How do you solve $sqrt(5x-9)-1=sqrt( 3x-6)$ and find any extraneous solutions?

Question

2876 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-06T11:32:43

$x=5$ or $x=2$

We have $x\geq 2$ since the radicand must be non negative.
Writing your equation in the form
$sqrt(5x-9)=sqrt(3x-6)+1$

squaring we get

$5x-9=3x-6+1+2sqrt(3x-6)$
collecting like Terms and isolating the square root

$x-2=sqrt(3x-6)$
since $x\geq 2$ we can square again

$x^2-7x+10=0$
solving this quadratic equation by the quadratic formula we get

$x_(1,2)=7/2pmsqrt(49/4-40/4)$

so

$x_1=5$

$x_2=2$
checking both Solutions we get that both fulfill the equation above.

Answer 2 · 2018-07-06T19:05:30

$x_1=2$ and $x_2=5$

$sqrt(5x-9)-1=sqrt(3x-6)$

$5x-9+1-2sqrt(5x-9)=3x-6$

$2x-2=2sqrt(5x-9)$

$x-1=sqrt(5x-9)$

$(x-1)^2=5x-9$

$x^2-2x+1=5x-9$

$x^2-7x+10=0$

$(x-2)*(x-5)=0$

So, $x_1=2$ and $x_2=5$ are solutions of this equation.

How do you solve sqrt(5x-9)-1=sqrt( 3x-6)sqrt(5x-9)-1=sqrt( 3x-6) and find any extraneous solutions?