How do you solve #63= - 1+ k ^ { \frac { 3} { 2} }#?

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Answer 1 · 2017-02-21T07:00:15

Change #k^(3/2)# to #k#

#k =16#

#k^(3/2)# can be written as #root2( k)^3#

#63 = -1+k^(3/2)#

Add 1 to both sides;

#63+1= k^(3/2)#

#64=k^(3/2)#

To remove the cube take the cube root of 64, which is 4.

#root3 64 = root3 (k^(3/2)#

#4 = k^(1/2)#

And then to remove the power of #1/2# from #k# you take

#4^2# and #(k^(1/2))^2#

Which gives you the answer:

#16=k#