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

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Answer 1 · 2016-05-31T21:20:43

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

Not sure where the extraneous solutions come from!

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)#