How do you simplify $sqrt(5x-4)-sqrt(x+8)=2$ ?

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Answer 1 · 2016-09-02T16:04:14

The solution set is ${8}$ .

Isolate one of the $sqrt$ 's.

$sqrt(5x - 4) = 2 + sqrt(x + 8)$

Square both sides of the equation:

$(sqrt(5x - 4))^2= (2 + sqrt(x + 8))^2$

$5x - 4 = 4 + 4sqrt(x+ 8) + x + 8$

$4x - 16 = 4sqrt(x + 8)$

$4(x - 4) = 4sqrt(x + 8)$

$x - 4 = sqrt(x + 8)$

Square again:

$(x - 4)^2 = (sqrt(x + 8))^2$

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

$x^2 - 9x + 8 = 0$

$(x -8)(x - 1) = 0$

$x = 8 and 1$

Checking in the original equation, you will find that $x = 8$ works while $x = 1$ is extraneous.

Hopefully this helps!

How do you simplify sqrt(5x-4)-sqrt(x+8)=2sqrt(5x-4)-sqrt(x+8)=2?