Simplification of Radical Expressions

Add yours
How To Approximate The Value Of The Square Root Of 2

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

  • 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!

  • 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.

  • 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.

  • This key question hasn't been answered yet. Answer question
  • This key question hasn't been answered yet. Answer question

Questions