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Presleyram
 3 years ago
Algebra II Problems
(work shown)
Presleyram
 3 years ago
Algebra II Problems (work shown)

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Presleyram
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1355298946753:dw

nubeer
 3 years ago
Best ResponseYou've already chosen the best response.0u can start by squaring both sides.

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0Before running off it it smart to take a look at two things: 1. The number under the radical, 6x+2 2. The result of the square root, x1 If you think about it, both numbers must be >=0! 6x+2 >=0 because there is no (real) square root of a negative number. x+1 >=0 because it's a square root and that is >=0 by definition.

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0This means that you answers are restricted two ways: 1. 6x+2>=0, so 6x>=2, x>=2/6, x>=1/3 2. x1>=0, so x>=1 BOTH must hold, so we combine them: x>=1. Whatever "solutions" you may find, they MUST be >= 1. Now square everything in sight: (x1)² = 6x+2 Work out the brackets Everything to the left side of the =sign Factorize, or use abcformula Check the solutions! Only if they are >= 1 they are really solutions. Reason for this nitpicking: by squaring everything, you lose the information about the square root and you could get fake solutions...

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0Oops, typo: 6x>=2, so x>= 1/3 Makes no difference: x >=1 anyway.

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0@nubeer: I wouldn't *start* with squaring, see above reasoning...

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.0yw! Have you got the answer?
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