Question

If `8sqrt(4z^2 - 43) = 40`, what is the value of `z`?

Answer 1

Assumption: the question is: `" "8sqrt(4z^2)-43=40`

`z=+83/16 ->5 3/16`

Explanation:

`color(brown)("The objective is to manipulate the equation such that you have a")` `color(brown)("single z. This is to be on one side of the equals sign and everything")` `color(brown)("else on the other side.")`

This is done in stages. First you have all the terms with z on the LHS of = and all the terms without z on the other side.
Then you manipulate the LHS side until the only thing left is the single z.

`color(blue)("Step 1 - Isolate "sqrt(4z^2)`

Add 43 to both sides giving

`8sqrt(4z^2)+0=40+43`

Divide both sides by 8. Same as multiply by `1/8` to get rif og the 8 from `8sqrt(4z^2)`

`8/8xxsqrt(4z^2)=83/8`

But `8/8=1`

`sqrt(4z^2)=83/8`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Step 2 - Isolate z")`

But `4 = 2^2 -> 4z^2 = 2^2z^2`

Write as:`" "sqrt(2^2z^2)=83/8`

Taking the root

`=>+-2z=83/8`

divide both side by 2 to get rid of the 2 from `2z`

`+-z=83/16-> 5 3/16`

After testing `z=+83/16`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: Consider the left hand side only for `z=+83/16`

`8sqrt(4z^2)-43`

`8xx(2z)-43`

`8xx2xx83/16-43`

But `2xx8=16`

`cancel(16)xx83/(cancel(16))-43`

`83-43`

`40`

Thus LHS=RHS so

`color(red)(" The equation will not works for "z=-83/16)`

Answer 2

`z = +-sqrt(17)`

Explanation:

`8sqrt(4z^2 - 43) = 40`

Isolate the square-root.

`sqrt(4z^2 - 43) = 5`

`(sqrt(4z^2 - 43))^2 = 5^2`

`4z^2 - 43 = 25`

`4z^2 = 25 + 43`

`4z^2 = 68`

`z^2 = 17`

`z = +-sqrt(17`

Checking in the original equation, both solutions work.