Question

How do you solve `sqrt(6-b)=sqrt(b+10)` and check your solution?

Answer 1

See a solution process below:

Explanation:

First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

`(sqrt(6 - b))^2 = (sqrt(b + 10))^2`

`6 - b = b + 10`

Next, add `color(red)(b)` and subtract `color(blue)(10)` from each side of the equation to isolate the `b` term while keeping the equation balanced:

`6 - b + color(red)(b) - color(blue)(10) = b + 10 + color(red)(b) - color(blue)(10)`

`6 - color(blue)(10) - b + color(red)(b) = b + color(red)(b) + 10 - color(blue)(10)`

`-4 - 0 = 1b + color(red)(1b) + 0`

`-4 = (1 + color(red)(1))b`

`-4 = 2b`

Now, divide each side of the equation by `color(red)(2)` to solve for `b` while keeping the equation balanced:

`-4/color(red)(2) = (2b)/color(red)(2)`

`-2 = (color(red)(cancel(color(black)(2)))b)/cancel(color(red)(2))`

`-2 = b`

`b = -2`

To validate the answer we will substitute `color(red)(-2)` for each occurrence of `color(red)(b)` in the original equation and determine if both sides of the equation are equal:

`sqrt(6 - color(red)(b)) = sqrt(color(red)(b) + 10)` becomes:

`sqrt(6 - color(red)(-2)) = sqrt(color(red)(-2) + 10)`

`sqrt(6 + color(red)(2)) = sqrt(color(red)(-2) + 10)`

`sqrt(8) = sqrt(8)`

Both sides of the equation are equal so the answer is correct.