CTG1983 Posted January 4, 2006 Share Posted January 4, 2006 Is Sqrt[(X-3)^2]=3-X ? i)X !=3 (!= means NOT EQUAL TO) ii) -XIXI >0 (I I means modulus) Quote Link to comment Share on other sites More sharing options...
ChhotiAankhe_BadeSapane Posted January 4, 2006 Share Posted January 4, 2006 I think for all values the equation is true. Sqrt[(X-3)^2]=3-X (x-3)^2 = (3-x)^2 Whatever you put it will be equal! Is this correct question or am I missing something? Quote Link to comment Share on other sites More sharing options...
arjmen Posted January 5, 2006 Share Posted January 5, 2006 Is Sqrt[(X-3)^2]=3-X ? i)X !=3 (!= means NOT EQUAL TO) ii) -XIXI >0 (I I means modulus)Vote B The stem can be restated as: Is (3-X) >= 0, since Sqrt[(X-3)^2] is the principal root of (X-3)^2 and is non-negative I: Clearly insuff II: -X*|X| > 0 => X 0 => 3-X > 0 Suff Quote Link to comment Share on other sites More sharing options...
CTG1983 Posted January 5, 2006 Author Share Posted January 5, 2006 Vote B The stem can be restated as: Is (3-X) >= 0, since Sqrt[(X-3)^2] is the principal root of (X-3)^2 and is non-negative I: Clearly insuff II: -X*|X| > 0 => X 0 => 3-X > 0 Suff arjmen, can u plz make the bold part a bit clearer? Wht's principal root? Quote Link to comment Share on other sites More sharing options...
ChhotiAankhe_BadeSapane Posted January 5, 2006 Share Posted January 5, 2006 Agree with Arj! CTG principle root is the positive root. For example sqrt(9) = +/- 3. +3 is principle root. Quote Link to comment Share on other sites More sharing options...
jlt Posted January 30, 2006 Share Posted January 30, 2006 can anyone pls clarify this a little bit more? thx. Quote Link to comment Share on other sites More sharing options...
800Bob Posted January 30, 2006 Share Posted January 30, 2006 sqrt[(x-3)^2] will equal x-3 when x-3 >= 0, that is, when x >= 3. sqrt[(x-3)^2] will equal 3-x when x-3 So the question is asking simply: "Is x less than or equal to 3?" Statement 2 alone is sufficient because it says that x is negative. Quote Link to comment Share on other sites More sharing options...
hitzs Posted January 30, 2006 Share Posted January 30, 2006 Hi arjmen/Bob I couldn't make out on how exactly you came to equation Is (3-X) >= 0 ? Can you plz explain it a bit more. From the question I should check that Is IX-3I = 3-X ? Using this equation I was able to find the answer, but I'm interseted in knowing your way. Thanks a lot Quote Link to comment Share on other sites More sharing options...
800Bob Posted January 31, 2006 Share Posted January 31, 2006 I couldn't make out on how exactly you came to equation Is (3-X) >= 0 ? Can you plz explain it a bit more.Take a positive number, square it, and then take the principal square root of the result... you will end up with the number you started with. Example: Take 5. Square it: 5^2 = 25. Take the principal square root of the result: sqrt(25) = 5. You end up with the number you started with. On the other hand, start with a negative number, square it, and then take the principal square root of the result... you will end up with the opposite of the number you started with. Example: Take -5. Square it: (-5) = 25. Take the principal square root of the result: sqrt(25) = 5. You end up with the opposite of the number you started with. So, to generalize: sqrt(x^2) = x if x > 0 sqrt(x^2) = -x if x To put it even more simply: sqrt(x^2) = |x|. So, turning to the question at hand... sqrt(x-3) = x-3 if x-3 > 0 sqrt(x-3) = -(x-3) = 3-x if x-3 Quote Link to comment Share on other sites More sharing options...
hitzs Posted February 1, 2006 Share Posted February 1, 2006 Thanx a lot Bob !! You made my funda crystal clear...explaination was really great. Quote Link to comment Share on other sites More sharing options...
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