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Which of the following represents all the possible values of 29 May 2010, 06:40
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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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E
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Bunuel
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Which of the following represents all the possible values of 29 May 2010, 07:01
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perseverant
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!
Show more

\(3x=|x^2 -10|\).

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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Which of the following represents all the possible values of 10 Jul 2013, 09:09
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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)
Show more

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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Which of the following represents all the possible values of 29 May 2010, 07:17
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Thank you very much :-D
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Which of the following represents all the possible values of 30 May 2010, 09:03
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But why didn't we check for 3x = -(x^2)-10 ???? Can someone explain???
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Which of the following represents all the possible values of 30 May 2010, 09:43
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amitjash
But why didn't we check for 3x = -(x^2)-10 ???? Can someone explain???
Show more

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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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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Which of the following represents all the possible values of 06 Jul 2013, 12:14
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jjack0310
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
Show more

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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Which of the following represents all the possible values of 06 Jul 2013, 18:19
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Bunuel
jjack0310
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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Right. I didnt do x^4. My mistake.
Kept it at x^2

Thanks
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Which of the following represents all the possible values of Updated on: 09 Jul 2013, 15:52
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.
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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)
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Which of the following represents all the possible values of 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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Which of the following represents all the possible values of 08 Jul 2013, 23:19
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WholeLottaLove

(also, if 3x is positive why couldn't I square both sides?)
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Check here: which-of-the-following-represents-all-the-possible-values-of-94984.html#p1243493
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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
WholeLottaLove
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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Which of the following represents all the possible values of 19 Jun 2014, 02:43
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Bunuel
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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

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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RHS = an absolute value then 3x>=0 => x>=0 => Only E is appropriate
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Which of the following represents all the possible values of 29 Oct 2020, 08:21
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3x=|x^2 -10|

We know that RHS will always be positive, this means LHS should also be positive.

The only way LHS can be positive is when x is positive.
The only option with all positive values is E.

Correct answer is E.

Correct me if my reasoning is flawed.
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Guys excuse me could we just go straight to E by eliminating all answers that include a negative number ?
perseverant
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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Which of the following represents all the possible values of 26 Nov 2024, 23:52
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Nikosliberis
Guys excuse me could we just go straight to E by eliminating all answers that include a negative number ?
perseverant
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
Show more

Yes, x cannot be negative because it would make the left-hand side negative, which cannot equal the right-hand side, as it is an absolute value and, by definition, always non-negative.
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