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P = the product of all x-values

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by Brent@GMATPrepNow » Mon Jan 09, 2017 12:03 pm
Here's a 650-level (ish) question I just created. See how you do.
P = 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
Answer: C
Last edited by Brent@GMATPrepNow on Mon Jan 09, 2017 12:37 pm, edited 2 times in total.
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by DavidG@VeritasPrep » Mon Jan 09, 2017 12:21 pm
Brent@GMATPrepNow wrote:Here's a 650-level (ish) question I just created. See how you do.
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x - 12)
Which of the following is true?

A) P < -14
B) -14 ≤ P < -4
C) -4 ≤ P < 4
D) 4 ≤ P < 14
E) 14 ≤ P
Answer: C
I enjoy these Brent Originals.

A hint to would-be question tacklers: The GMAT (and Brent) often design a question to both appear complicated and address a simple fundamental concept at the same time.
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by Brent@GMATPrepNow » Mon Jan 09, 2017 12:36 pm
DavidG@VeritasPrep wrote:
Brent@GMATPrepNow wrote:Here's a 650-level (ish) question I just created. See how you do.
P = the product of all x-values that satisfy the equation (x²)^(x² - 2x + 1) = x^(3x² + x - 12)
Which of the following is true?

A) P < -14
B) -14 ≤ P < -4
C) -4 ≤ P < 4
D) 4 ≤ P < 14
E) 14 ≤ P
Answer: C
I enjoy these Brent Originals.

A hint to would-be question tacklers: The GMAT (and Brent) often design a question to both appear complicated and address a simple fundamental concept at the same time.
Slight problem with my original question.
I've now edited it to remove the - 12 part.

Cheers,
Brent
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by Brent@GMATPrepNow » Tue Jan 10, 2017 10:31 am
Brent@GMATPrepNow wrote:Here's a 650-level (ish) question I just created. See how you do.
P = 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
Answer: C
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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by Shameilia » Tue Jan 10, 2017 7:19 pm
I factored out the quadratic equation in the exponents. I got the following:
(2X-2)(X-1) (1,1)
(3X+6)(X-2) (-2, 2)

From there I multiplied the 1 x 1 x -2 x 2 = -4.

I selected answer C.

Was my path to the correct answer by chance?
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by Matt@VeritasPrep » Fri Jan 20, 2017 12:43 am
Shameilia wrote:I factored out the quadratic equation in the exponents. I got the following:
(2X-2)(X-1) (1,1)
(3X+6)(X-2) (-2, 2)

From there I multiplied the 1 x 1 x -2 x 2 = -4.

I selected answer C.

Was my path to the correct answer by chance?
(Note: I'm going to break this up into parts, since it has a few neat twists and also is a bit too long for one post.)

There was some luck involved there, but as they say in poker, better lucky than good!

You could start by dividing both sides by the right hand side, giving

(x²)^(x² - 2x + 1) / x^(3x² + x + 8) = 1

Now let's distribute the 2 in the numerator on the left:

(x)^(2x² - 4x + 2) / x^(3x² + x + 8) = 1

Then we'll apply our exponent property, xᵃ/xᵇ = xᵃ�ᵇ:

x^(2x² - 4x + 2 - (3x² + x + 8)) = 1

x^(-x² - 5x - 6) = 1

Since the right hand side, 1, is the same as x�, we can say

x^(-x² - 5x - 6) = x�

and that

-x² - 5x - 6 = 0

0 = x² + 5x + 6

0 = (x + 2) * (x + 3)

So it looks like x = -2 and x = -3 are the solutions, but did you notice my mistake?
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by Matt@VeritasPrep » Fri Jan 20, 2017 12:45 am
Back in the line above in blue, I casually divided by both sides, without recognizing that I was dividing by a variable! If I do this, I'm (perhaps unwittingly) assuming that x ≠ 0, because if x IS 0, then I've divided by 0 and broken my equation.

Going back to the beginning, I can test to see if x = 0 is a solution by plugging it into the equation. Lo and behold, x = 0 DOES satisfy the equation (it gives 0 = 0), so x = 0 IS a solution that I've missed, and now I've got

x = 0
x = -2
x = -3

But wait, it gets worse!
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by Matt@VeritasPrep » Fri Jan 20, 2017 12:45 am
I made another assumption above in orange, that 1 = x� and not simply 1 to some other power. Since this is also a possibility, I need to test that as well:

x^(-x² - 5x - 6) = 1

If x = 1, then 1 to any power (including -x² - 5x - 6) will give 1, so I could have x = 1. And if x = -1, then -1 to any EVEN power will equal 1. If x = -1, then -x² - 5x - 6 is even, so x = -1 is one more solution. Yeesh!

So, all told, my solutions are

x = 1
x = 0
x = -1
x = -2
x = -3

... and now we can see why Brent only asked for the PRODUCT of the solutions :D
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