Question

how many integer x satisfies the equation (x^2-x-1)^x+2 = 1; how many integer x satisfies the equation (x^2-x-1)^x+2 = 1

Answer 1

Answer:

x = -2 & -1

Step-by-step explanation:

(x^2-x-1)^x+2 = 1

$(x^2 - x -1)^{(x+2)} = 1$

There can be 3 cases where equation Holds true

case 1 when

x + 2 = 0

=> x = -2

Case 2 When

x² - x -1 = 0

=> x = (1 +/- √5 )/2  -  Not integers

Case 3  When

x² - x -1 = 1  and x +2 = 1

=> x² - x - 2 = 0    and x = -1

=> x² -2x + x -2 = 0    and x = -1

=> x(x-2) + 1 (x-2) = 0   and x = -1

=> (x + 1) ( x-2) = 0     and x = -1

=> x = -1 & x = 2   and x = -1

-1 is common

lets verify solution

x = -2

(4 + 2 -1)^0 = 5^0 = 1

x = -1

(1 + 1 - 1)^1 = 1^1  = 1