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04 Aug 2015, 08:10
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\(\sqrt{2x^{2} - x - 9} = x + 1\)
Since I cannot simplify \(2x^{2} - x - 9\) I would first of all square the whole equation:
\(|2x^{2} - x - 9| = (x + 1)^{2}\)
Now, I would normally split into 2 cases, solve for x and re-check every solution:
1) \(2x^{2} - x - 9 x\geq{3} 0\)
2) \(2x^{2} - x - 9 x\leq{5} 0\)
... however this takes a lot of time.
In the OE, the total value is ignored:
\(\sqrt{2x^{2} - x - 9} = x + 1\)
\(2x^{2} - x - 9 = (x + 1)^{2}\)
and after solving the equation, the solutions are plugged into the original equation to check for errors.
My actual question is: Can I always ignore the absolute value and, at the end, only plug in the solutions in the original equation? Is this "plug-in check" sufficient?
Since I cannot simplify \(2x^{2} - x - 9\) I would first of all square the whole equation:
\(|2x^{2} - x - 9| = (x + 1)^{2}\)
Now, I would normally split into 2 cases, solve for x and re-check every solution:
1) \(2x^{2} - x - 9 x\geq{3} 0\)
2) \(2x^{2} - x - 9 x\leq{5} 0\)
... however this takes a lot of time.
In the OE, the total value is ignored:
\(\sqrt{2x^{2} - x - 9} = x + 1\)
\(2x^{2} - x - 9 = (x + 1)^{2}\)
and after solving the equation, the solutions are plugged into the original equation to check for errors.
My actual question is: Can I always ignore the absolute value and, at the end, only plug in the solutions in the original equation? Is this "plug-in check" sufficient?
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ENGRTOMBA2018
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04 Aug 2015, 08:51
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nicok06
Can you please format the question properly as per the forum guidelines? (the links are in my signature!). Post your doubt as a new question and provide options so that the forum members can try the question and start a discussion. If not, I will have to delete this topic altogether as it does not meet the posting guidelines.
05 Aug 2015, 10:24
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nicok06
Nicok06 -
While all numbers have two square roots, when the radical sign is used, only the principal (positve) square root is considered. Thus, in your problem above, there is no need to consider the absolute value. A good way to remember is: when x^2 = 16, x = 4 and -4, but \sqrt{16} = 4 only.
Hope this helps.
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!
