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

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Answer 1 · 2016-09-25T09:56:36

$x=8$

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

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

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

$6x+5=53$

Next, we subtract $5$ from both sides.

$6x=48$

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

$x=8$

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

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

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

How do you solve sqrt(6x+5) = sqrt53sqrt(6x+5) = sqrt53 and find any extraneous solutions?