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Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]

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06 Jul 2013, 09:53

Expert's post

jjack0310 wrote:

Bunuel wrote:

jjack0310 wrote:

Sorry. It is not clear.

Can you explain what was wrong with the way I was approaching the problem?

I mean other than the part that you marked red, what was I doing wrong? Do I have to solve theproblem using the solution that you mentioned?

If x > -2, how is |x + 2| = (x - 2)? Is there an identity that I am missing? If I plug in, X = -1, |x + 2| = 1, but (x - 2) = -3

Why the discrepancy? What identity am I missing?

There was a typo: When \(x\leq{-2}\), then \(|x+2|=-(x+2)\) When \(x>{-2}\), then \(|x+2|=(x+2)\).

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

For our question, when x>-2 (when x+2>0), |x+2|=x+2.

Hope it's clear.

Got it.

Thanks,

Final question, why are there two possibilities for when x = 0? Is that correct? or a typo?

No, that's not a typo: |0|=0=-0. _________________

x<1 x^2 - 4x + 6 = 3 - |x - 1| x^2 - 4x + 6 = 3 - -(x-1) x^2 - 4x + 6 = 3 - (-x+1) x^2 - 4x + 3 = + x - 1 x^2 - 5x + 4 = 0 (x-1)(x-4) = 0 x=1, x=4 Neither 1 or 4 fall within the range of x<1 INVALID

The product is (2)

(C)

(Bunuel, could you explain to me how we know where the greater than or equals sign goes in these problems? I see that in your solution you had x>=1, but I don't know why that is as opposed to say, x<=1)

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]

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19 Jun 2014, 02:57

Bunuel wrote:

carcass 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

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 \(x\geq{1}\), 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.

Dear Bunuel,

We do not need to consider two situation of |x - 1|.

What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]

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11 Aug 2015, 18:03

paul16 wrote:

Hello everyone,

I am so close to understanding this question, but the one thing I do not understand is why the positive of |x+2| is >= and the negative of |x+2| is just <?

Sorry if its a dumb question

Paul

I have this question too. Typically it doesnt matter if we use the +ve or -ve absolute value function since the value is zero anyways. But for this case, we need to use the value to include/exclude values, so is >= and < the standard?

Re: What is the product of all the solutions of x^2 - 4x + 6=3 [#permalink]

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11 Aug 2015, 18:15

Expert's post

ggurface wrote:

paul16 wrote:

Hello everyone,

I am so close to understanding this question, but the one thing I do not understand is why the positive of |x+2| is >= and the negative of |x+2| is just <?

Sorry if its a dumb question

Paul

I have this question too. Typically it doesnt matter if we use the +ve or -ve absolute value function since the value is zero anyways. But for this case, we need to use the value to include/exclude values, so is >= and < the standard?

1 thing, the text in red is not always true. |x| can be any value other than 0 as well.

Coming back to your question, the reason we include "=" with ">" sign because |x| = x for x\(\geq\)0 while |x| =-x for x<0

The convention is to always include "+" with ">" and not with "<" as the NATURE of |x| remains the same for x\(\geq\) 0, while the nature changes for |x| (we need to put a negative infront of 'x') when x<0.

For example, |4| = 4 or |0| = 0 but |-5| = -(-5) = 5

\(x+2=0\) --> \(x=-2\); OR \((x+2)^2-1=0\) --> \((x+2)^2=1\) --> \(x=-1\) or \(x=-3\).

The product of the roots: \((-2)*(-1)*(-3)=-6\).

Answer: A.

Hope it's clear.

Hi Bunuel,

When I tried to follow the same method I get different answer for the below question.

x^2 + 4x + 7 = |x + 2| + 3.

When x <= 0 then |x + 2| = - (x+2 ) = -x - 2.

Then x^2 + 4x + 7 = - x - 2 + 3.

=> x^2 + 4x + 7 = - x + 1. => x^2 + 5x + 6 = 0 => ( x +2 ) ( x + 3) = 0 => x = -2 and - 3 . ( Since x < = 0 both the numbers are possible ).

When x > 0 then |x + 2| = x + 2.

Then x^2 + 4x + 7 = x + 2 + 3. => x^2 + 4x + 7 = x + 5. => x^2 + 3x + 2 = 0 . =>( x + 2 ) ( x + 1 ) = 0. x = - 2 and - 1 ( Since x > 0 both numbers doesn't fit in the range ).

Then suitable numbers would be -2 and -3 and the product should be 6.

\(x+2=0\) --> \(x=-2\); OR \((x+2)^2-1=0\) --> \((x+2)^2=1\) --> \(x=-1\) or \(x=-3\).

The product of the roots: \((-2)*(-1)*(-3)=-6\).

Answer: A.

Hope it's clear.

Hi Bunuel,

When I tried to follow the same method I get different answer for the below question.

x^2 + 4x + 7 = |x + 2| + 3.

When x <= 0 then |x + 2| = - (x+2 ) = -x - 2.

Then x^2 + 4x + 7 = - x - 2 + 3.

=> x^2 + 4x + 7 = - x + 1. => x^2 + 5x + 6 = 0 => ( x +2 ) ( x + 3) = 0 => x = -2 and - 3 . ( Since x < = 0 both the numbers are possible ).

When x > 0 then |x + 2| = x + 2.

Then x^2 + 4x + 7 = x + 2 + 3. => x^2 + 4x + 7 = x + 5. => x^2 + 3x + 2 = 0 . =>( x + 2 ) ( x + 1 ) = 0. x = - 2 and - 1 ( Since x > 0 both numbers doesn't fit in the range ).

Then suitable numbers would be -2 and -3 and the product should be 6.

Please clarify me if I am missing anything...

The transition point (value) must be the point for which the expression in modulus changes its sign. For |x + 2| it's -2 not 0. So, you should consider the ranges when x<=-2 and when x>-2. Check highlighted text in my solution. _________________

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