https://www.gmatprepnow.com/module/gmat- ... video/1034
Hi.I was going through this question and had a little bit of confusion.
The first statement of question says that √x2 = 4. The problem is:
1) If I square both sides the result is x2 = 16. Then x could be either 4 or -4. Therefore, the statement is INSUFFICIENT.
2) However, the square on left side of equation could be cancelled with the square root as well. Then x MUST be 4. Therefore, the statement is SUFFICIENT.
The same problem is with the second statement. Could you please guide?
Thanks.
Regards,
Fahad
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Data Sufficiency - Roots
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Jim@StratusPrep
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You can't just 'cancel' the square and the root. You have to treat the operations separately and handle them in order.
When you have x^2 = 16, there are 2 possible answers 4 and -4
When you have x^2 = 16, there are 2 possible answers 4 and -4
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Hi Fahad!
√x² actually works out to |x|, not x.
For instance, if x = -2, then √x² = √(-2)² = √4 = 2. So if x is negative, the √x² transforms a negative into a positive, and we're left with |x|, not x itself.
√x² actually works out to |x|, not x.
For instance, if x = -2, then √x² = √(-2)² = √4 = 2. So if x is negative, the √x² transforms a negative into a positive, and we're left with |x|, not x itself.
Jim@StratusPrep and Matt@VeritasPrep: Ok but what rule is being applied here? PEDMAS? Because we have to solve the Parenthesis first? And the square root falls in the category of Exponents?
1) But if this is true then for the first statement where "√x^2 = 4". The square root falls in the exponent category and ^2 is also an exponent. "On what rules" do we decide which to solve first.
2) In statement 2 we can say that PEDMAS applies as we are solving Parenthesis first and then squaring it. So x could not be negative.
1) But if this is true then for the first statement where "√x^2 = 4". The square root falls in the exponent category and ^2 is also an exponent. "On what rules" do we decide which to solve first.
2) In statement 2 we can say that PEDMAS applies as we are solving Parenthesis first and then squaring it. So x could not be negative.
