How do you solve $sqrt(3x-8) = 2$ and find any extraneous solutions?

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Answer 1 · 2016-05-29T02:04:32

$x = 4$

As we have $sqrt(3x-8)$ in the equation, we will require that $3x-8 >= 0$ , that is, $x >= 8/3$ .

$sqrt(3x-8) = 2$

$=>(sqrt(3x-8))^2 = 2^2$

$=> 3x-8 = 4$

(Note that typically $(sqrt(a))^2 = sqrt(a^2) = |a|$ , but we already required $3x-8>=0$ , meaning $|3x-8| = 3x-8$ )

$=> 3x = 12$

$:. x = 4$

Checking our answer, we find that

$sqrt(3(4)-8) = sqrt(4) = 2$

as desired.

How do you solve sqrt(3x-8) = 2sqrt(3x-8) = 2 and find any extraneous solutions?