Question

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

Answer 1

Neither

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

(x-1)^5=32

Answer 8

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

Answer 9

x-5=2

Answer 10

so x=2+5

Answer 11

x=7

Answer 12

how did you get 2

Answer 13

what is 32^(1/5)

Answer 14

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

Answer 15

got it when I put in calculator

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

did you catch my mistake above?

Answer 20

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

Answer 21

No I didn't but I was checking it

Answer 22

x-1=2

Answer 23

x=1+2=3

Answer 24

It is x-1+2=3 LOL

Answer 25

so it isn't 7

Answer 26

it is 3

Answer 27

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

Answer 28

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