If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of

Question

GMATBusters’ Quant Quiz Question -3 
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If  M = (x^2-7x+11)  and  N = (x^2-13x+42) , what is the number of roots of the equation  M^N = 1 ?

A. 3
B. 4
C. 5
D. 6
E. 7

GMATBusters · Answer

The Official Solution is as follows: https://gmatclub.com/forum/download/file.php?id=52759 WhatsApp Image 2020-02-03 at 6.49.55 PM (2).jpeg

AndreV · Answer

Factorizing
N=(x-6)(x-7)
for M^N=1, we have the following possibilities:
1) N=0 and M is different than 0
N=0, then (x-6)(x-7)=0
here x=6 or x=7
2) M=1 v M=-1
x^2-7x+11=1
x^2-7x+10=0
(x-2)(x-5)=0
here x=2 or x=5
^2-7x+11=-1 and N is even
x^2-7x+12=0
(x-3)(x-4)=0
here x=3 or x=4. In both cases N is even
Number of roots=6

merajul · Answer

M^N = 1 => M^N = 1^0 => N=0

Now N= (x^2-13x+42)=0 will derive to x = 6 or 7

that means M^N = 1 => (x^2-7x+11)^6 =1 or (x^2-7x+11)^7 =1

So the number of roots should be 7

ITGMAT · Answer

sol; 
we can see from the question that multiple roots are possible. 
M^N=1 the equation can be satisfied by M=1,-1 or N=0 or integer

case-1: 
lets consider if M=1,

x^2-7x+11= 01
x^2-7x =10=0 
(x-2)(x-5)=0
x= 2, 5

N= 0 >>>>X^2-13x+42= 0
(x-6)(x-7)= 0,
x= 6,7 
so roots = 2,5,6,7
case-2: 
now consider other condition
M= -1, then N=even,
x^2-7x+11= -1
x^2-7x+12= 0
(x-3)(x-4)=0
x= 3,4
hence no. of total roots = 6 
Answer is D.

Archit3110 · Answer

not sure
given
x^2-7x+11^( x^2-13x+11) = 1
we get
x^2-7x+11 ^ ( (x-6)*(x-7) = 1
total possible roots would be 4 
IMO B

J2S2019 · Answer

Ans is B
Attachment is given for answer

Kinshook · Answer

Given: M = (x^2−7x+11) and N = (x^2−13x+42)
Asked: What is the number of roots of the equation M^N= 1?
A. 3
B. 4
C. 5
D. 6
E. 7

When N=0; M^N=1 
N=0 when x=6 or x=7; 2 roots

When M=1; M^N=1
x^2-7x+11 = 1; x^2 -7x+10 = 0; x = 2 or x=5; 2 roots

When M=-1; M^N=1 when N is even
x^2-7x+11 = -1; x^2 -7x+12= 0; x = 4 or x=3
When x=4; N=6; 1 root
When x=3; N= 12; 1 root

There are total 6 roots of the equation.

IMO D

pyunika · Answer

Here we can divide solution into three parts.
Anything raised to 0 is 1 so equating equation n to zero gives us two roots.
2nd part= 1 to the power of anything will give us 1 so equating equation M to 1 we get two solutions for M
3rd part = -1 to any even power of anything will give us 1 so equating equation M to -1 we get two solutions for M.
So total 6 solutions.

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If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of

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If M = (x^2 - 7x + 11) and N = (x^2 - 13x + 42), what is the number of

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