so

Please boxing,

your replies are very delightful. However, please as a test and not as a screenshot.

Thank you so much for your collaboration. The board is cleaner that way.

regards

Short & Quick way to get the Answer is :

3x has to be less than 10 as we can't find out the Sqaure root of a a negative number so x can only take values of 0,1,2 & exclude non negative solutions so 2 is rejected. Answer b

Answer: B
Sure root of 10 - 3x is greater than x and x is a non-negative integer.
First, as x is non-negative, thus |x| equals x in A.
We try numbers. The minimum value for x can be 0.
If x = 0, square root of 10-3x equals 10, which is greater than x.
If x = 1, 10-3 is greater than 1
If x = 2, 10- 6 is not greater than 2.
So, x can be either 0 or 1 and always less than 2.
The answer is B.

However, it is not mentioned that x has to be an integer. x can be just less than 10/3 or 3 for that matter.

Given: x>0
and
\sqrt{(10-3x)} > x gives -5<x<2

Ignoring negative solutions. 0<x<2 and |x| would be less than 2

Hey guys, I understand your solution, but could you point out the fallacy or mistake in my solution.

sqrt(10-3x) > x
x>=0 so sqrt(10-3x)>0
then 10-3x >0
Thus x < (10/3)

So D because the absolute value of x can be greater than 2 or smaller than 2.

Perhaps it is a simple typo, but what you should have (x+5)(x-2) as the factor, then it would mean that -5>x>2.

You are wrong in replacing X with 0.
$\sqrt{10-3x}>x$..
So the LHS and RHS are interconnected, as increase in X will increase both sides ..

You cannot simply replace X with 0...
That would ok if you were simply finding out if some term is negative or positive

Okay, here is a solution that is purely algebraic ( boxing has a mistake in inequality direction).

sqrt (10-3x) > x
10-3x > x^2
0 > x^2 + 3x - 10
x^2 + 3x - 10 < 0
(x+5)(x-2) < 0

There are two options, x+5 < 0 and x- 2 > 0 || or || x+5 > 0 & x-2< 0
Why? Because each option, when multiplied, would still give you a negative.
First option: x< -5 and x > 2, x is nonnegative so you can throw out first inequality out. For the second inequality, plug this in the criteria 10-3x>x^2 and you will see it doesn't work. Thus, throw it out as well.
Second option: x>-5 and x<2, again, here we throw out the negative solution x>-5.

So x < 2.

Thus B.

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