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

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

Hope this helps.

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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pushpitkc , Bunuel, @niks18
can you please explain, looking at Bunuel`s solution in which case when testing values do i need use more or equal sign VS just more / less sign

\(x<1\) VS \(x\geq{1}\) and why in case TWO we dont use this \(x>1\) instead of \(x\geq{1}\)

or why in case one \(x<1\) we exclude "less or equal" unlike case 2

on the other hand in the post about important properties both signs (more than / less than) are combined with "equal" sign

https://gmatclub.com/forum/absolute-val ... 75002.html

Hey dave13

Bunuel has already clarified your doubt here
https://gmatclub.com/forum/what-is-the- ... l#p1183663

The basic idea is that you can have both the possibilities \(x>1\) and \(x \leq 1\) or \(x \geq1\) and \(x<1\)

Hope this helps you!

hello generis can you please explain how Bunuel got this \(x<1\), I think I am doing something wrong ,

if Negative I rewrite it as \(|x - 1|<0\) so I get \(-(x-1)< 0\), ---> \(-x+1<0\) --> \(-x<-1\) ---> divide each side by \(-1\) and get I \(x>1\)

same issue with positive case...

i followed your explanation here https://gmatclub.com/forum/what-is-the- ... s#p2081311

"When
(x+2) is positive (one of two "original conditions")
x+2>0
x>−2
)"



.

dave13 , you are really close. You made two small missteps, easily corrected.
The stricken part is not correct, but I can understand why you did what you did. There is some confusion because ...

We actually have 3 different but related things going on:

--- specifying the negative case: \((x-1)<0\)

--- defining a condition, an allowable range for the solutions: (\(x>1\))

--- in simplest form, writing what we will actually plug into the equation: \((1-x)\)


He got that inequality from "If (x-1) < 0 ..." Work backward from what he wrote
\(x<1\)
Subtract 1 from both sides:
\(x-1<(1-1)\), thus:
\(x-1<0\)

In other words, when Bunuel writes \(x<1\), that statement is derived from \(x-1<0\) (the "negative" case). He adds 1 to both sides to get \(x<1\)

That \(x<1\) is sort of SEPARATE. \(x<1\) gives us a range for allowed solutions. Write that condition and ignore it until we have an answer.

You wrote,

Almost. Take the absolute value bars off and replace with parentheses, thus \((x-1)<0\)

We are really evaluating what is inside the bars.

What is inside? The expression x - 1
So we write (x-1), not |x-1|
In more familiar lingo, perhaps, we "remove the [absolute value] brackets."

AND: We cannot write |x-1| = -1
Absolute value cannot be negative because it is a distance from 0, and distances are not negative. The expression inside the absolute bars can be negative.

You wrote:
Almost. Separate it into two parts
\((x-1)<0\): "I am considering the case in which the expression is negative"

\(-(x-1)\): "After I rewrite it a little, I will insert THIS term into the equation."

Steps, Part I:

Step 1) write down that you are considering the negative case, i.e., \((x-1)<0\).

Step 2) separately, decipher what we will actually insert into the equation
--put the negative sign in front of the expression, just as you did: \(-(x-1)\)

Put the expression in simplest terms
\(-(x-1)=\)
\((-x+1)\). Now switch the numbers' order and get
\((1-x)\)
THAT term goes into the equation

Step 3) Before we put that term into the equation, find the condition.
From \((x-1)<0\):
\(x-1+1<0+1\)
\(x<1\)
We ignore that term for a bit

Step 4) Plug #2 -- the whole thing-- into the equation
\(x^2 - 4x + 6 = 3 - |x - 1|\)
\(x^2 - 4x + 6 = 3 - (1-x)\)
\(x^2 - 4x + 6 = 3 - 1+x\)
\(x^2 - 5x + 4 = 0\)

Step 5) Factor to get the roots and solve:
\((x-1)(x-4)=0\)
\((x-1)=0\), so \(x=1\) OR
\((x-4)=0\), so \(x=4\)

Step 6: Check those answers against the "condition," the range we allow from Step 3.
Do the solutions 1 and 4 fall in the allowed range of \(x<1\)? No. 1 and 4 are not smaller than 1. Those solutions are not valid.

Part II -- For the non-negative case (the EXPRESSION is non-negative)

1) Write "x is greater than or equal to 0" in mathematical terms

2) write, separately, what you will actually put into the equation. Put a positive sign in front of the expression.(The "result" when we put a negative sign changed the inside. This one shouldn't.)

3) Write, separately, the "condition" part (get \(x\) on one side). Ignore till the end

4) Plug #2 into the equation

5) Get everything on LHS, RHS = 0

6) factor the quadratic and solve (set each factor equal to 0)

7) check your solutions against the allowed range for THIS PART's Step 3. Are those solutions in your allowed range from #3?

Last step: if the solutions are valid, do what the prompt asks.

As mentioned, we are doing three different but related things...

\((x-1)<0\) announces: I am considering the negative case

\(x<1\) says: here is the allowable range for my solution, and

\(-(x-1)=\)
\((1-x)\) is what actually gets plugged into the equation.

Let me know how it goes. I hope that helps.
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Oh yes. Thanks.

So basically the whole point is that we can't just exclude 1. The value of 1 needs to be covered in either of the ranges.

Got it!

x^2 - 4x + 6 = 3- |x-1|
x^2 - 4x + 3 = - |x-1|
|x-1| = - (x^2 - 4x + 3)
now , since anything in mod always has to be positive
LHS >= 0
- (x^2 - 4x + 3) >= 0
(x^2 - 4x + 3) <= 0
(x-3)(x-1) <= 0...................(1)

NOW ,

x^2 - 4x + 6 = 3- |x-1|

1st Case
x^2 - 4x + 6 = 3- x + 1
x^2 - 3x + 2 = 0
(x-2)(x-1) = 0
x=1 , 2
checking these values in equation (1)
both values satisfy thus we have two solutions.... x = 1 and x=2

2nd Case


x^2 - 4x + 6 = 3- |x-1|
x^2 - 4x + 3 = x -1
x^2 - 5x +4 = 0
(x-4)(x-1)= 0
Values are x=1 , 4
Putting these values in equation (1)
x=4 is not a solution...

Hence the only two acceptable values are x=1 and x=2
Thus the product is 2

Bunuel can you explain why you discarded -4 and -1 as solutions? Haven't seen this anywhere
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We got \(x=1\) or \(x=4\) when we considered the case when x < 1. These values are not less than 1, so they are not valid.
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This includes two cases :
When |x- 1| is positive the equation can be written as :
|x-1| = x-1
\(x^2-4x+6\ =\ 3\ -\ \left(x-1\right)\)
\(x^2-4x+6\ =\ 4\ -\ x\)
\(x^2-3x+2\ =\ 0\)
The possible values for x are 1, 2
If |x-1| is negative : \(\left|x-1\right|\) = 1-x
\(x^2-4x+6\ =\ 3+x-1\)
\(x^2-5x+4\ =\ 0\)
The possible values of x that satisfy this condition are 1, 4.
But since x must be negative. Both the cases fail.
Hence the possible values are x = 1, x = 2.
The product is 2.


After solving we get 2 valuees for x= 2,4 we might be tempted to multiply and get the answer

However when we put 4 the eqn reads 6= 0 which is not possible

THerefore the answer should be 2 IMO C

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