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P = the product of all x-values that satisfy the  [#permalink]

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Difficulty:   75% (hard)

Question Stats: 54% (02:25) correct 46% (02:10) wrong based on 105 sessions

HideShow timer Statistics P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
Which of the following is true?

A) P < -14
B) -14 ≤ P < -4
C) -4 ≤ P < 4
D) 4 ≤ P < 14
E) 14 ≤ P

*kudos for all correct solutions

_________________

Originally posted by GMATPrepNow on 09 Jan 2017, 13:01.
Last edited by GMATPrepNow on 09 Jan 2017, 13:32, edited 1 time in total.
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GMAT 1: 780 Q51 V46 Re: P = the product of all x-values that satisfy the  [#permalink]

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With complicated equations, especially those that include ugly exponents, it can be helpful to consider easy cases first, often 0 and 1.

Here, 1 satisfies the equation (zero doesn't since we end up with zero raised to a negative, which is the same as dividing by zero). So our answer needs to include 1.

C is the only option that includes 1, so that should be our answer.
CEO  V
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Re: P = the product of all x-values that satisfy the  [#permalink]

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With complicated equations, especially those that include ugly exponents, it can be helpful to consider easy cases first, often 0 and 1.

Here, 1 satisfies the equation (zero doesn't since we end up with zero raised to a negative, which is the same as dividing by zero). So our answer needs to include 1.

C is the only option that includes 1, so that should be our answer.

Argh! You're totally right about zero raised to a negative.
I have edited the question to reflect the intent of my question (the correct answer is still C though!)
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Re: P = the product of all x-values that satisfy the  [#permalink]

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GMATPrepNow wrote:
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
Which of the following is true?

A) P < -14
B) -14 ≤ P < -4
C) -4 ≤ P < 4
D) 4 ≤ P < 14
E) 14 ≤ P

*kudos for all correct solutions

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

So, the correct answer is C

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

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I simplified equation, results two x values
$$(x^{2}) ^{x^{2} - 2x + 1}$$ = $$x^{3x^{2} + x + 8}$$
On simplifying
$$0 = x^{2} + 5x + 6$$
x = -2 ; x = -3
??
Could anyone pls explain.Thanks
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CEO  V
Joined: 12 Sep 2015
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Re: P = the product of all x-values that satisfy the  [#permalink]

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kanusha wrote:
I simplified equation, results two x values
$$(x^{2}) ^{x^{2} - 2x + 1}$$ = $$x^{3x^{2} + x + 8}$$
On simplifying
$$0 = x^{2} + 5x + 6$$
x = -2 ; x = -3
??
Could anyone pls explain.Thanks

That's great work, but you need to keep going.
We also need to test whether the equation also holds true when the base (x) equals 0, 1 and -1
See my earlier post for this.

Once we've checked those values as well, we see that all possible solutions are x = -2, -3, 0, 1 and -1
So, P = (-2)(-3)(0)(1)(-1) = 0
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Re: P = the product of all x-values that satisfy the  [#permalink]

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GMATPrepNow wrote:
GMATPrepNow wrote:
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
Which of the following is true?

A) P < -14
B) -14 ≤ P < -4
C) -4 ≤ P < 4
D) 4 ≤ P < 14
E) 14 ≤ P

*kudos for all correct solutions

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

So, the correct answer is C

Cheers,
Brent

But that give both C&E as answer

Posted from my mobile device
Intern  B
Joined: 17 May 2018
Posts: 7
Re: P = the product of all x-values that satisfy the  [#permalink]

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Hi,

If I simplify the equations above, I will arrive at: X^2+5x+6 = 0

The product of the solutions can be determined by the rule that the product equals = c/a when the equation is in the form ax^2+bx+c=0. Using this will lead us to the product being 6.

Can anyone from the experts tell what went wrong here please?
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Joined: 12 Sep 2015
Posts: 3787
Re: P = the product of all x-values that satisfy the  [#permalink]

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UdayPratapSingh wrote:
GMATPrepNow wrote:
GMATPrepNow wrote:
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x + 8)
Which of the following is true?

A) P < -14
B) -14 ≤ P < -4
C) -4 ≤ P < 4
D) 4 ≤ P < 14
E) 14 ≤ P

*kudos for all correct solutions

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

So, the correct answer is C

Cheers,
Brent

But that give both C&E as answer

Posted from my mobile device

If P = 0, answer choice E doesn't work, since it says P is GREATER THAN or equal to 14

Cheers,
Brent
_________________ Re: P = the product of all x-values that satisfy the   [#permalink] 30 Dec 2018, 10:35
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