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MGMAT Quant. Question - Equations, Inequalities and VICS
[#permalink]
22 Aug 2010, 04:07
1
Kudos
Hi all,
In the problem set at the end of chapter 5 of the E's,I's and VICS book, Q.2 throws a bit of a curveball concerning the 'dangerous base' of the expression within the sq. root...
This is the question:
Q.2 - \(If g(x) = 3x + \sqrt{x}\), what's the value of \(g(d^2 + 6d + 9)\)?
Well the answer comes up with the following intermediate solution: \(3d^2 + 18d + 27 + \sqrt{(d + 3)^2}\)
... then it goes on to solve for when \(\sqrt{(d + 3)^2}\) is NEGATIVE and POSITIVE... The thing is, when square rooting a squared expression like this, can't you just simply remove the exponent from the brackets and drop the square root/radical sign, thus arriving at \((d + 3)\) ?
I can't find out the reason for why they do this... Why do you need to treat the \(\sqrt{(d + 3)^2}\) as if it were a solved expression BEFORE square rooting it when you could just drop the sq. root sign AND the exponent ?
Thanks!
Archived Topic
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Re: MGMAT Quant. Question - Equations, Inequalities and VICS
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22 Aug 2010, 04:28
1
Kudos
I've also just realised something else - In Ch. 6 of the same book, page 98 (4th edition) it says \(\sqrt{x^2}\) = |x| --> so surely, \(\sqrt{(d + 3)^2}\) = |(d + 3)| which is positive?
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
gmatclubot
Re: MGMAT Quant. Question - Equations, Inequalities and VICS [#permalink]