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x is a nonnegative number and the square root of (10 – 3x) [#permalink]
17 Sep 2017, 03:55
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x is a nonnegative 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 nonnegative number and the square root of (10 – 3x) [#permalink]
21 Sep 2017, 08:35
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We need to solve the equation \(sqrt(103x)>x\). Squaring both sides we get \(103x>x^2\) that can be rewritten as \(x^2+3x10<0\). The left hand side can be rewritten as \((x5)(x+2)\), 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 nonnegative number and the square root of (10 – 3x) [#permalink]
28 Feb 2018, 14:56
IlCreatore wrote: We need to solve the equation \(sqrt(103x)>x\). Squaring both sides we get \(103x>x^2\) that can be rewritten as \(x^2+3x10<0\). The left hand side can be rewritten as \((x5)(x+2)\), 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 nonnegative number and the square root of (10 – 3x) [#permalink]
04 Mar 2018, 02:11
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Re: x is a nonnegative 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 nonnegative 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 nonnegative 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 nonnegative integer. First, as x is nonnegative, thus x equals x in A. We try numbers. The minimum value for x can be 0. If x = 0, square root of 103x equals 10, which is greater than x. If x = 1, 103 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 nonnegative 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 nonnegative number and the square root of (10 – 3x) [#permalink]
04 Apr 2018, 07:41
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Given: x>0 and \sqrt{(103x)} > 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 nonnegative 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(103x) > x x>=0 so sqrt(103x)>0 then 103x >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 nonnegative number and the square root of (10 – 3x) [#permalink]
12 Dec 2018, 07:52
IlCreatore wrote: We need to solve the equation \(sqrt(103x)>x\). Squaring both sides we get \(103x>x^2\) that can be rewritten as \(x^2+3x10<0\). The left hand side can be rewritten as \((x5)(x+2)\), 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)(x2) as the factor, then it would mean that 5>x>2.



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Re: x is a nonnegative 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(103x) > x x>=0 so sqrt(103x)>0 then 103x >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. \(\sqrt{103x}>x\).. 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 nonnegative 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 (103x) > x 103x > x^2 0 > x^2 + 3x  10 x^2 + 3x  10 < 0 (x+5)(x2) < 0
There are two options, x+5 < 0 and x 2 > 0  or  x+5 > 0 & x2< 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 103x>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.




Re: x is a nonnegative number and the square root of (10 – 3x)
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