GMATPrepNow wrote:
What is the sum of all solutions to the equation x^(2x² + 4x – 6) = x^(x² + 8x +6) ?
A) -4
B) -3
C) 3
D) 4
E) 5
* Kudos for all correct solutions
IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ -1For example, if we have 1^x = 1^y, we cannot conclude that x = y, since 1^x equals 1^y FOR ALL values of x and y. For example, 1² = 1³, but we can't conclude that 2 = 3.
So, let's first see what happens when the base (x) equals 0, 1 and -1
If x =
0, then we have:
0^(2(
0²) + 4(
0) – 6) =
0^(
0² + 8(
0) + 6)
Simplify:
0^(-6) =
0^6
Evaluate: 0 = 0
So, x =
0 is one solution to the equation (yes, I know that x =
0 does not change the SUM of the solutions. I just want to show all of the possible considerations)
If x =
1, then we have:
1^(2(
1²) + 4(
1) – 6) =
1^(
1² + 8(
1) + 6)
Simplify:
1^0 =
1^15
Evaluate: 1 = 1
So, x =
1 is another solution to the equation
If x =
-1, then we have:
(-1)^[2
(-1)² + 4(
-1) – 6] =
(-1)^[
(-1)² + 8(
-1) + 6]
Simplify:
(-1)^(-8) =
(-1)^(-1)
Evaluate: 1 = -1
So, x =
-1 is NOT a solution to the equation
Now let's assume that x ≠ 0, x ≠ 1 and x ≠ -1 and look for other x-values that satisfy the given equation.
Given: x^(2x² + 4x – 6) = x^(x² + 8x + 6)
Since the bases are the same, we can write: 2x² + 4x – 6 = x² + 8x + 6
Rearrange to get: x² - 4x – 12 = 0
Factor to get: (x - 6)(x + 2) = 0
So, x =
6 and x =
-2 are also solutions to the equation.
So, the solutions are x =
0, x =
1, x =
6, and x =
-20 +
1 +
6 + (
-2) = 5
Answer: E
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