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n2739178
Joined: 12 May 2010
Last visit: 05 Jul 2013
Posts: 233
Given Kudos: 12
Location: United Kingdom
Concentration: Entrepreneurship, Technology
Schools: LBS '14 Oxford '14 Judge Sloan '14 INSEAD '14
GMAT Date: 10-22-2011
GPA: 3
WE:Information Technology (Internet and New Media)
22 Aug 2010, 04:07
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Bookmarks
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!
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
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
n2739178
Joined: 12 May 2010
Last visit: 05 Jul 2013
Posts: 233
Given Kudos: 12
Location: United Kingdom
Concentration: Entrepreneurship, Technology
Schools: LBS '14 Oxford '14 Judge Sloan '14 INSEAD '14
GMAT Date: 10-22-2011
GPA: 3
WE:Information Technology (Internet and New Media)
22 Aug 2010, 04:28
Kudos
Bookmarks
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?
08 Jan 2012, 11:44
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Bookmarks
I am facing the same predicament...I was first convinced by the explanation at https://www.manhattangmat.com/forums/if ... 13498.html but then page 98 contradicts the same.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
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!
