dave13 wrote:
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.
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 , 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 \(x<1\)
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 \(|x - 1|<0\)
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:
\(-(x-1)< 0\)
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.
close
09 Jul 2013, 15:25
