How do you solve $4= sqrt(h) - 6$ and find any extraneous solutions?

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Answer 1 · 2017-06-10T02:01:30

See a solution process below:

First, add $color(red)(6)$ to each side of the equation to isolate the $h$ term while keeping the equation balanced:

$4 + color(red)(6) = sqrt(h) - 6 + color(red)(6)$

$10 = sqrt(h) - 0$

$10 = sqrt(h)$

Now, square each side of the equation to solve for $h$ while keeping the equation balanced:

$10^2 = (sqrt(h))^2$

$100 = h$

$h = 100$

There are no extraneous solutions for $h$ .

However, if we substitute $100$ for $h$ we get:

$4 = +-sqrt(100) - 6$

$4 = +-10 - 6$

The solution: $4 = +10 - 6$ is valid.

The solution: $4 = -10 - 6$ is not valid.

But there is only one solution for $h$ which is $100$ .

How do you solve 4= sqrt(h) - 64= sqrt(h) - 6 and find any extraneous solutions?