How do you solve $1+sqrt(3x+1)=x$ and find any extraneous solutions?

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Answer 1 · 2016-06-09T11:20:25

First, the argument under the root must be non-negative $(>=0)$

$3x+1>=0->x>=-1/3$

Subtract $1$ from both sides:
$sqrt(3x+1)=x-1$

Now we square both sides:
$(sqrt(3x+1))^2=(x-1)^2->$

$3x+1=x^2-2x+1->$

This can be normalised to:
$x^2-5x=0->x(x-5)=0->$

$x=0orx=5$

The first solution is false, as $1+sqrt(3xx0+1)!=0$
The other one gives $1+sqrt(3xx5+1)=5$

Conclusion: $x=5$

How do you solve 1+sqrt(3x+1)=x1+sqrt(3x+1)=x and find any extraneous solutions?