How do you solve $sqrt[x+2]-sqrt[6x]=-1$ and find any extraneous solutions?

Question

1714 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-20T01:16:20

$sqrt(x+2)=sqrt(6x)-1$

$(sqrt(x+2))^2=(sqrt(6x)-1)^2$

$x+2=6x-2sqrt(6x)+1$

$-5x+1=-2sqrt(6x)$

$((-5x+1)/-2)^2=(sqrt(6x))^2$

$(25x^2-10x+1)/4=6x$

$(25x^2-10x+1)=24x$

$25x^2-34x+1=0$

$x=(34(+/-)sqrt(34^2-4*25*1))/(2*25)$

$x=0.03, 1.33$ and 0.03 does not work when you plug it back into the equation to solve for x, so it is extraneous

How do you solve sqrt[x+2]-sqrt[6x]=-1sqrt[x+2]-sqrt[6x]=-1 and find any extraneous solutions?