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#.