perplexus.info :: Just Math : Quadratic Powered Quadratic

Quadratic Powered Quadratic (Posted on 2024-01-18)

How many different integers satisfy the equation

(x²-5x+5)^(x²-11x+30)=1?

No Solution Yet Submitted by Danish Ahmed Khan

Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)

Puzzle Solution

| Comment 1 of 2

(-1)^0 =1

1^0=1

Hence,

Either, x^2-5x+5=-1 => x^2-5 x+6 =0=> (x-2)(x-3)=0=> x=2, 3

Or, x^2--11x+30=0=0=> (x-5)(x-6) = 0=> x=5,6

Or, x^2--5x+5= 1=> x^2 -5x +4 = 0=> (x-1)(x-4)=0=> x= 1, 4

Therefore, x ε{1,2,3,4,5,6}

Consequently, precisely six distinct integers satisfy the equation.

Edited on January 18, 2024, 6:50 am

Posted by K Sengupta on 2024-01-18 06:48:16

Please log in:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (14)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info