Nina1987 wrote:What is the product of all the solutions of x^2 - 4x + 6 = 3 - |x - 1| ?

(A) -8
(B) -4
(C) 2
(D) 4
(E) 8

When an equation has absolute value on ONLY ONE SIDE, plug any possible solutions back into the original equation to confirm that they are valid.

x² - 4x + 3 = -|x-1|.

Case 1: signs unchanged
x² - 4x + 3 = -(x-1)
x² - 4x + 3 = -x + 1
x² - 3x + 2 = 0
(x-1)(x-2) = 0.
Possible solutions: x=1, x=2.

If we plug x=1 into x² - 4x + 3 = -|x-1|, we get:
1² - (4*1) + 3 = -|1-1|.
0 = 0.
Thus, x=1 is a valid solution.

If we plug x=2 into x² - 4x + 3 = -|x-1|, we get:
2² - (4*2) + 3 = -|2-1|.
-1 = -1.
Thus, x=2 is a valid solution.

Case 2: signs changed in the absolute value
x² - 4x + 3 = -(-x+1)
x² - 4x + 3 = x - 1
x² - 5x + 4 = 0
(x-1)(x-4) = 0.
Possible solutions: x=1, x=4.

If we plug x=4 into x² - 4x + 3 = -|x-1|, we get:
4² - (4*4) + 3 = -|4-1|.
3 = -3.
Doesn't work.
Thus, x=4 is NOT a valid solution.

Only x=1 and x=2 are valid solutions.
Product of these two solutions = 1*2 = 2.

The correct answer is C.

Great solution by NYGuru. Thanks much. But I was looking for a quicker approach rather than the conventional one. That's how I solved but ended up using more than 4mins. So would much appreciate quicker solutions. Here is one solution that kind of obviates plugging back to check validity of the roots: (pasted from gmatclub):

If x<1, then |x−1|=−(x−1)=1−x, so in this case we'll have x^2−4x+6=3−(1−x) --> x^2−5x+4=0 --> x=1 or x=4 --> discard both solutions since neither is in the range x<1.

If x≥1, then |x−1|=x−1, so in this case we'll have x^2−4x+6=3−(x−1) --> x^2−3x+2=0 --> x=1 or x=2.

Therefore, the product of the roots is 1*2=2.

Answer: C.

Any other more efficient solutions? Is there a way we can know which are invalid roots w/o actually plugging them back? Thanks

Nina1987 wrote:Great solution by NYGuru. Thanks much. But I was looking for a quicker approach rather than the conventional one. That's how I solved but ended up using more than 4mins. So would much appreciate quicker solutions. Here is one solution that kind of obviates plugging back to check validity of the roots: (pasted from gmatclub):

If x<1, then |x−1|=−(x−1)=1−x, so in this case we'll have x^2−4x+6=3−(1−x) --> x^2−5x+4=0 --> x=1 or x=4 --> discard both solutions since neither is in the range x<1.

Any other more efficient solutions? Is there a way we can know which are invalid roots w/o actually plugging them back? Thanks

The portion in red illustrates a line of reasoning that can be used to discard invalid roots.
For most students, the easiest way to verify the validity of a root will be to plug the root back into the original equation.

This problem is more complex than a typical GMAT problem.
Do not be too concerned that you required extra time to solve it.

How about this for a clever approach:

We know the minimum of x² - 4x + 6 is 2. (You can find the minimum by doing -b/2a, or -(-4)/2*1.)

So we have x² - 4x + 6 ≥ 2.

Since the left hand side and right hand side are the same, we must have 3 - |x - 1| ≥ 2. But |x - 1| is nonnegative, so we must have 1 ≥ |x - 1| ≥ 0, or 2 ≥ x ≥ 0.

Now think of what the equations will look like. x² - 4x + 6 is a parabola and 3 - |x - 1| is going to be two straight lines that meet in a kind of upside down v. So there should be TWO solutions, as the graphs will meet at two points. (They could also meet at ONE point, but then asking for the product of the solutions would be strange.)

... but either way, since the one/two solutions are between 0 and 2, the only possible answer is C! (D would require both solutions to be 2, which of course is illogical.)

GMATGuruNY wrote:

Nina1987 wrote:Great solution by NYGuru. Thanks much. But I was looking for a quicker approach rather than the conventional one. That's how I solved but ended up using more than 4mins. So would much appreciate quicker solutions. Here is one solution that kind of obviates plugging back to check validity of the roots: (pasted from gmatclub):

If x<1, then |x−1|=−(x−1)=1−x, so in this case we'll have x^2−4x+6=3−(1−x) --> x^2−5x+4=0 --> x=1 or x=4 --> discard both solutions since neither is in the range x<1.

Any other more efficient solutions? Is there a way we can know which are invalid roots w/o actually plugging them back? Thanks

The portion in red illustrates a line of reasoning that can be used to discard invalid roots.

NYGuru: Yes I know this an alternate way. Hence I pasted it here. I have also kinda of gotten hold of it now and should be able to apply to a similar problems fairly quickly (hopefully I know incorrect usage as per GMAT SC ). But pasted it here so that you can suggest any OTHER such SIMILAR approaches.

GMATGuruNY wrote: This problem seems far more complex than an official GMAT problem.
Do not be too concerned that you required extra time to solve it.

This is a proper GMAT question. I ran into it in a GMAT Focus test. I think my estimated score was 49-51. This got me worried since it was an official question and I am hitting for Q50/51. I'd not even solve an unofficial question let alone worry about it Thanks for you reply again NYGuru

Indeed it is a really clever approach. Thanks for wolfram links too. Shows you've gone an extra mile to make the solution accessible to me. Im predominantly a visual learner so Really appreciated.

Matt@VeritasPrep wrote:How about this for a clever approach:

We know the minimum of x² - 4x + 6 is 2. (You can find the minimum by doing -b/2a, or -(-4)/2*1.)

Can you elaborate a little on this minimum value formula or direct me to a resource?

Matt@VeritasPrep wrote: 1 ≥ |x - 1| ≥ 0, or 2 ≥ x ≥ 0.

Is there a typo? should it be
2 ≥ x ≥ 1?

Thanks

Nina1987 wrote:Can you elaborate a little on this minimum value formula or direct me to a resource?

Sure! Here's a link.

Matt@VeritasPrep wrote: 1 ≥ |x - 1| ≥ 0, or 2 ≥ x ≥ 0.

Is there a typo? should it be
2 ≥ x ≥ 1?

Thanks

Nope, 1 ≥ |x - 1| ≥ 0 is the same as 2 ≥ x ≥ 0.

1 ≥ |x - 1| ≥ 0 is essentially "the distance between x and 1 is between zero and one units". So x can be at most one unit from 1, i.e. anywhere from 0 to 2, inclusive. To see this, you can try plugging in numbers in our range (from 0 to 2) and numbers outside our range, then checking the results.

Wow!! This is amazing- I am learning so much just from this one example. Almost like in Thoreau's words 'One strike at the roots for a thousand hacking at the branches'

Sure! Here's a link.

That link was really really helpful!

Matt@VeritasPrep wrote: 1 ≥ |x - 1| ≥ 0, or 2 ≥ x ≥ 0.

Is there a typo? should it be
2 ≥ x ≥ 1?

Thanks

Nope, 1 ≥ |x - 1| ≥ 0 is the same as 2 ≥ x ≥ 0.

1 ≥ |x - 1| ≥ 0 is essentially "the distance between x and 1 is between zero and one units". So x can be at most one unit from 1, i.e. anywhere from 0 to 2, inclusive. To see this, you can try plugging in numbers in our range (from 0 to 2) and numbers outside our range, then checking the results.

[/quote]
Wow again! Never felt so clear about modulus and inequality. One more question how would you explain the following inequality? 1 ≥ |x + 1| ≥ 0
Can't thank you enough Matt

Nina1987 wrote:Wow!! This is amazing- I am learning so much just from this one example. Almost like in Thoreau's words 'One strike at the roots for a thousand hacking at the branches'

So glad to hear that! "It's not what you look at that matters, it's what you see."

Matt, Can you pls help me with the following as well?

Wow again! Never felt so clear about modulus and inequality. One more question how would you explain the following inequality? 1 ≥ |x + 1| ≥ 0
Can't thank you enough Matt

Nina1987 wrote:Matt, Can you pls help me with the following as well?

Wow again! Never felt so clear about modulus and inequality. One more question how would you explain the following inequality? 1 ≥ |x + 1| ≥ 0
Can't thank you enough Matt

Sure!

|x - y| is the distance between x and y. |x + y| is really |x - (-y)|, so this represents the distance between x and -y.

1 ≥ |x + 1| ≥ 0, in words, is

"the distance on the number line from x and -1 is between zero and one units, inclusive"

If the distance is zero units, then x = -1 itself. If the distance is one unit, then x = 0 or -2. Since x can be anywhere in that region, this reduces to

-2 ≤ x ≤ 0