How do you solve $7- sqrt(x-6)=3$ and find any extraneous solutions?

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Answer 1 · 2017-02-28T20:31:13

First, subtract $color(red)(7)$ from each side of the equation to isolate the square root expression while keeping the equation balanced:

$-color(red)(7) + 7 - sqrt(x - 6) = -color(red)(7) + 3$

$0 - sqrt(x - 6) = -4$

$-sqrt(x - 6) = -4$

Next, multiply each side of the equation by $color(red)(-1)$ to eliminate the negative terms while keeping the equation balanced:

$color(red)(-1) xx -sqrt(x - 6) = color(red)(-1) xx -4$

$sqrt(x - 6) = 4$

Then, square both sides of the equation to eliminate the square root function while keeping the equation balanced:

$(sqrt(x - 6))^2 = 4^2$

$x - 6 = 16$

Now, add $color(red)(6)$ to each side of the equation to solve for $x$ while keeping the equation balanced:

$x - 6 + color(red)(6) = 16 + color(red)(6)$

$x - 0 = 22$

$x = 22$

An extraneous solution would be $sqrt(x - 6) = -4$

How do you solve 7- sqrt(x-6)=37- sqrt(x-6)=3 and find any extraneous solutions?