7)/(3x^2)) =2/3root3 (27x^5)# #2/3 root3 (3^3·x^3·x^2) = 2xroot3 (x^2)# I hope you can find it useful, and here is a link to solve this ones with different indices.
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There are two common ways to simplify radical expressions, depending on the denominator.
Using the identities
#\sqrt{a}^2=a# and#(a-b)(a+b)=a^2-b^2# , in fact, you can get rid of the roots at the denominator.Case 1: the denominator consists of a single root. For example, let's say that our fraction is
#{3x}/{\sqrt{x+3}}# . If we multply this fraction by#{\sqrt{x+3}}/{\sqrt{x+3}}# , we won't change its value (since of course#{\sqrt{x+3}}/{\sqrt{x+3}}=1# , but we can rewrite it as follows:
#{3x}/{\sqrt{x+3}} \cdot {\sqrt{x+3}}/{\sqrt{x+3}}= \frac{3x\sqrt{x+3}}{\sqrt{x+3}^2}# , and finally obtain
#\frac{3x\sqrt{x+3}}{x+3}# Case 2: the denominator consists of a sum/difference of roots. If we multiply by the difference/sum of the roots, we'll have the same result as above. For example, if you have
#\frac{\cos(x)}{\sqrt{x}+\sqrt{\sin(x)}# You'll multiply numerator and denominator by the difference
#\sqrt{x}-\sqrt{\sin(x)# , and obtain#\frac{\cos(x)}{\sqrt{x}+\sqrt{\sin(x))} \cdot \frac{\sqrt{x}-\sqrt{\sin(x)}}{\sqrt{x}-\sqrt{\sin(x)}}# which is#\frac{\cos(x)(\sqrt{x}-\sqrt{\sin(x)})}{\sqrt{x}^2-\sqrt{\sin(x)}^2}# which finally equals
#\frac{\cos(x) (\sqrt{x}-\sqrt{\sin(x)})}{x-\sin(x)}# Of course, when working with radicals, you always need to pay attention and make sure that the argument of the root is positive, otherwise you will write things that have no meaning!