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How many real numbers x satisfy ||x-3|-4|=4?

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Math Revolution GMAT Instructor
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How many real numbers x satisfy ||x-3|-4|=4?  [#permalink]

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New post 04 Dec 2017, 01:36
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[GMAT math practice question]

How many real numbers \(x\) satisfy \(||x-3|-4|=4\)?

A. 1
B. 2
C. 3
D. 4
E. 5

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How many real numbers x satisfy ||x-3|-4|=4?  [#permalink]

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New post 04 Dec 2017, 05:35
3
MathRevolution wrote:
[GMAT math practice question]

How many real numbers \(x\) satisfy \(||x-3|-4|=4\)?

A. 1
B. 2
C. 3
D. 4
E. 5


\(||x-3|-4|=4\)

\(=>|x-3|-4=4\) or \(|x-3|-4=-4\)

\(=>|x-3|=8\) or \(|x-3|=0\)

\(=>x-3=8\) or \(x-3=-8\) or \(x=3\)

\(=>x=11\) or \(x=-5\) or \(x=3\)

Hence \(3\) real values satisfy the equation

Option C
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Re: How many real numbers x satisfy ||x-3|-4|=4?  [#permalink]

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New post 04 Dec 2017, 06:14
1
1
MathRevolution wrote:
[GMAT math practice question]

How many real numbers \(x\) satisfy \(||x-3|-4|=4\)?

A. 1
B. 2
C. 3
D. 4
E. 5





if you have POSITIVE values on both sides, you can square both sides..


\(||x-3|-4|=4............(|x-3|-4)^2=4^2.........(x-3)^2+4^2-2*4*|x-3|=4^2.........(x-3)^2=8|x-3|\)
you can square both sides again
\((x-3)^2=8|x-3|.............(x-3)^4=8^2*(x-3)^2..............(x-3)^4-8^2*(x-3)^2=0.................(x-3)^2((x-3)^2-8^2)=0.....................(x-3)^2(x-3-8)(x-3+8)=0............(x-3)^2(x-11)(x+5)=0\)
so roots of equation or the different values of x are 3,11,-5
ans 3 different values
C
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How many real numbers x satisfy ||x-3|-4|=4?  [#permalink]

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New post 06 Dec 2017, 01:09
2
=>

\(| |x-3| - 4 | = 4\)
\(⇔ |x-3| - 4= ± 4\)
\(⇔ |x-3| = 4 ± 4\)
\(⇔ |x-3| = 8\) or \(|x-3| = 0\)
\(⇔ x-3 = ±8\) or \(x-3 = 0\)
\(⇔ x = 3 ± 8\) or \(x = 3\)
\(⇔ x = 11, x = -5\) or \(x = 3\)

The equation has 3 solutions.

Therefore, the answer is C.

Answer : C
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How many real numbers x satisfy ||x-3|-4|=4?  [#permalink]

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New post 27 Mar 2018, 04:58
1
MathRevolution wrote:
[GMAT math practice question]

How many real numbers \(x\) satisfy \(||x-3|-4|=4\)?

A. 1
B. 2
C. 3
D. 4
E. 5


Open the outer Modulus and make the RHS + or -

|x-3|-4=+or-4. Now isolate the remaining Mod.

|x-3|=4+-4. So |x-3|=8. Or |x-3|=0. Both 0 and 8 are non negative, so no need to discard either formula. Now use both formulas to retrieve solutions.

(1) |x-3|=8 x=-5 or 11

(2) |x-3|=0 x=3

So 3 solutions. Good question.
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How many real numbers x satisfy ||x-3|-4|=4?   [#permalink] 27 Mar 2018, 04:58
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