How do you solve $\sqrt{5 x - 6} = 2$ $sqrt[5x-6]=2$ and find any extraneous solutions?

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How do you solve $\sqrt{5 x - 6} = 2$ $sqrt[5x-6]=2$ and find any extraneous solutions?

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Answer 1 · 2018-04-23T19:15:27

$x = 2$ $x=2$

Explanation:

$square both sides$ $color(blue)"square both sides"$

$note that \sqrt{a} \times \sqrt{a} = {(\sqrt{a})}^{2} = a$ $"note that "sqrtaxxsqrta=(sqrta)^2=a$

${(\sqrt{5 x - 6})}^{2} = 2^{2}$ $(sqrt(5x-6))^2=2^2$

$\Rightarrow 5 x - 6 = 4$ $rArr5x-6=4$

$add 6 to both sides$ $"add 6 to both sides"$

$5xcancel(-6)cancel(+6)=4+6$

$rArr5x=10$

$"divide both sides by 5"$

$(cancel(5) x)/cancel(5)=10/5$

$rArrx=2$

$color(blue)"As a check"$

Substitute this value into the left side of the equation and if equal to the right side then it is the solution.

$sqrt(10-6)=sqrt4=2=" right side"$

$rArrx=2" is the solution"$

$"There are no extraneous solutions"$

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How do you solve $\sqrt{5 x - 6} = 2$ $sqrt[5x-6]=2$ and find any extraneous solutions?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

How do you solve sqrt[5x-6]=2√5x−6=2sqrt[5x-6]=2 and find any extraneous solutions?

1 Answer

Explanation:

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How do you solve $\sqrt{5 x - 6} = 2$ $sqrt[5x-6]=2$ and find any extraneous solutions?