How do you solve #(x+9)^"2/3"=4#?

1 Answer
Apr 3, 2018

Answer:

See below.

Explanation:

To solve this problem, we need to do something with that #2/3# power first.

Since the #2/3# encompasses an expression using addition, the power does not distribute. So we have to do a little manipulation:

#(x+9)^(2/3)=4 => root(3)((x+9)^(2))=4#

Now that we have that, we can start to simplify this a little.

Since the cube root encompasses the entirety of the left side, we can cube the whole thing to get rid of it.

#color(blue)(Note:"What you do to one side, you must do to the other".#

We should now have this:

#root(3)((x+9)^(2))=4#

#=>(root(3)((x+9)^(2)))^3=4^3#

#=>(x+9)^2=64#

From here we need to put the #(x+9)^2# into standard form, and solve from there:

#(x+9)^2=64#

#=>x^2+18x+81=64#

#=>x^2+18x+15=0#

To solve this, we can use the quadratic formula, which is defined as:

#-color(red)(b)+-sqrt(color(red)(b)^2-4color(blue)(a)color(green)(c))/(2color(blue)(a))#

#color(blue)(a=1)# #color(red)(b=18)# #color(green)(c=15)#

#=(-color(red)(18)+-sqrt(color(red)(18)^2-4(color(blue)(1))(color(green)(15))))/(2(color(blue)(1))#

#=-(18+-sqrt(324-60))/(2)#

#=(-18+-2sqrt(66))/2#

#=-9+-sqrt(66)#

Hope this helped!