What is the sum of all integer values of x such that
(x^2−17x+71)^(x^2−34x+240)=1?
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.
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.
To find the sum of all integer values of x that satisfy the equation (x^2 - 17x + 71)^(x^2 - 34x + 240) = 1, we first need to consider the various cases where the equation holds true.
When a number raised to any power equals 1, it means that the base number can be either 1 or -1.
So, we have two possible cases to consider:
Case 1: (x^2 - 17x + 71) = 1 and (x^2 - 34x + 240) is even
Case 2: (x^2 - 17x + 71) = -1 and (x^2 - 34x + 240) is odd
Let's solve each case individually:
Case 1: (x^2 - 17x + 71) = 1 and (x^2 - 34x + 240) is even
Subtracting 1 from both sides of the first equation, we get:
x^2 - 17x + 70 = 0
Factoring the equation, we have:
(x - 10)(x - 7) = 0
Setting each factor equal to zero, we get two possible solutions:
x - 10 = 0 -> x = 10
x - 7 = 0 -> x = 7
Now, let's consider the second equation (x^2 - 34x + 240). Since we want it to be even, we need the discriminant (b^2 - 4ac) to be a perfect square.
For this equation, the discriminant is:
b^2 - 4ac = (-34)^2 - 4(1)(240) = 1156 - 960 = 196
It is a perfect square (14^2 = 196), so we can proceed.
Now we solve for x:
x^2 - 34x + 240 = 0
Factoring the equation, we have:
(x - 10)(x - 24) = 0
Setting each factor equal to zero, we get two additional solutions:
x - 10 = 0 -> x = 10
x - 24 = 0 -> x = 24
So, for Case 1, the possible solutions are x = 7, 10, 24.
Case 2: (x^2 - 17x + 71) = -1 and (x^2 - 34x + 240) is odd
Adding 1 to both sides of the first equation, we get:
x^2 - 17x + 72 = 0
Factoring the equation, we have:
(x - 8)(x - 9) = 0
Setting each factor equal to zero, we get two possible solutions:
x - 8 = 0 -> x = 8
x - 9 = 0 -> x = 9
Now, let's consider the second equation (x^2 - 34x + 240). Since we want it to be odd, we need the discriminant (b^2 - 4ac) to be an odd number.
For this equation, the discriminant is:
b^2 - 4ac = (-34)^2 - 4(1)(240) = 1156 - 960 = 196
It is an even number, so there are no solutions for Case 2.
Therefore, the sum of all integer solutions of x is:
7 + 10 + 24 + 8 + 9 = 58.