How do you solve #4x^(3/2) - 5 = 103# and find any extraneous solutions?

1 Answer
May 31, 2016

Answer:

#x=2color(white)(.)root(3)(99)#

Not sure where the extraneous solutions come from!

Explanation:

Add 5 to both sides

#4x^(3/2)=108#

Write as : # 4sqrt(x^3)=108#

Divide both sides by 4

#sqrt(x^3)=27#

square both sides

#x^3=792" "#

cube root both sides

#x=root(3)(792) = root(3)(2^3xx3^2xx11)" "larr" prime factors"#

#x=2color(white)(.)root(3)(99)#