How do you solve $sqrt(x + 2)+4 = x$ ?

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Answer 1 · 2015-08-05T17:53:24

$x = 7$

Start by isolating the radical on one side of the equation by adding $-4$ to both sides

$sqrt(x+2) + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = x-4$

$sqrt(x+2) = x-4$

This means that, in order to be considered a valid solution, the value(s) of $x$ must satisfy the condition

$x-4>=0 <=> x>=4$

Next, square both sides of the equation to get rid of the square root

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

$x+2 = x^2 - 8x + 16$

Rearrange this equation to get all the terms on one side

$x^2 - 9x + 14 = 0$

You can find the two solutions to this quadratic by using the quadratic formula

$x_(1,2) = (-(-9) +- sqrt( (-9)^2 - 4 * 1 * (-14)))/(2 * 1)$

$x_(1,2) = (9 + sqrt(25))/2 = (9 +-5)/2 = {(x_1 = (9+5)/2 = color(green)(7)), (x_2 = (9-5)/2 = cancel(color(red)(2))) :}$

The only valid solution will be $x=7$ ; $x=2$ will be an extraneous solution, since it does not satisfy the aforementioned condition $x>=4$ .

How do you solve sqrt(x + 2)+4 = xsqrt(x + 2)+4 = x?