How do you simplify #(\frac { 9p ^ { 2} } { 16q ^ { 4} } ) ^ { 1/ 2}#?

1 Answer
George C.
Jan 30, 2017

#((9p^2)/(16q^4))^(1/2) = (3abs(p))/(4q^2)#

Explanation:

Note that in general:

#sqrt(x^2) = abs(x)#

If in addition you know that #x >= 0# then:

#sqrt(x^2) = x#

In our example, we are not told if #p >= 0#, but we know that #q^2 >= 0#.

So:

#((9p^2)/(16q^4))^(1/2) = sqrt((3p)^2/(4q^2)^2)#

#color(white)(((9p^2)/(16q^4))^(1/2)) = sqrt(((3p)/(4q^2))^2)#

#color(white)(((9p^2)/(16q^4))^(1/2)) = abs((3p)/(4q^2))#

#color(white)(((9p^2)/(16q^4))^(1/2)) = (3abs(p))/(4q^2)#

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