SOLUTION: Solve each equation. Check for extraneous solutions.
[]means square root..
[2x-1]=x-8
(4x+3)^2/3=(16x+44)^1/3
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Question 181376: Solve each equation. Check for extraneous solutions.
[]means square root..
[2x-1]=x-8
(4x+3)^2/3=(16x+44)^1/3
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Square both sides:
Collect like terms and put into standard form:
Note that 
and 
, so:
(x - 13)= 0 \ \ \Rightarrow \ \ x = 5 \text{ or }\math x = 13)
Check for extraneous roots:
 - 1}=5 - 8 \ \ \Rightarrow \ \ sqrt{9} = -3)
False, therefore 
is extraneous.
 - 1}=13 - 8 \ \ \Rightarrow \ \ sqrt{25} = 5)
True, therefore 
is valid.
===================================================================
^{2/3}=(16x+44)^{1/3})
Raise each side to the third power:
^2=(16x+44))
Expand the binomial:
Collect like terms, leave in standard form:
Note that 
, 
, and (-5) + (4)(7) = 8)
so:
(4x + 7)= 0 \ \ \Rightarrow \ \ x = \frac {5}{4} \text{ or }\math x = \frac {-7}{4})
Check for extraneous roots:
+3)^{2/3}=(16(\frac{5}{4})+44)^{1/3} \ \ \Rightarrow \ \ 8^{2/3} = 64^{1/3} )
, True: 
is a valid root.
+3)^{2/3}=(16(\frac{-7}{4})+44)^{1/3} \ \ \Rightarrow \ \ (-4)^{2/3} = 16^{1/3} )
, True: 
is a valid root.
John
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