Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 0808:00 PM EDT
-10:00 PM EDT
Master the GMAT with expert live instruction, a personalized study plan, and real-time support. Includes 40 hours of online classes plus 6 months of access to the TTP GMAT OnDemand video course. Mon/Wed June 8, 2026 →August 12, 2026 8:00pm-10:00pm EST
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..- Jun 10
10:00 AM PDT
-11:00 AM PDT
Scoring 715 on the GMAT Focus Edition requires more than just learning formulas, memorizing concepts, or solving hundreds of questions. In this episode, Nishant shares how he improved his GMAT preparation by focusing on application of concepts, and more. 
Jun 1111:00 AM EDT
-01:00 PM EDT
TTP GMAT OnDemand gives serious students 400+ hours of expert video instruction, the full TTP course, AI support, weekly office hours, and a 715+ score guarantee—all built for elite GMAT score improvement.
26 May 2023, 10:30
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
26% (02:31) correct
74%
(01:56)
wrong
based on 115
sessions
History
Date
Time
Result
Not Attempted Yet
What is the number of roots of the equation \((x^2-5x+5)^{(x^2-5x-6)}=1\)?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
To determine the number of roots of the equation, we need to consider two cases:
Case 1: The exponent (x^2 - 5x - 6) equals zero.
In this case, we have:
x^2 - 5x - 6 = 0
Using the quadratic formula, we can solve for x:
x = (5 ± √(5^2 - 4 * 1 * (-6))) / (2 * 1)
x = (5 ± √(25 + 24)) / 2
x = (5 ± √49) / 2
x = (5 ± 7) / 2
The solutions for this case are x = 6 and x = -1.
Case 2: The base (x^2 - 5x + 5) equals one.
In this case, we have:
x^2 - 5x + 5 = 1
Rearranging the equation:
x^2 - 5x + 4 = 0
Factoring the quadratic equation:
(x - 1)(x - 4) = 0
The solutions for this case are x = 1 and x = 4.
Therefore, the equation (x^2 - 5x + 5)^(x^2 - 5x - 6) = 1 has a total of four roots: x = 6, x = -1, x = 1, and x = 4.
Hence, option C.
×--‐---------×--------------×
Update:-
Correct choice should be E, as I failed to consider the 3rd case where the base = -1 and the exponent = even numbers. Just follow the gmatophobia's explanation.
Thanks!
Posted from my mobile device
Case 1: The exponent (x^2 - 5x - 6) equals zero.
In this case, we have:
x^2 - 5x - 6 = 0
Using the quadratic formula, we can solve for x:
x = (5 ± √(5^2 - 4 * 1 * (-6))) / (2 * 1)
x = (5 ± √(25 + 24)) / 2
x = (5 ± √49) / 2
x = (5 ± 7) / 2
The solutions for this case are x = 6 and x = -1.
Case 2: The base (x^2 - 5x + 5) equals one.
In this case, we have:
x^2 - 5x + 5 = 1
Rearranging the equation:
x^2 - 5x + 4 = 0
Factoring the quadratic equation:
(x - 1)(x - 4) = 0
The solutions for this case are x = 1 and x = 4.
Therefore, the equation (x^2 - 5x + 5)^(x^2 - 5x - 6) = 1 has a total of four roots: x = 6, x = -1, x = 1, and x = 4.
×--‐---------×--------------×
Update:-
Correct choice should be E, as I failed to consider the 3rd case where the base = -1 and the exponent = even numbers. Just follow the gmatophobia's explanation.
Thanks!
Posted from my mobile device
26 May 2023, 13:17
Kudos
Bookmarks
Bunuel
\(a^{b}=1\) under the following three conditions -
1) a = 1; b can be any value
2) b = 0; a can be any value except 0
3) a = -1; b has to be an even integer
Let's check all of these conditions for the above equation
\((x^2-5x+5)^{(x^2-5x-6)}=1\)
Case 1: \(x^2-5x+5 = 1\)
\(x^2 - 5x + 4 = 0\)
\(x^2 - 4x - x + 4 = 0\)
\((x-1)(x-4) = 0\)
- x = 1
- x = 4
Case 2: \(x^2-5x-6 = 0 \) AND \(x^2-5x+5 \neq 0\)
\(x^2-6x+x-6 = 0 \)
\((x+1)(x-6) = 0 \)
x = -1
Let's verify whether \(x^2-5x+5=0\)
\(1+5+5 = 11\)
\(11 \) → As the value is not equal to zero, x = -1 is a valid value.
x = 6
Let's verify whether \(x^2-5x+ 5 = 0\)
\(36-30+5\)
\(11 \) → As the value is not equal to zero, x = 6 is a valid value.
Case 3: \(x^2-5x+5 = -1 \) AND \(x^2-5x-6 = \text{even}\)
\(x^2-5x+6=0\)
\(x^2-3x-2x+6=0\)
\((x-3)(x-2)=0\)
x = 3 Let's verify whether \(x^2-5x-6 = \text{even}\)
odd - odd - even = even → As \(x^2-5x-6\) is even, x = 3 is a valid value.
x = 2 Let's verify whether \(x^2-5x-6 = \text{even}\)
even - even - even = even → As \(x^2-5x-6\) is even, x = 2 is a valid value.
Hence, a total of 6 values are possible.
Option E
