(t-5)^2=81 do I take 5 from both sides or add it?

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- anonymous

(t-5)^2=81 do I take 5 from both sides or add it?

- chestercat

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- anonymous

Neither

- myininaya

to get rid of the square, you must square root both sides

- anonymous

no what you do first is you hvae to square root both sides . only then you can add 5 to both sides.
so t - 5 = - or + sqrt(81)
so t - 5 = + or negative 9
so t - 5 = 9 and therefore t = 14
and t - 5 = -9 and therefore t = -4
so t = 14 and -4 are both valid answers

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- anonymous

I see how you got it now I guess with similar problems I thought I did the same thing to this one too. How would I do one with 1/2(x-1)^5=16

- anonymous

1/2(x-1)=16^(1/5)

- myininaya

first multiply by the reciprocal of 1/2 on both sides so you have grouping symbol by itself on one of the equation then you will have (x-1)^5=2*16

- myininaya

(x-1)^5=32

- myininaya

now since this is raised to the 5th power with take the 5th power of both sides giving us x-5=32^(1/5)

- myininaya

x-5=2

- myininaya

so x=2+5

- myininaya

x=7

- anonymous

how did you get 2

- myininaya

what is 32^(1/5)

- myininaya

since 32^(1/5)=[2*2*2*2*2]^(1/5)=2

- anonymous

got it when I put in calculator

- anonymous

is this right 3z^2 =48
z=16

- myininaya

z^2=48/3=16 so z=plus or minus 4

- myininaya

hey but what happens when you check x=7 in the question we did before this one

- myininaya

did you catch my mistake above?

- myininaya

it was suppose to be x-1=(32)^(1/5)

- anonymous

No I didn't but I was checking it

- myininaya

x-1=2

- myininaya

x=1+2=3

- anonymous

It is x-1+2=3 LOL

- anonymous

so it isn't 7

- anonymous

it is 3

- myininaya

so you got it good! i have to go eat. yes it is 3

- anonymous

thank you boy I could use your help all the time have a good supper

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