Question

How do you solve `(5x+4)^(1/2)-3x=0` and find any extraneous solutions?

Answer 1

`x=1" is a solution",x=-4/9" is extraneous"`

Explanation:

`(5x+4)^(1/2)-3x=0`

`rArrsqrt(5x+4)=3x`

`color(blue)"square both sides"`

`rArr5x+4=9x^2`

`rArr9x^2-5x-4=0larrcolor(blue)"in standard form"`

`rArr(x-1)(9x+4)=0`

`x-1=0rArrx=1`

`9x+4=0rArrx=-4/9`

`color(blue)"check these values in the original equation"`

`x=1to(5+4)^(1/2)-3=3-3=0" True"`

`x=-4/9to(-20/9+36/9)^(1/2)-3(-4/9)=4/3+4/3=8/3`

`rArrx=1" is a solution"`

`x=-4/9" is an extraneous solution"`