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x is a non-negative number and the square root of (10 – 3x) [#permalink]
 17 Sep 2017, 03:55
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x is a non-negative number and the square root of (10 – 3x) is greater than x.
Quantity A |
Quantity B |
\(|x|\) |
\(2\) |
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
21 Sep 2017, 08:35
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We need to solve the equation \(sqrt(10-3x)>x\). Squaring both sides we get \(10-3x>x^2\) that can be rewritten as \(x^2+3x-10<0\). The left hand side can be rewritten as \((x-5)(x+2)\), thus the solution of the equation is \(-5<x<2\).
Since x is non negative, among the solutions we must consider only those between 0 and 2, excluded. Thus, no matter the absolute value, column A is always smaller than 2.
Answer B
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
28 Feb 2018, 14:56
IlCreatore wrote: We need to solve the equation \(sqrt(10-3x)>x\). Squaring both sides we get \(10-3x>x^2\) that can be rewritten as \(x^2+3x-10<0\). The left hand side can be rewritten as \((x-5)(x+2)\), thus the solution of the equation is \(-5<x<2\).
Since x is non negative, among the solutions we must consider only those between 0 and 2, excluded. Thus, no matter the absolute value, column A is always smaller than 2.
Answer B Good explanation. Yes I will go with B as well.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
04 Mar 2018, 02:11
so
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
05 Mar 2018, 14:47
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
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
29 Mar 2018, 00:39
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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
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
01 Apr 2018, 17:21
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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.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
04 Apr 2018, 07:38
Yiddo_Bhushan wrote: 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 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.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
04 Apr 2018, 07:41
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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
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
07 Dec 2018, 15:23
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.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
12 Dec 2018, 07:52
IlCreatore wrote: We need to solve the equation \(sqrt(10-3x)>x\). Squaring both sides we get \(10-3x>x^2\) that can be rewritten as \(x^2+3x-10<0\). The left hand side can be rewritten as \((x-5)(x+2)\), thus the solution of the equation is \(-5<x<2\).
Since x is non negative, among the solutions we must consider only those between 0 and 2, excluded. Thus, no matter the absolute value, column A is always smaller than 2.
Answer B 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.
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
13 Dec 2018, 00:16
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QuantumWonder wrote: 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. 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
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Re: x is a non-negative number and the square root of (10 – 3x) [#permalink]
14 Dec 2018, 13:15
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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Re: x is a non-negative number and the square root of (10 – 3x)
[#permalink]
14 Dec 2018, 13:15
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