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P = the product of all x-values that satisfy the

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P = the product of all x-values that satisfy the [#permalink] New post 09 Jan 2017, 12:02
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60% (02:23) correct 40% (03:27) wrong based on 5 sessions
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
What is the value of P?

Answer:
[Reveal] Spoiler:
0

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Re: P = the product of all x-values that satisfy the [#permalink] New post 23 Jan 2017, 01:35
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(x²)^(x² - 2x + 1) = x^(3x² + x + 8)
(x)^(2x² - 4x + 2) = x^(3x² + x + 8)
then, exponents are equal now,
2x² - 4x + 2 = 3x² + x + 8
-X^2 - 5X - 6 = 0
X^2 + 5X + 6 = 0
(X+2)(X+3)=0
thn, X= -2, and X= -3.
Also, X=0, X=1, and X= -1
P= -2*-3*0*1*-1=0
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Re: P = the product of all x-values that satisfy the [#permalink] New post 28 Feb 2018, 13:00
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GreenlightTestPrep wrote:
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
What is the value of P?

Answer:
[Reveal] Spoiler:
0


IMPORTANT: If b^x = b^y, then x = y, as long as b ≠ 0, b ≠ 1 and b ≠ -1
For 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.
So, although 1² = 1³, we can't then conclude that 2 = 3.


So, let's first see what happens when the base (x) equals 0.

If x = 0, then we get: (0²)^(0² - 2(0) + 1) = 0^(3(0²) + 0 + 8)
Simplify: 0^1 = 0^8
Evaluate: 0 = 0
Perfect! We know that x = 0 is one solution to the equation.
This means the PRODUCT of all of the solutions will be ZERO, regardless of the other solutions.
In other words, P = 0

Answer: 0

Cheers,
Brent
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Re: P = the product of all x-values that satisfy the   [#permalink] 28 Feb 2018, 13:00
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