Which of the following represents all the possible values of

Thank you very much

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?

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

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)

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

Check here: which-of-the-following-represents-all-the-possible-values-of-94984.html#p1243493

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!

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

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.

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.

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.

Hi

Could you explain how you got \(3x = -x^2 + 10\)

Also, why you think E is the right answer?

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

Scenario 1:

\(3x = x^2 - 10\)
\(0 = x2 - 3x - 10\)
\(x = 5, -2\)

Scenario 2:

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

Since the absolute value can not be negative, x can not be -5 or -2. To confirm this, lets plug -5 into the initial equation:

\(3(-5)=|-5^2 -10|\)
\(-15 = 15\)
As you can see, -15 is not equal to 15.

The possible values of x are 2 and 5. Answer is E.

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.