Question

How do you solve `2(y-7)^2=64`?

Answer 1

`y=12.6568`

Explanation:

Consider the given equation #2(y-7)^2=64$

Now, taking the given equation, let us divide it by `2` in both sides, hence we shall have
`{2(y-7)^2}/2=64/2\impliescancel2(y-7)^2/cancel2=cancel64^32/cancel2`

So, we end up with `(y-7)^2=32`

Now, `2^1=2`, and `2^2=4`, and `2^3=8`, and so on we see that `2^5=32`, making the equation
`(y-2)^2=2^5`

Let's now take the square root on both sides of the equation, giving us
`sqrt{(y-7)^2}=sqrt{2^5`

Now, the square root of any square is the original value itself, so
`sqrt{(y-7)^2}=y-7`
But, what about the right-hand side of the equation?
`2^5=2^2*2^2*2`
We know that the square root of the product of any two real numbers is equal to the product of the square root of those numbers, (except when both are negative), so
`sqrt{2^5}=sqrt{2^2*2^2*2}=sqrt{2^2}*sqrt{2^2}*sqrt{2}=2*2*sqrt2`
We can write the square root of any value as value raised to the power of half, i.e `m=n^{1/2}`
So, that means `sqrt2=2^{1/2`
That ends with `sqrt{2^5}=4sqrt2`

So that makes the original equation as `y-7=4sqrt2`

If you now add `7` to both sides of the equation and use a calculator to simplify `sqrt2`, you'll end up with the answer.