Respuesta :
Answer: x = 4 is not an extraneous solution.
Step-by-step explanation: [tex]2(2x)^{\frac{1}{3} }+1=5[/tex]
Subtracting 1 on both the sides of the equation, we get
= [tex]2(2x)^{\frac{1}{3} }+1-1=5-1[/tex]
= [tex]2(2x)^{\frac{1}{3} }=4[/tex]
Dividing the equation by 2, we get
= [tex](2x)^{\frac{1}{3} }=2[/tex]
Cubic both sides of the equation, we get
= [tex]2x=8[/tex]
= [tex]x=4[/tex]
Check for extraneous solution
Putting x = 4 in the original equation [tex]2(2x)^{\frac{1}{3} }+1=5[/tex], we get
= [tex]2(2(4))^{\frac{1}{3} }+1=5[/tex]
= [tex]2(8)^{\frac{1}{3} }+1=5[/tex]
= [tex]2(2)+1=5[/tex]
= [tex]5=5[/tex]
⇒ LHS = RHS
∴ x = 4 is not an extraneous solution.
Know more about extraneous solution: - https://brainly.com/question/4563555
