Hints:
The base is
(x^2-17x+71) which does not have rational factors, because 71 has only two factors, 71 and 1, each of which is not a factor to the expression. This means that all integer values of x will make the base non-zero.
The exponent, (x^2-34x+240), has two integral factors, which means that when x=x1 or x=x2, the exponent vanishes.
For these values, base^exponent always equal 1.
What is the sum of all integer values of x such that
(x^2−17x+71)^(x^2−34x+240)=1?
2 answers
Consider a = (x^2 -17x +71) and b = (x^2 -34x +240).
case 1: let a = 1, b may be anything.
x^2 -17x +71 = 1, which gives x = 7 or 10.
case 2: let b =0 and a non-zero.
x^2 -34x +240 = 0, gives x = 10 or 24.
case 3 : let a = -1 and b = even.
x^2 -17x +71 = -1, gives x = 9 or 8.
here only x = 8 is valid because, substituting 8 in the equation gives even value {(-1) ^ even = 1}.
thus x = 7,10,24,8 satisfy the given problem. Sum of 7 + 10 + 24 + 8 = 49.
case 1: let a = 1, b may be anything.
x^2 -17x +71 = 1, which gives x = 7 or 10.
case 2: let b =0 and a non-zero.
x^2 -34x +240 = 0, gives x = 10 or 24.
case 3 : let a = -1 and b = even.
x^2 -17x +71 = -1, gives x = 9 or 8.
here only x = 8 is valid because, substituting 8 in the equation gives even value {(-1) ^ even = 1}.
thus x = 7,10,24,8 satisfy the given problem. Sum of 7 + 10 + 24 + 8 = 49.