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Intern  Joined: 24 Apr 2010
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Which of the following represents all the possible values of  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 58% (01:44) correct 42% (01:49) wrong based on 673 sessions

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Which of the following represents all the possible values of x that are solutions to the equation 3x=|x^2 -10| ?

A. -5, -2, and 0
B. -5, -2 , 2, and 5
C. -5 and 2
D. -2 and 5
E. 2 and 5
Math Expert V
Joined: 02 Sep 2009
Posts: 58312
Re: equation with absolut value  [#permalink]

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5
4
perseverant wrote:
Which of the following represents all the possible values of x that are solutions to the equation $$3x=|x^2 -10|$$?

A. -5, -2, and 0
B. -5, -2, 2, and 5
C. -5 and 2
D. -2 and 5
E. 2 and 5

Could someone explain how to get to the solution?
Thanks!

$$3x=|x^2 -10|$$.

First of all: as RHS (right hand side) is absolute value, which is never negative (absolute value $$\geq{0}$$), LHS also must be $$\geq{0}$$ --> $$3x\geq{0}$$ --> $$x\geq{0}$$.

$$3x=x^2-10$$ --> $$x=-2$$ or $$x=5$$. First values is not valid as $$x\geq{0}$$.

$$3x=-x^2+10$$ --> $$x=-5$$ or $$x=2$$. First values is not valid as $$x\geq{0}$$.

So there are only two valid solutions: $$x=5$$ and $$x=2$$.

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Intern  Joined: 24 Apr 2010
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Re: equation with absolut value  [#permalink]

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Thank you very much Manager  Joined: 17 Mar 2010
Posts: 130
Re: equation with absolut value  [#permalink]

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But why didn't we check for 3x = -(x^2)-10 ???? Can someone explain???
Math Expert V
Joined: 02 Sep 2009
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Re: equation with absolut value  [#permalink]

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amitjash wrote:
But why didn't we check for 3x = -(x^2)-10 ???? Can someone explain???

We checked for two cases:

1. $$3x=x^2-10$$ --> $$x=-2$$ or $$x=5$$. First values is not valid as $$x\geq{0}$$.
and
2. $$3x=-(x^2-10)$$ --> $$x=-5$$ or $$x=2$$. First values is not valid as $$x\geq{0}$$.

What case does $$3x = -(x^2)-10$$ represent?
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Intern  Joined: 04 May 2013
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Re: Which of the following represents all the possible values of  [#permalink]

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I squared both the LHS and RHS.
Get totally weird answer. (Roots come out to be -5 and 5/2)

What is wrong with that approach?
Clearly something is because those two roots are not options
Math Expert V
Joined: 02 Sep 2009
Posts: 58312
Re: Which of the following represents all the possible values of  [#permalink]

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jjack0310 wrote:
I squared both the LHS and RHS.
Get totally weird answer. (Roots come out to be -5 and 5/2)

What is wrong with that approach?
Clearly something is because those two roots are not options

Probably math:

$$(3x)^2=(x^2-10)^2$$ --> $$x^4-29 x^2+100 = 0$$ --> solve for $$x^2$$: $$x^2=4$$ or $$x^2=25$$ --> $$x=2$$ or $$x=5$$ (discarding negative roots $$x=-2$$ or $$x=-5$$, since from 3x=|x^2 -10| it follows that x cannot be negative).

Hope it's clear.
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Re: Which of the following represents all the possible values of  [#permalink]

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Bunuel wrote:
jjack0310 wrote:
I squared both the LHS and RHS.
Get totally weird answer. (Roots come out to be -5 and 5/2)

What is wrong with that approach?
Clearly something is because those two roots are not options

Probably math:

$$(3x)^2=(x^2-10)^2$$ --> $$x^4-29 x^2+100 = 0$$ --> solve for $$x^2$$: $$x^2=4$$ or $$x^2=25$$ --> $$x=2$$ or $$x=5$$ (discarding negative roots $$x=-2$$ or $$x=-5$$, since from 3x=|x^2 -10| it follows that x cannot be negative).

Hope it's clear.

Right. I didnt do x^4. My mistake.
Kept it at x^2

Thanks
Senior Manager  Joined: 13 May 2013
Posts: 406
Re: Which of the following represents all the possible values of  [#permalink]

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Which of the following represents all the possible values of x that are solutions to the equation 3x=|x^2 -10| ?

x>3.33:
Positive:
3x=x^2-10
-x^2+3x+10=0
x^2-3x-10=0
(x-5)*(x+2)=0
x=5, x=-2
x=5 falls in the range of x>3.33. X=2 does not.

x<3.33
Negative:
3x=-(x^2-10)
3x=-x^2+10
x^2+3x-10=0
(x+5)*(x-2)=0
x=-5, x=2
Both -5 and 2 fall within the range of x<3.33

Solutions = -5,2,5

In other words, in a similar problem I would check the results of the positive and negative case and see if they fell within the range I was testing. For example, for x>3.33, we get two solutions (x=5, x=-2) but only x=5 is valid. Why is that? I see that you are using 3x (and therefore, x) as the metric for what values are valid and what ones arent) but it doesn't seem like thats the case in other problems)

Originally posted by WholeLottaLove on 08 Jul 2013, 20:42.
Last edited by WholeLottaLove on 09 Jul 2013, 15:52, edited 2 times in total.
Intern  Joined: 27 Jun 2013
Posts: 2
Re: Which of the following represents all the possible values of  [#permalink]

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2
I think no need to calculate at all...
We are solving for LHS = RHS and we know RHS can not be negative (x^2-10) can be negative but |x^2-10| is always positive.
So the RHS should also be positive for equality.
RHS>0
3x>0
HENCE...X>0 and only option E satisfies  Math Expert V
Joined: 02 Sep 2009
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Re: Which of the following represents all the possible values of  [#permalink]

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WholeLottaLove wrote:
(also, if 3x is positive why couldn't I square both sides?)

Check here: which-of-the-following-represents-all-the-possible-values-of-94984.html#p1243493
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Re: Which of the following represents all the possible values of  [#permalink]

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3
1
WholeLottaLove wrote:
Which of the following represents all the possible values of x that are solutions to the equation 3x=|x^2 -10| ?

x>3.33:
Positive:
3x=x^2-10
-x^2+3x+10=0
x^2-3x-10=0
(x-5)*(x+2)=0
x=5, x=-2
x=5 falls in the range of x>3.33. X=2 does not.

x<3.33
Negative:
3x=-(x^2-10)
3x=-x^2+10
x^2+3x-10=0
(x+5)*(x-2)=0
x=-5, x=2
Both -5 and 2 fall within the range of x<3.33

Solutions = -5,2,5

In other words, in a similar problem I would check the results of the positive and negative case and see if they fell within the range I was testing. For example, for x>3.33, we get two solutions (x=5, x=-2) but only x=5 is valid. Why is that? I see that you are using 3x (and therefore, x) as the metric for what values are valid and what ones arent) but it doesn't seem like thats the case in other problems)

Your intervals are not correct:

$$3x=|x^2 -10|$$, $$x^2 -10>0$$ if $$x>\sqrt{10}$$ and if $$x<-\sqrt{10}$$.

So if $$x>\sqrt{10}$$ or $$x<-\sqrt{10}$$ => positive
$$3x=x^2-10$$ that has two solutions: $$x=5$$ (possible) and $$x=-2$$ not possible because it's outside the range we are considering.

If $$-\sqrt{10}<x<\sqrt{10}$$ => negative
$$3x=-x^2+10$$ that has two solutions: $$x=2$$ (possible) and $$x=-5$$ not possible because it's outside the range we are considering.

So overall x can be 2 or 5.

"Why is that? " you anlyze parts of the function each time, so you have to pay attention to which ranges you are considering. If a given solution falls within that range, then it's solution, but it it falls out, it's not valid.

Hope it clarifies
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Senior Manager  Joined: 13 May 2013
Posts: 406
Re: Which of the following represents all the possible values of  [#permalink]

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Ahhh...I wasn't considering that for x^2, the range might be greater than x or less than negative x. I am used to solving for, say, |x-3| which is a bit more straightforward. Thanks a lot for the explanation it cleared everything up!

Zarrolou wrote:
WholeLottaLove wrote:
Which of the following represents all the possible values of x that are solutions to the equation 3x=|x^2 -10| ?

x>3.33:
Positive:
3x=x^2-10
-x^2+3x+10=0
x^2-3x-10=0
(x-5)*(x+2)=0
x=5, x=-2
x=5 falls in the range of x>3.33. X=2 does not.

x<3.33
Negative:
3x=-(x^2-10)
3x=-x^2+10
x^2+3x-10=0
(x+5)*(x-2)=0
x=-5, x=2
Both -5 and 2 fall within the range of x<3.33

Solutions = -5,2,5

In other words, in a similar problem I would check the results of the positive and negative case and see if they fell within the range I was testing. For example, for x>3.33, we get two solutions (x=5, x=-2) but only x=5 is valid. Why is that? I see that you are using 3x (and therefore, x) as the metric for what values are valid and what ones arent) but it doesn't seem like thats the case in other problems)

Your intervals are not correct:

$$3x=|x^2 -10|$$, $$x^2 -10>0$$ if $$x>\sqrt{10}$$ and if $$x<-\sqrt{10}$$.

So if $$x>\sqrt{10}$$ or $$x<-\sqrt{10}$$ => positive
$$3x=x^2-10$$ that has two solutions: $$x=5$$ (possible) and $$x=-2$$ not possible because it's outside the range we are considering.

If $$-\sqrt{10}<x<\sqrt{10}$$ => negative
$$3x=-x^2+10$$ that has two solutions: $$x=2$$ (possible) and $$x=-5$$ not possible because it's outside the range we are considering.

So overall x can be 2 or 5.

"Why is that? " you anlyze parts of the function each time, so you have to pay attention to which ranges you are considering. If a given solution falls within that range, then it's solution, but it it falls out, it's not valid.

Hope it clarifies
Intern  Joined: 27 Mar 2014
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Re: Which of the following represents all the possible values of  [#permalink]

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Bunuel wrote:
perseverant wrote:
Which of the following represents all the possible values of x that are solutions to the equation $$3x=|x^2 -10|$$?

A. -5, -2, and 0
B. -5, -2, 2, and 5
C. -5 and 2
D. -2 and 5
E. 2 and 5

Could someone explain how to get to the solution?
Thanks!

$$3x=|x^2 -10|$$.

First of all: as RHS (right hand side) is absolute value, which is never negative (absolute value $$\geq{0}$$), LHS also must be $$\geq{0}$$ --> $$3x\geq{0}$$ --> $$x\geq{0}$$.

$$3x=x^2-10$$ --> $$x=-2$$ or $$x=5$$. First values is not valid as $$x\geq{0}$$.

$$3x=-x^2+10$$ --> $$x=-5$$ or $$x=2$$. First values is not valid as $$x\geq{0}$$.

So there are only two valid solutions: $$x=5$$ and $$x=2$$.

RHS = an absolute value then 3x>=0 => x>=0 => Only E is appropriate
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Re: Which of the following represents all the possible values of  [#permalink]

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BIG RULE for roots problems - **TEST THE ROOTS!!!**

3x = abs(x^2 - 10)

Scenario (1)
-3x = x^2 - 10
x^2 + 3x - 10 = 0
(x-2)(x+5)=0
x=2, -5

Scenario (2)
3x = x^2 -10
x^2 -3x-10=0
(x+2)(x-5)=0
x=-2,5

-5 and -2 don't work because we can't have a negative sign on the LHS of the equation, thus our only answers are 2,5

E.
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Re: Which of the following represents all the possible values of  [#permalink]

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Team/ Bunnel,

I understand this is old problem and all aspects has been already discussed. I do understand the absolute value concept applied here which helps to get rid of negative value answer choices. However I am getting now confused here after going through another question and explanation on absolute value.

Refer - https://gmatclub.com/forum/what-is-the- ... 40003.html
https://gmatclub.com/forum/what-is-the-value-of-x-1-x-4-2-x-240003.html

In above we can |X| = 4 is considered as X= 4 and -4. Going by this why we can not consider
3X=|X^2-10| as
3X= X^2-10 and 3X=- [ X^2-10] which gives answer choice B as correct answer.
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Re: Which of the following represents all the possible values of  [#permalink]

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When x^2-10 > 0 => x^2 > 10 => x > sqrt(10) and x < - sqrt(10), then we get x^2-3x-10=0 which gives us 2 solutions -> x = 5 or x = -2. However x = -2 is invalid solution since it doesn't satisfy the condition x < -sqrt(10).

Similarly when x^2-10 < 0 => x^2 < 10 => x < sqrt(10) and x > - sqrt(10), then we get x^2+3x-10=0 which gives us 2 solutions -> x = -5 or x = 2. However x = -5 is invalid solution since it doesn't satisfy the condition x > -sqrt(10).

So the possible solutions for the equation are x = 2 and x =5. Hence, Option E is the right choice.
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Re: Which of the following represents all the possible values of  [#permalink]

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Bunuel wrote:
perseverant wrote:
Which of the following represents all the possible values of x that are solutions to the equation $$3x=|x^2 -10|$$?

A. -5, -2, and 0
B. -5, -2, 2, and 5
C. -5 and 2
D. -2 and 5
E. 2 and 5

Could someone explain how to get to the solution?
Thanks!

$$3x=|x^2 -10|$$.

First of all: as RHS (right hand side) is absolute value, which is never negative (absolute value $$\geq{0}$$), LHS also must be $$\geq{0}$$ --> $$3x\geq{0}$$ --> $$x\geq{0}$$.

$$3x=x^2-10$$ --> $$x=-2$$ or $$x=5$$. First values is not valid as $$x\geq{0}$$.

$$3x=-x^2+10$$ --> $$x=-5$$ or $$x=2$$. First values is not valid as $$x\geq{0}$$.

So there are only two valid solutions: $$x=5$$ and $$x=2$$.

Hi

Could you explain how you got $$3x = -x^2 + 10$$

Also, why you think E is the right answer? Re: Which of the following represents all the possible values of   [#permalink] 20 Jun 2019, 11:39
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