Question 181376
<font face="Times New Roman" size="+2">


*[tex \Large \text{          }\math sqrt{2x - 1}=x - 8]


Square both sides:


*[tex \Large \text{          }\math 2x - 1=x^2 -16x + 64]


Collect like terms and put into standard form:


*[tex \Large \text{          }\math x^2 - 18x + 65 = 0]


Note that *[tex \Large -5 \times -13 = 65] and *[tex \Large -5 + -13 = -18], so:


*[tex \Large \text{          }\math (x - 5)(x - 13)= 0 \ \ \Rightarrow \ \ x = 5 \text{ or }\math x = 13]


Check for extraneous roots:


*[tex \Large \text{          }\math x = 5  \ \ \Rightarrow \ \ sqrt{2(5) - 1}=5 - 8 \ \ \Rightarrow \ \ sqrt{9} = -3]  False, therefore *[tex \Large x = 5] is extraneous.


*[tex \Large \text{          }\math x = 13  \ \ \Rightarrow \ \ sqrt{2(13) - 1}=13 - 8 \ \ \Rightarrow \ \ sqrt{25} = 5]  True, therefore *[tex \Large x = 13] is valid.

===================================================================



*[tex \Large \text{          }\math (4x+3)^{2/3}=(16x+44)^{1/3}]


Raise each side to the third power:


*[tex \Large \text{          }\math (4x+3)^2=(16x+44)]


Expand the binomial:


*[tex \Large \text{          }\math 16x^2 + 24x + 9 = 16x + 44]


Collect like terms, leave in standard form:


*[tex \Large \text{          }\math 16x^2 + 8x - 35 = 0]


Note that *[tex \Large -5 \times 7 = -35], *[tex \Large 4^2 = 16], and *[tex \Large (4)(-5) + (4)(7) = 8] so:


*[tex \Large \text{          }\math (4x - 5)(4x + 7)= 0 \ \ \Rightarrow \ \ x = \frac {5}{4} \text{ or }\math x = \frac {-7}{4}]


Check for extraneous roots:


*[tex \Large \text{          }\math (4( \frac{5}{4} )+3)^{2/3}=(16(\frac{5}{4})+44)^{1/3} \ \ \Rightarrow \ \ 8^{2/3} = 64^{1/3} ], True: *[tex \Large x = \frac {5}{4}] is a valid root.


*[tex \Large \text{          }\math (4( \frac{-7}{4} )+3)^{2/3}=(16(\frac{-7}{4})+44)^{1/3} \ \ \Rightarrow \ \ (-4)^{2/3} = 16^{1/3} ], True: *[tex \Large x = \frac {-7}{4}] is a valid root.




John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>