Question 182180
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31.  I think you mean:


*[tex \LARGE \text{          }\math sqrt{10 + 3 \sqrt{x}}=sqrt{x}]


Square both sides:


*[tex \LARGE \text{          }\math 10 + 3 sqrt{x}= x ]


Add -10 to both sides:


*[tex \LARGE \text{          }\math 3 sqrt{x}= x - 10]


Square both sides:


*[tex \LARGE \text{          }\math 9x = x^2 - 20x + 100]


*[tex \LARGE \text{          }\math  x^2 - 29x + 100 = 0]


*[tex \LARGE \text{          }\math  (x - 4)(x - 25) = 0]


*[tex \LARGE \text{          }\math  x = 4 \text{ or } \math x = 25]


Check answers:


*[tex \LARGE \text{          }\math sqrt{10 + 3 \sqrt{4}}=?sqrt{4}]


*[tex \LARGE \text{          }\math sqrt{16}\neq sqrt{4}]


*[tex \LARGE \text{          }\math  x = 4 ] is extraneous


*[tex \LARGE \text{          }\math sqrt{10 + 3 \sqrt{25}}=?sqrt{25}]


*[tex \LARGE \text{          }\math sqrt{25}=sqrt{25}]


*[tex \LARGE \text{          }\math  x = 25 ] is valid.


33. 


*[tex \LARGE \text{          }\math  (3x +1)^{\frac{1}{2}} = 4]


This is exactly the same as:


*[tex \LARGE \text{          }\math  sqrt{3x +1} = 4]


Just square both sides and solve.


39.  Two possible interpretations:


*[tex \LARGE \text{(1)          }\math  x^{3 \over 2} - 3x^{1 \over 2} = 0\ \ \Rightarrow\ \ x^{3 \over 2} = 3x^{1 \over 2} \cr \text{      }\math\ \ \Rightarrow\ \ x^3 = 3x \ \ \Rightarrow\ \ x^2 = 3\ \ \Rightarrow\ \ x = \pm sqrt{3}]


*[tex \LARGE \text{(2)          }\math  x^{3 \over 2} - (3x)^{1 \over 2} = 0\ \ \Rightarrow\ \ x^{3 \over 2} = (3x)^{1 \over 2} \cr \text{      }\math\ \ \Rightarrow\ \ x^3 = 9x \ \ \Rightarrow\ \ x^2 = 9\ \ \Rightarrow\ \ x = \pm 3]



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