+0

# Plato Classroom Question #18

0
52
4
+102

Simplify $$\sqrt[4]{256a^2b^4}$$.

$$\color{green}\boxed{{\color{navy}x}~{\color{lightblue}x}^{\color{navy}y}~{\color{black}\frac{\color{navy}x}{\color{navy}y}}~\color{black}\sqrt{\color{navy}x}~~{\color{navy}+~~-~\times}}$$

$$\color{red}\boxed{{\color{black}\sqrt[\color{navy}y]{\color{lightblue}x}}}$$

SpaceModo  Jan 26, 2018
edited by SpaceModo  Jan 26, 2018
edited by SpaceModo  Jan 26, 2018
Sort:

#1
+11278
0

4 x 4 x 4 x 4 = 256

=4ba^(1/2)

I do not know what the rest of your question means.....

ElectricPavlov  Jan 26, 2018
#2
+91775
+2

I beg to differ just a little EP

By convention the answer must be positive and we do not know if a or b is pos or neg.

$$\sqrt[4]{256a^2b^4}\\ =\sqrt[2]{16|a|b^2}\\ =4\sqrt{|a|}|b|\\ =|\;4b\sqrt{|a|}\;|\\ or\\ =|\;4b|a|^{1/2}\;|\\$$

Melody  Jan 27, 2018
#3
+11278
+1

Agree !   Melody is correct.... !      I was thinking about  a^2   and b^4   both being POSITIVE results...but that does not tell us if a or b are positive themselves !

ElectricPavlov  Jan 28, 2018
#4
+91775
0