BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Algebra and absolute value question I'm stuck on

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pm
Thanked: 104 times
Followed by:1 members

by scoobydooby » Sat Aug 15, 2009 11:15 am
is it 8?

(x^2) - 4x + 6 = 3 - |x-1|
=>(x^2) - 4x + 6-3=-|x-1|
=>x^2 -4x+3=-|x-1|
=>(x-1)(x-3)=-|x-1|
squaring both sides

(x-1)^2 (x-3)^2=(x-1)^2
=>(x-1)^2*[(x-3)^2-1]=0
=>(x-1)^2*[x^2+9-6x-1]=0
=>(x-1)^2*(x-4)(x-2)=0
=>x=1 or x=4 or x=2

product of all possible solutions: 1*4*2=8
Top
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Sat Aug 08, 2009 5:31 pm

by beboppin » Sat Aug 15, 2009 12:35 pm
scoobydooby wrote:is it 8?

(x^2) - 4x + 6 = 3 - |x-1|
=>(x^2) - 4x + 6-3=-|x-1|
=>x^2 -4x+3=-|x-1|
=>(x-1)(x-3)=-|x-1|
squaring both sides

(x-1)^2 (x-3)^2=(x-1)^2
=>(x-1)^2*[(x-3)^2-1]=0
=>(x-1)^2*[x^2+9-6x-1]=0
=>(x-1)^2*(x-4)(x-2)=0
=>x=1 or x=4 or x=2

product of all possible solutions: 1*4*2=8
Hmm..that is what the book said too. But if you plug 4 in, the equation does not work, right? 6 = 0?
Top
Legendary Member
Posts: 1035
Joined: Wed Aug 27, 2008 10:56 pm
Thanked: 104 times
Followed by:1 members

by scoobydooby » Sat Aug 15, 2009 9:11 pm
yes you are right. 4 doesnt work when put back in the original equation.
1 and 2 are the only possible solutions.

which book was the question from?
Top
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Sat Aug 08, 2009 5:31 pm

by beboppin » Sun Aug 16, 2009 6:36 am
scoobydooby wrote:yes you are right. 4 doesnt work when put back in the original equation.
1 and 2 are the only possible solutions.

which book was the question from?
Total GMAT Math. I do want to caveat that this book has been great otherwise.

So is the method of assuming an absolute value term is positive and then assuming it's negative something I should avoid looking to do? I'd never gone down that route before and am wondering why it doesn't work in this case, assuming that it doesn't.
Top
Legendary Member
Posts: 752
Joined: Sun May 17, 2009 11:04 pm
Location: Tokyo
Thanked: 81 times
GMAT Score:680

by tohellandback » Sun Aug 16, 2009 9:02 am
scoobydooby wrote:is it 8?

(x^2) - 4x + 6 = 3 - |x-1|
=>(x^2) - 4x + 6-3=-|x-1|
=>x^2 -4x+3=-|x-1|
=>(x-1)(x-3)=-|x-1|
squaring both sides

(x-1)^2 (x-3)^2=(x-1)^2
=>(x-1)^2*[(x-3)^2-1]=0
=>(x-1)^2*[x^2+9-6x-1]=0
=>(x-1)^2*(x-4)(x-2)=0
=>x=1 or x=4 or x=2

product of all possible solutions: 1*4*2=8
sccobydooby,
you gotta eliminate the answer here in this line:
(x-1)(x-3)=-|x-1|
|x-1| is always>=0
so -|x-1|<=0
so (x-1)(x-3)<=0
so 1<=x<=3 (for any values "inside" the roots on the number lines, the result is negative for any quadratic equation)
from the answer choices you have got finally, only x=1 and 2 satisfy
answer should be 2
The powers of two are bloody impolite!!
Top
Post new topic Post Reply
• Page 1 of 1