SOLUTION: I need to find z when z^6=64. I know the answer is 2 but what are the steps to solve for z?

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Question 151647This question is from textbook PreAlgebra
: I need to find z when z^6=64. I know the answer is 2 but what are the steps to solve for z? This question is from textbook PreAlgebra

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
z%5E6=64 Start with the given equation.



z%5E6=2%5E6 Rewrite 64 as 2%5E6.


From here you can see that the answer is 2, but let's keep going

root%286%2Cz%5E6%29=root%286%2C2%5E6%29 Take the 6th root of both sides. This will eliminate the exponent "6"



z=2 Take the 6th root of z%5E6 to get "z". Take the 6th root of 2%5E6 to get 2


So the solution is z=2



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Here's another way to solve for "z"


z%5E6=64 Start with the given equation.


z%5E6-64=0 Subtract 64 from both sides.


z%5E6-2%5E6=0 Rewrite 64 as 2%5E6.


%28z%5E3%29%5E2-%282%5E3%29%5E2=0. Rewrite z%5E6 as %28z%5E3%29%5E2. Rewrite 2%5E6 as %282%5E3%29%5E2


Now let A=z%5E3 and B=2%5E3


So we now have

A%5E2-B%5E2=0


%28A%2BB%29%28A-B%29=0 Factor using the difference of squares formula


%28z%5E3%2B2%5E3%29%28z%5E3-2%5E3%29=0 Plug in A=z%5E3 and B=2%5E3


%28%28z%2B2%29%2A%28z%5E2-2%2Az%2B4%29%29%28z%5E3-2%5E3%29=0 Factor the first binomial z%5E3%2B2%5E3 using the sum of cubes formula


%28%28z%2B2%29%2A%28z%5E2-2%2Az%2B4%29%29%28%28z-2%29%2A%28z%5E2%2B2%2Az%2B4%29%29=0 Factor the second binomial z%5E3-2%5E3 using the sum of cubes formula


%28z%2B2%29%28z-2%29%28z%5E2-2%2Az%2B4%29%28z%5E2%2B2%2Az%2B4%29=0 Rearrange the terms



Now set each factor equal to zero:

z%2B2=0, z-2=0, z%5E2-2%2Az%2B4=0 or z%5E2%2B2%2Az%2B4=0


So our first two solutions are z=-2 or z=2


Now let's solve z%5E2-2%2Az%2B4=0

z%5E2-2z%2B4=0 Start with the given equation.


Notice we have a quadratic equation in the form of az%5E2%2Bbz%2Bc where a=1, b=-2, and c=4


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%28-2%29+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%284%29+%29%29%2F%282%281%29%29 Plug in a=1, b=-2, and c=4


z+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%281%29%284%29+%29%29%2F%282%281%29%29 Negate -2 to get 2.


z+=+%282+%2B-+sqrt%28+4-4%281%29%284%29+%29%29%2F%282%281%29%29 Square -2 to get 4.


z+=+%282+%2B-+sqrt%28+4-16+%29%29%2F%282%281%29%29 Multiply 4%281%29%284%29 to get 16


z+=+%282+%2B-+sqrt%28+-12+%29%29%2F%282%281%29%29 Subtract 16 from 4 to get -12


z+=+%282+%2B-+sqrt%28+-12+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


z+=+%282+%2B-+2i%2Asqrt%283%29%29%2F%282%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


z+=+%282%29%2F%282%29+%2B-+%282i%2Asqrt%283%29%29%2F%282%29 Break up the fraction.


z+=+1+%2B-+sqrt%283%29%2Ai Reduce.


z+=+1%2Bsqrt%283%29%2Ai or z+=+1-sqrt%283%29%2AiBreak up the expression.


So the next two solutions are are z+=+1%2Bsqrt%283%29%2Ai or z+=+1-sqrt%283%29%2Ai


which approximate to z=1%2B1.732%2Ai or z=1-1.732%2Ai


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Now let's solve z%5E2%2B2%2Az%2B4=0




z%5E2%2B2z%2B4=0 Start with the given equation.


Notice we have a quadratic equation in the form of az%5E2%2Bbz%2Bc where a=1, b=2, and c=4


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%282%29+%2B-+sqrt%28+%282%29%5E2-4%281%29%284%29+%29%29%2F%282%281%29%29 Plug in a=1, b=2, and c=4


z+=+%28-2+%2B-+sqrt%28+4-4%281%29%284%29+%29%29%2F%282%281%29%29 Square 2 to get 4.


z+=+%28-2+%2B-+sqrt%28+4-16+%29%29%2F%282%281%29%29 Multiply 4%281%29%284%29 to get 16


z+=+%28-2+%2B-+sqrt%28+-12+%29%29%2F%282%281%29%29 Subtract 16 from 4 to get -12


z+=+%28-2+%2B-+sqrt%28+-12+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


z+=+%28-2+%2B-+2i%2Asqrt%283%29%29%2F%282%29 Simplify the square root (note: If you need help with simplifying square roots, check out this solver)


z+=+%28-2%29%2F%282%29+%2B-+%282i%2Asqrt%283%29%29%2F%282%29 Break up the fraction.


z+=+-1+%2B-+sqrt%283%29%2Ai Reduce.


z+=+-1%2Bsqrt%283%29%2Ai or z+=+-1-sqrt%283%29%2Ai Break up the expression.


So the next two solutions are z+=+-1%2Bsqrt%283%29%2Ai or z+=+-1-sqrt%283%29%2Ai


which approximate to z=-1%2B1.732%2Ai or z=-1-1.732%2Ai




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Answer:

So altogether, we have the 6 solutions


z=-2, z=2, z+=+1%2Bsqrt%283%29%2Ai , z+=+1-sqrt%283%29%2Ai, z+=+-1%2Bsqrt%283%29%2Ai or z+=+-1-sqrt%283%29%2Ai