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What is the product of all the solutions of x^2 - 4x + 6=3 09 Jul 2013, 15:25
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What is the product of all the solutions of x^2 - 4x + 6 = 3 - |x - 1| ?
x^2 - 4x + 6 = 3 - |x - 1|
x^2 - 4x + 3 = - |x - 1|

x>1
x^2 - 4x + 6 = 3 - |x - 1|
x^2 - 4x + 3 = - (x-1)
x^2 - 4x + 3 = -x+1
x^2 - 3x +2 = 0
(x-1)(x-2)=0
x=1, x=2
2 falls within the range of x>1
x=2

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)

Thanks!
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What is the product of all the solutions of x^2 - 4x + 6=3 11 Aug 2015, 18:03
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paul16
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
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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?
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What is the product of all the solutions of x^2 - 4x + 6=3 11 Aug 2015, 18:15
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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?
Show more

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.
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What is the product of all the solutions of x^2 - 4x + 6=3 10 Jun 2016, 17:54
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Bunuel
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I've seen similar question which reads:
What is the product of all the solutions of x^2 + 4x + 7 = |x + 2| + 3 ?
A. -6
B. -2
C. 2
D. 6
E. 12

OA:
Show Spoiler
A

Try it. I'll provide solution for this question later, if necessary.

if |x+2|>=-2 then |x+2|= (x+2)

eqation becomes (x+2)(x+1)=0

x=-2,-1

if |x+2|<-2 then |x+2|=-(x+2)

equation becomes (x+2) (x+3) =0
x=-2,-3 ( can't be -2 since x<-2)

product of the solution -

-2*-1*-3= -6 Ans

I guess you meant the following:

When \(x\leq{-2}\), then \(|x+2|=-(x-2)\).
When \(x>{-2}\), then \(|x+2|=(x-2)\).

Complete solution:

\(x^2 + 4x + 7 = |x + 2| + 3\) --> \(x^2 + 4x + 4 = |x + 2|\) --> \((x+2)^2=|x+2|\) --> \((x+2)^4=(x+2)^2\) --> \((x+2)^2((x+2)^2-1)=0\):

\(x+2=0\) --> \(x=-2\);
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.
Show more


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...
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What is the product of all the solutions of x^2 - 4x + 6=3 11 Jun 2016, 00:43
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msk0657
Bunuel
guerrero25
I've seen similar question which reads:
What is the product of all the solutions of x^2 + 4x + 7 = |x + 2| + 3 ?
A. -6
B. -2
C. 2
D. 6
E. 12

OA:
Show Spoiler
A

Try it. I'll provide solution for this question later, if necessary.

if |x+2|>=-2 then |x+2|= (x+2)

eqation becomes (x+2)(x+1)=0

x=-2,-1

if |x+2|<-2 then |x+2|=-(x+2)

equation becomes (x+2) (x+3) =0
x=-2,-3 ( can't be -2 since x<-2)

product of the solution -

-2*-1*-3= -6 Ans

I guess you meant the following:

When \(x\leq{-2}\), then \(|x+2|=-(x-2)\).
When \(x>{-2}\), then \(|x+2|=(x-2)\).

Complete solution:

\(x^2 + 4x + 7 = |x + 2| + 3\) --> \(x^2 + 4x + 4 = |x + 2|\) --> \((x+2)^2=|x+2|\) --> \((x+2)^4=(x+2)^2\) --> \((x+2)^2((x+2)^2-1)=0\):

\(x+2=0\) --> \(x=-2\);
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...
Show more

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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What is the product of all the solutions of x^2 - 4x + 6=3 Updated on: 20 Jun 2018, 13:54
Originally posted by dave13 on 20 Jun 2018, 11:51.
Last edited by dave13 on 20 Jun 2018, 13:54, edited 1 time in total.
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Bunuel
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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.
Show more


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
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What is the product of all the solutions of x^2 - 4x + 6=3 20 Jun 2018, 13:53
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dave13
Bunuel
carcass
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.


pushpitkc , Bunuel
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 \(x\geq{1}\)


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
Show more

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!
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What is the product of all the solutions of x^2 - 4x + 6=3 21 Jun 2018, 07:15
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Bunuel
carcass
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.
Show more



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
)"



.
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What is the product of all the solutions of x^2 - 4x + 6=3 21 Jun 2018, 18:45
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dave13
Bunuel
carcass
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.
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 (1)\(|x - 1|<0\)

(2) 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
Show more
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: \((1-x)\)

Quote:
how Bunuel got this \(x<1\)
Show more
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,
Quote:
if Negative I rewrite it as \(|x - 1|<0\)
Show more
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:
Quote:
\(-(x-1)< 0\)
Show more
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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What is the product of all the solutions of x^2 - 4x + 6=3 28 Jan 2019, 06:50
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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!
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What is the product of all the solutions of x^2 - 4x + 6=3 08 Aug 2020, 07:45
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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
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What is the product of all the solutions of x^2 - 4x + 6=3 04 Feb 2021, 13:03
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Bunuel
carcass
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.
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Bunuel can you explain why you discarded -4 and -1 as solutions? Haven't seen this anywhere
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What is the product of all the solutions of x^2 - 4x + 6=3 04 Feb 2021, 13:15
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CEdward
Bunuel
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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.

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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What is the product of all the solutions of x^2 - 4x + 6=3 24 Feb 2022, 05:18
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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.

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What is the product of all the solutions of x^2 - 4x + 6=3 25 Feb 2022, 06:55
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carcass
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

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Edited the answer choice C from -2 to 2 and the OA. It's C not E.
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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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What is the product of all the solutions of x^2 - 4x + 6=3 10 Sep 2025, 06:23
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