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What is the product of all the roots on the equation (x-2)^2=|x-2|?

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What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 28 Sep 2017, 02:34
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[GMAT math practice question]

What is the product of all the roots on the equation (x-2)^2=|x-2|?

A. 2
B. -2
C. 3
D. -3
E. 6

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What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post Updated on: 30 Sep 2017, 02:36
MathRevolution wrote:
[GMAT math practice question]

What is the product of all the roots on the equation (x-2)^2=|x-2|?

A. 2
B. -2
C. 3
D. -3
E. 6


Case 1: If \(x-2>0\), then \((x-2)^2=(x-2)\)

or \((x-2)^2-(x-2)=0 =>(x-2)(x-2-1)=0\)

Hence the roots are \(x-2=0\), or \(x=2\) and \(x-3=0\), or \(x=3\)

Case 2 If \(x-2<0\), then \((x-2)^2=-(x-2)\)

or \((x-2)^2+(x-2)=0 =>(x-2)(x-2+1)=0\)

Hence the roots are \(x-2=0\), or \(x=2\) and \(x-1=0\), or \(x=1\)

So, combining Case 1 & Case 2, the roots of the equation are \(1\), \(2\) and \(3\)

Product of roots \(=1*2*3=6\)

Option E

-------------------------------------------------------
Another method

\((x-2)^2=|x-2|\), square both sides to get \((x-2)^4=(x-2)^2\)

or \((x-2)^4-(x-2)^2=0 =>(x-2)^2(x-3)(x-1)=0\)

So roots of the equation are \((x-2)^2=0\), \(x-3=0\) and \(x-1=0\)

Hence \(x = 1\), \(2\) and \(3\)

So Product of roots \(= 1*2*3\)

Originally posted by niks18 on 28 Sep 2017, 02:52.
Last edited by niks18 on 30 Sep 2017, 02:36, edited 1 time in total.
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Re: What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 28 Sep 2017, 08:01
Hi MathRevolution,

I could work out one of the integer the x = 2, however couldn't work out how to get the -3. Where does the the -1 come from in the operation (x-2-1)?

(x−2)2−(x−2)=0=>(x−2)(x−2−1)=0

Thank you,

Nick
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Re: What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 28 Sep 2017, 08:05
nmargot wrote:
Hi MathRevolution,

I could work out one of the integer the x = 2, however couldn't work out how to get the -3. Where does the the -1 come from in the operation (x-2-1)?

(x−2)2−(x−2)=0=>(x−2)(x−2−1)=0

Thank you,

Nick


Hi nmargot

in the expression \((x-2)^2-(x-2)\), I have taken \((x-2)\) as common
So \((x-2)^2-1*(x-2) = (x-2)(x-2-1)\)

Hope this is clear
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Re: What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 29 Sep 2017, 07:04
for x<2
the equation becomes: \((x-2)^2 = (2-x)\)
on solving: x=1 or x=2(not possible, since x<2)

for x>2
the equation can be rewritten as: \((x-2)^2 = (x-2)\)
on solving: x=2(not possible, since x>2) or x=3

for x=2
\((2-2)^2 = (2-2)\); which satisfies the equation.
Hence the roots are 1,2,3
product = 6
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Re: What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 29 Sep 2017, 21:44
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naman90 wrote:
for x<2
the equation becomes: \((x-2)^2 = (2-x)\)
on solving: x=1 or x=2(not possible, since x<2)

for x>2
the equation can be rewritten as: \((x-2)^2 = (x-2)\)
on solving: x=2(not possible, since x>2) or x=3

for x=2
\((2-2)^2 = (2-2)\); which satisfies the equation.
Hence the roots are 1,2,3
product = 6


Hi naman90 and MathRevolution

I have two fundamental disconnect here with the solution, Kindly explain

1.) Any 2nd degree polynomial will have only two roots, but here we are getting \(3\) roots. Is this possible?

2.) the equation is of the form \(a^2=|a|\), or \(a^2=±a\). Now we know that \(a^2\) or LHS is always positive, so RHS cannot be negative, hence \(a^2≠-a\). Hence the only solution is \(a^2=a\). But as per the solution provided \(a^2=-a\) is yielding as possible root \(1\). what am I missing here?
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What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 30 Sep 2017, 00:17
hi niks18
although i am no math expert but here is my take on the points you raised
1. introducing modulus to an equation increases the number of possible solutions. take |x| = 2 for example. even though it is a 1st degree equation, it has 2 roots (2 and -2)
2. when you say "Now we know that \(a^2\) or LHS is always positive, so RHS cannot be negative, hence \(a^2≠−a\)", you are making an assumption that a>0. you also have to solve the equation for the case when a<0.
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Re: What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 30 Sep 2017, 02:36
naman90 wrote:
hi niks18
although i am no math expert but here is my take on the points you raised
1. introducing modulus to an equation increases the number of possible solutions. take |x| = 2 for example. even though it is a 1st degree equation, it has 2 roots (2 and -2)
2. when you say "Now we know that \(a^2\) or LHS is always positive, so RHS cannot be negative, hence \(a^2≠−a\)", you are making an assumption that a>0. you also have to solve the equation for the case when a<0.


Thanks naman90 for highlighting a key fact that i completely missed.
Actually the equation is a 4th degree polynomial and hence will have 4 roots = 1,2 (twice) & 3
because to remove mod we square it, hence the equation becomes a 2nd degree polynomial, hence two roots. For ex \(|x|=2\), \(x^2=4\)
Thanks again, was lucky to get the answer, but will have to edit the solution. :-)
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What is the product of all the roots on the equation (x-2)^2=|x-2|? [#permalink]

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New post 01 Oct 2017, 18:17
=>

(x-2)^2=|x-2|
⬄ |x-2|^2=|x-2|
⬄ |x-2|^2-|x-2| = 0
⬄ |x-2| ( |x-2| - 1 ) = 0
⬄ |x-2| = 0 or |x-2| = 1
⬄ x = 2 or x-2 = ±1
⬄ x = 2 or x = 2 ±1
⬄ x = 2 or x = 3 or x = 1
1 * 2 * 3 = 6

Ans: E
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What is the product of all the roots on the equation (x-2)^2=|x-2|?   [#permalink] 01 Oct 2017, 18:17
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