Question

Solve equation
(X^2-7x+11)^(x^2-7x+6)=1
I have found solutions
X=1 x=6 x=5 and x=2 but
there are 2 more solutions I can’t find

Answer 1

3,4 are two else because-1 to the power 2n can also be an solution and by taking

Answer 2

Answer:

$1, 2, \ \bold{\underline{\underline{3, 4}}},\ 5, 6$

Step-by-step explanation:

$(x^2 - 7x + 11)^{x^2 - 7x + 6} = 1$

$As \ n^0 = 1, \\ \\ If \ x^2 - 7x + 6 = 0,\ x^2 - 7x + 11\ can have many values. So take \ x^2 - 7x + 6 = 0\ and find its solutions. Don't mention about \ x^2 - 7x + 11. \\ \\ \\ x^2 - 7x + 6 = 0 \\ \\ = x^2 - x - 6x + 6 = 0 \\ \\ = x(x - 1) - 6(x - 1) = 0 \\ \\ = (x - 1)(x - 6) = 0 \\ \\ \\ \therefore\ x = \bold{1}\ \ ;\ \ x = \bold{6}$

$As \ 1^n = 1, \\ \\ If \ x^2 - 7x + 11 = 1,\ x^2 - 7x + 6\ can have many values. So take \ x^2 - 7x + 11 = 1\ and find its solutions. Don't mention about \ x^2 - 7x + 6. \\ \\ \\ x^2 - 7x + 11 = 1 \\ \\ x^2 - 7x + 10 = 0 \\ \\ = x^2 - 2x - 5x + 10 = 0 \\ \\ = x(x - 2) - 5(x - 2) = 0 \\ \\ = (x - 2)(x - 5) = 0 \\ \\ \\ \therefore\ x = \bold{2}\ \ ;\ \ x = \bold{5}$

$Now look at this carefully. \\ \\ x^2 - 7x + 6 = (x - 1)(x - 6) \\ \\ When \ x\ is odd, \ x - 1\ becomes even and \ x - 6\ becomes odd. So the product becomes, even \ \times\ odd = even. \\ \\ When \ x\ is even, \ x - 1\ becomes odd and \ x - 6\ becomes even. So the product becomes odd \times\ even = even. \\ \\ \therefore\ Whether \ x\ is odd or even, \ x^2 - 7x + 6\ becomes even always.$

$As \ (-1)^{2n} = 1, \\ \\ If \ x^2 - 7x + 11 = -1,\ x\ in \ x^2 - 7x + 6\ can have many values. So \ x^2 - 7x + 6 \ can have many even values. So take\ x^2 - 7x + 11 = -1\ and find its solutions. Don't mention about \ x^2 - 7x + 6.$

$We're going to find in which cases \ x^2 - 7x + 11\ becomes \ -1. \\ \\ \\ x^2 - 7x + 11 = -1 \\ \\ x^2 - 7x + 12 = 0 \\ \\ = x^2 - 3x - 4x + 12 = 0 \\ \\ = x(x - 3) - 4(x - 3) = 0 \\ \\ = (x - 3)(x - 4) = 0 \\ \\ \\ \therefore\ x = \bold{3}\ \ ;\ \ x = \bold{4}$

$So, the remaining solutions are got. \\ \\ \\ \therefore\ The solutions are 1, 2, 3, 4, 5, 6.$

$Here, we can't do it with the concept of \ 0^0 = 1 \\ \\ Because, \\ \\ As \ 0^0 = 1, \\ \\ x^2 - 7x + 11 = x^2 - 7x + 6 = 0 \\ \\ \therefore\ By cutting \ x^2 - 7x\ from both sides, \\ \\ 11 = 6 \\ \\ ???$

$Hope this may be helpful. \\ \\ Please\ mark my answer as the \ \bold{brainliest}\ if this may be helpful. \\ \\ Thank you. Have a nice day. \\ \\ \\ \#adithyasajeevan$