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My way: standard procedure: 1st: \(x+10=2x+8\) \(x=2\) OR 2nd:\(-(x+10)=2x+8\) \(x=-6\) the condition \(x=i\) is met: Insufficient.

This approach is NOT correct.

Always be sure to plug back the solutions for equations with modulus sign the values that you have found are x = 2 and x = -6 ONLY x=2 satisfies the given condition For X = -6

LHS = |x+10| = 4 RHS = 2x + 8 = -4 SO x = -6 is NO GOOD

This is happening because you have found the roots of \((x+10)^2\) = \((2x+8)^2\) _________________

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

We have to analyze the two cases so 1)Range: \(x+10>0, x>-10\) \(x+10=2x+8\) \(x=2\) \(-2\) is in the range we are analyzing (>-10) so -2 is a solution

2)Range: \(x+10<0,x<-10\) \(-x-10=2x+8\) \(x=-6\) \(-6\) is OUT of the range we are considering (<-10) so is NOT a possible solution

Let me know if it's clear _________________

It is beyond a doubt that all our knowledge that begins with experience.

This could ALSO be solved YOUR way but you have missed the definition of the modulus sign |x|= x WHEN x >or = 0 |x| = -x WHEN x<0

Now lets come back to the problem,

|x+10| = 2x+8

SO x+10 = 2x+8 when x+10> 0 So it gives x=2 Now lets check: is 2+10 > 0 YES. So this is GOOD Next -(x+10) = 2x+8 when x+10<0 it gives x=-6 Lets check: is -6+10 < 0 NO. So this is what you call an extraneous root. Does no good. _________________

the sums in parentheses must have opposite signs, so:

\(\frac{3}{2}<x<3\)

consider the condition \(x=i\):

\(x=2\) Sufficient.

2) is not clear:

\(|x+10|=2x+8\)

My way: standard procedure: 1st: \(x+10=2x+8\) \(x=2\) OR 2nd:\(-(x+10)=2x+8\) \(x=-6\) the condition \(x=i\) is met: Insufficient.

The original explanation: (2) \(|x+10|=2x+8\). The left hand side (LHS) is an absolute value, which is always non-negative, hence RHS must also be non-negative: \(2x+8\geq0\) giving us \(x\geq-4\). Now, for this range \(x+10\) is positive, hence\(|x+10|=x+10\). So, \(|x+10|=2x+8\) can be written as \(x+10=2x+8\), solving for \(x\) gives \(x=2\). Sufficient.

Noting that \(2x+8\geq0\) excludes the negative value and leaves off only one value. But what the heck is the mechanics behind this problem that makes the the good old way of solving inequalities insufficiently precise here?

Merging similar topics.

The way you call "standard procedure" is not complete.

When expanding \(|x+10|=2x+8\) you should consider x<-10 range and x>=-10 range:

\(x<-10\) --> \(-(x+10)=2x+8\) --> \(x=-6\). Discard this solution since it's not in the range \(x<-10\). \(x\geq{-10}\) --> \(x+10=2x+8\) --> \(x=2\).

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Re: What is the value of integer x ? I- 2x^2 + 9 < 9x II- |x+10 [#permalink]

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12 Jun 2014, 11:59

What is the value of integer x ?

I- 2x^2 + 9 < 9x II- |x+10| = 2x+8

I took some time but solved it my way.

IMO D.

Stmt I. we can make equation \(2x^2 -9x + 9 < 0\)

solve for X = 1.5, 3 -- 2 is the only integer in between satisfies the relation.

Sufficient.

II. Square both side. \((|x+10|) ^2=(2x+8)^2\)

\(x^2 + 4x - 12 = 0\)

solve for X = 2, -6 -- substitute values back to original equation, only 2 satisfies the relation.

Sufficient. _________________

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Re: What is the value of integer x? (M27-14) [#permalink]

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29 Mar 2016, 05:04

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