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

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Math Revolution GMAT Instructor
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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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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
[Reveal] Spoiler: OA

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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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28 Sep 2017, 02:52
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 $$= 1*2*3$$

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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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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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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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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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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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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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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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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