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

1 Answer
Jan 27, 2018

Answer:

#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"#