How do you solve $\sqrt{6 x + 5} = \sqrt{53}$ $sqrt(6x+5) = sqrt53$ and find any extraneous solutions?

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How do you solve $\sqrt{6 x + 5} = \sqrt{53}$ $sqrt(6x+5) = sqrt53$ and find any extraneous solutions?

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Answer 1 · 2016-09-25T09:56:36

$x = 8$ $x=8$

Explanation:

First, we need to square both sides of the equation so that we can easily work with it.

${\sqrt{6 x + 5}}^{2} = {\sqrt{53}}^{2}$ $sqrt(6x+5)^2=sqrt(53)^2$

Doing this will turn the equation into something much more simple.

$6 x + 5 = 53$ $6x+5=53$

Next, we subtract $5$ $5$ from both sides.

$6 x = 48$ $6x=48$

Then, we divide both sides of the equation by $6$ $6$ .

$x = 8$ $x=8$

In order to determine whether or not $x = 8$ $x=8$ is extraneous, we need to plug it into the original equation.

$\sqrt{6 x + 5} = \sqrt{53}$ $sqrt(6x+5)=sqrt(53)$
$\sqrt{6 (8) + 5} = \sqrt{53}$ $sqrt(6(8)+5)=sqrt(53)$
$\sqrt{48 + 5} = \sqrt{53}$ $sqrt(48+5)=sqrt(53)$
$\sqrt{53} = \sqrt{53}$ $sqrt(53)=sqrt(53)$

Since the statement that resulted in plugging in $x = 8$ $x=8$ is true, we can confirm that $x = 8$ $x=8$ is not an extraneous solution.

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How do you solve $\sqrt{6 x + 5} = \sqrt{53}$ $sqrt(6x+5) = sqrt53$ and find any extraneous solutions?

1 Answer

Explanation:

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How do you solve √6x+5=√53sqrt(6x+5) = sqrt53 and find any extraneous solutions?

1 Answer

Explanation:

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How do you solve $\sqrt{6 x + 5} = \sqrt{53}$ $sqrt(6x+5) = sqrt53$ and find any extraneous solutions?