Can someone help me with this algebra question?

Question

(1/2t-1)^2=0

Answer 1

Take square root of both sides:

1/2t - 1 = 0

Add one to both sides:

1/2t = 1

Multiply both sides by 2:

t = 2

Answer 2

Take the squrae root of each side and solve
1/2t - 1 = 0 add 1 to each side
1/2t = 1 multiply each side by 2t
1 = 2t Divide by 2
1/2 = t
Check
1/2t - 1 = 1/(2x1/2) - 1 = 1/1 - 1 = 1 - 1 = 0 and 0^2 = 0
It's a little ambigous the way your'e written this, if you meant
(1/(2t-1))^2 = 0 then t would be different, but you'd work it out in a simliar manner, take the square root of each side and solve for t.
I understood it to mean (1/(2t) - 1)^2 = 0 while others understood it to mean ...........((1/2)t) - 1)^2 = 0 and got t = 2. So which is right? That depends on what your problem really says, like I said, it's not clear the way you wrote it. Put each of the answers back into the origional equation, as it was given to you, and find out which one works. that's the one that is correct.

Answer 3

Instead of having the hassle of working out (1/2t-1)^2, you take its square root. Square root 0 to get 0 too. Therefore you get:

1/2t - 1 = 0

You add 1 to both sides, therefore:

1/2t = 1

Therefore, the answer should be obvious as you take each side and multiply them by 2.

t = 2


Hope this helps (:

Answer 4

(1/2t-1)^2=0

=> (1-2t)^2/4t^2 =0

=> (1-2t)^2 = 0

(because according to the given expression if denominator equal to zero then the value equal to infinity)

=>1-2t = 0

2t = 1

t=1/2 = 0.5

Answer 5

(1/2t - 1)² = 0

I am assuming that when you say 1/2t, you mean that 2t is the denominator (bottom of the fraction) and 1 is the numerator (top) like this:
1
---
2t

Take the square root of both sides:
1/2t - 1 = 0
Solve for t:
1/2t = 1
2t = 1 (invert both sides, 1/1 = 1)
t = 1/2

Plug in t = 1/2 to the initial equation to double check.

Answer 6

Since you are solving for t, you can solve the eqn in two ways.

The first way is to FOIL out the binomial then solve for t. (which is not efficient b/c you can solve it right there and then)

The second is to set (1/2t-1)=0

The reason why is b/c 0^2 is 0.

Now just solve for t and there you go.

You should get t=2

Answer 7

(1/2t-1)^2=0
find square roots of L.H.S and R.H.S
1/2t-1=0
1/2t=0+1
1/2t=1
t=1*2
t=2

Answer 8

OK!

(1/2t-1)^2=0

Take each side to the 1/2 power.

(1/2t -1)= 0

1/2t=1

t=2

This is an order 2 polynomial, so it has two roots.
It happens that both roots=2.

Done!

Answer 9

Whats the question? Are you trying to solve for t? If so the answer is t=2. After you take the square root of both sides, you will end up with ½t-1=0. Then add 1 to both sides to get ½t=1. Finaly multiply 2 to both sides to get t=2.

Answer 10

(1/2t-1)²=0
1/4t² - 2.1/2t.1 + 1² = 0
1/4t² - t + 1 = 0
delta = 1² - 4.1/4.1
delta = 1 - 1 = 0

t' = t" = (1 +/- 0) : 2.1/4 = 2
Solution: There is only one root: {t elements of R | t' = t" = 2}
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