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22 May 2018, 18:25
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
69% (01:44) correct
31%
(01:53)
wrong
based on 156
sessions
History
Date
Time
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Not Attempted Yet
Which if the following has a root in common with x^2 - 6x + 5 = 0
A. x^2+ 1 = 0
B. x^2 - x-2 = 0
C. x^2 - 10x - 5 = 0
D. 2x^2 - 2 = 0
E. x^2 - 2x-3 = 0
A. x^2+ 1 = 0
B. x^2 - x-2 = 0
C. x^2 - 10x - 5 = 0
D. 2x^2 - 2 = 0
E. x^2 - 2x-3 = 0
FillFM
Joined: 22 Jun 2017
Last visit: 07 Dec 2019
Posts: 45
Given Kudos: 69
Location: Brazil
Schools: Rotman '21 McCombs '21 Foster '21 Arizona State'21 Carlson '21 Goizueta '21 Jones '21 Olin '21 Cox '21 Simon '21
GMAT 1: 600 Q48 V25
GPA: 3.5
WE:Engineering (Energy)
Products:
Schools: Rotman '21 McCombs '21 Foster '21 Arizona State'21 Carlson '21 Goizueta '21 Jones '21 Olin '21 Cox '21 Simon '21
GMAT 1: 600 Q48 V25
Posts: 45
22 May 2018, 19:35
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Option D
\(x^2-6x+5\) = \((x-5)\)\((x-1)\)
roots = 5 or 1
- now check/test each answer choice using x = 1 and x = 5 (In these situations, always check the answer choices from E to A: https://www.gmatprepnow.com/articles/handling-%E2%80%9Canalyze-answer-choices%E2%80%9D-questions)
Only D satisfies.
\(x^2-6x+5\) = \((x-5)\)\((x-1)\)
roots = 5 or 1
- now check/test each answer choice using x = 1 and x = 5 (In these situations, always check the answer choices from E to A: https://www.gmatprepnow.com/articles/handling-%E2%80%9Canalyze-answer-choices%E2%80%9D-questions)
Only D satisfies.
23 May 2018, 08:05
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Abhimanyu111
Factor the quadratic to find its roots
\(x^2 - 6x + 5 = 0\)
\((x - 1)(x - 5)= 0\)
\(x = 1\) or \(x = 5\)
Plug both roots into the answer choices. Scan the choices first.
Eliminate A (inevitable positive result, not 0); and
Eliminate C (with roots this small, inevitable negative result).*
FillFM 's link** and GMATPrepNow 's article are helpful. +1
The arithmetic here is easy. I started with B. This question took a little over half a minute.
Analysis of all choices for those who are curious:
A. \(x^2+ 1 = 0\)
\(x=1\): \((1+1) = 2\) NO
\(x=5\): \((25+1)=26\) NO
B. \(x^2 - x-2 = 0\)
\(x=1\): \((1-1-2) = -2\) NO
\(x=5\): \((25-5-2) = 18\) NO
C. \(x^2 - 10x - 5 = 0\)
\(x=1\): \((1-10-5) = -14\)
\(x=5\): \((25-50-5) = -30\)
D. \(2x^2 - 2 = 0\)
\(x=1\): \((2-2) = 0\) YES
Finished. We just need one root.
In case there is doubt, (D) cont.:
\(x=5\): \((50-2=48)\) NO. Does not matter. We need "a" root (one) in common. We have \(x = 1\)
E. \(x^2 - 2x-3 = 0\)
\(x=1\): \((1-2-3) = -4\)
\(x=5\): \((25-10-3) = 12\) NO
ANSWER D
*Eliminate A: Both roots are positive. Answer A has two positive terms and a plus sign. Positive roots cannot make that equation true.
Eliminate C: For roots 1 and 5, the squared \(a\) term is not great enough to overcome the negative coefficient of the \(b\) term
** FillFM links to an excellent article that recommends starting with E for "analyze the answer choices". (The stats are startling.) I'm a bit paranoid. GMAC adapts to public warnings.
