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

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

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Answer 1 · 2016-05-29T02:04:32

$x = 4$ $x = 4$

Explanation:

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

$\sqrt{3 x - 8} = 2$ $sqrt(3x-8) = 2$

$\Rightarrow {(\sqrt{3 x - 8})}^{2} = 2^{2}$ $=>(sqrt(3x-8))^2 = 2^2$

$\Rightarrow 3 x - 8 = 4$ $=> 3x-8 = 4$

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

$\Rightarrow 3 x = 12$ $=> 3x = 12$

$:. x = 4$

Checking our answer, we find that

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

as desired.

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

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Explanation:

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