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29 Aug 2022, 17:53
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
81% (02:43) correct
19%
(01:09)
wrong
based on 26
sessions
History
Date
Time
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Not Attempted Yet
John and Tom attempted to solve a quadratic equation. John made a mistake in writing down the constant term. He ended up with the roots (4, 3). Tom made a mistake in writing down the coefficient of x. He got the roots as (3, 2). What will be the exact roots of the original quadratic equation?
A. (6, 1)
B. (-3, -4)
C. (4, 3)
D. (-4, -3)
E. (-4, 3)
(adapted from gmatfree)
A. (6, 1)
B. (-3, -4)
C. (4, 3)
D. (-4, -3)
E. (-4, 3)
(adapted from gmatfree)
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29 Aug 2022, 22:42
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pintukr
Solution:
John got the roots: (4, 3). Therefore, the quadratic equation written by John \(= x^2-(sum of roots)x+(product of roots)=0\)
\(⇒x^2-7x+12=0\)
The constant term i.e., 12 is a mistake here
Tom got the roots: (3, 2). Therefore, the quadratic equation written by John \(= x^2-(sum of roots)x+(product of roots)=0\)
\(⇒x^2-5x+6=0\)
The coefficient of \(x\) i.e., -5 is a mistake here
Thus, we can say that the correct equation is \(x^2-7x+6=0\)
\(⇒x^2-x-6x+6=0\)
\(⇒x(x-1)-6(x-1)=0\)
\(⇒(x-1)(x-6)=0\)
The roots are (1, 6)
Hence the right answer is Option A
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 Problem Solving (PS) Forum
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