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Which of the following represents all the possible values of

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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| ?

A. -5, -2, and 0
B. -5, -2 , 2, and 5
C. -5 and 2
D. -2 and 5
E. 2 and 5
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Re: equation with absolut value [#permalink]

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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 \(x=2\).

Answer: E.
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Re: equation with absolut value [#permalink]

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New post 29 May 2010, 07:17
Thank you very much :-D

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New post 30 May 2010, 09:03
But why didn't we check for 3x = -(x^2)-10 ???? Can someone explain???

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New post 30 May 2010, 09:43
amitjash wrote:
But why didn't we check for 3x = -(x^2)-10 ???? Can someone explain???


Please read the solution.

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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New post 06 Jul 2013, 12:07
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

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New post 06 Jul 2013, 12:14
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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New post 06 Jul 2013, 18:19
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

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

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New post 08 Jul 2013, 20:42
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)

Last edited by WholeLottaLove on 09 Jul 2013, 15:52, edited 2 times in total.

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New post 08 Jul 2013, 21:23
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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
:lol: :lol:

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

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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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New post 10 Jul 2013, 09:25
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

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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 \(x=2\).

Answer: E.


RHS = an absolute value then 3x>=0 => x>=0 => Only E is appropriate

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New post 03 Dec 2016, 16:07
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] 03 Dec 2016, 16:07
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