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# Is sqrt(x-5)^2 = 5-x?

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Is sqrt(x-5)^2 = 5-x? [#permalink]  06 Sep 2010, 11:20
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Is $$sqrt(x-5)^2 = 5-x?$$ ?

(1) -x|x| > 0
(2) 5-x > 0

PS. Is always this true?: $$sqrt{x^2}$$ = lxl ?
[Reveal] Spoiler: OA

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Re: Inequalities with absolute value [#permalink]  06 Sep 2010, 11:30
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metallicafan wrote:
Is $$sqrt(x-5)^2 = 5-x?$$

(1) -xlxl > 0
(2) 5-x > 0

PS. Is always this true?: $$sqrt{x^2}$$ = lxl ?

Is $$\sqrt{(x-5)^2}=5-x$$?

Remember: $$\sqrt{x^2}=|x|$$.

So "is $$\sqrt{(x-5)^2}=5-x$$?" becomes: is $$|x-5|=5-x$$?

$$|x-5|=5-x$$ is true only for $$x\leq{5}$$, as in this case $$\{LHS=|x-5|=5-x\}=\{RHS=5-x\}$$. So we have that if $$x\leq{5}$$, then $$|x-5|=5-x$$ is true.

Basically question asks is $$x\leq{5}$$?

(1) $$-x|x| > 0$$ --> $$|x|$$ is never negative (positive or zero), so in order to have $$-x|x| > 0$$, $$-x$$ must be positive $$-x>0$$ --> $$x<0$$, so $$x$$ is less than 5 too. Sufficient.

(2) $$5-x>0$$ --> $$x<5$$. Sufficient.

Hope it helps.

P.S. Explanation of: $$\sqrt{x^2}=|x|$$.

The point here is that as square root function can not give negative result then $$\sqrt{some \ expression}\geq{0}$$.

So $$\sqrt{x^2}\geq{0}$$. But what does $$\sqrt{x^2}$$ equal to?

Let's consider following examples:
If $$x=5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=x=positive$$;
If $$x=-5$$ --> $$\sqrt{x^2}=\sqrt{25}=5=-x=positive$$.

So we got that:
$$\sqrt{x^2}=x$$, if $$x\geq{0}$$;
$$\sqrt{x^2}=-x$$, if $$x<0$$.

What function does exactly the same thing? The absolute value function: $$|x|=x$$, if $$x\geq{0}$$ and $$|x|=-x$$, if $$x<0$$. That is why $$\sqrt{x^2}=|x|$$.
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Re: Inequalities with absolute value [#permalink]  07 Sep 2010, 01:44
Bunuel wrote:
metallicafan wrote:
Is $$sqrt(x-5)^2 = 5-x?$$

(1) -xlxl > 0
(2) 5-x > 0

PS. Is always this true?: $$sqrt{x^2}$$ = lxl ?

Is $$\sqrt{(x-5)^2}=5-x$$?

Remember: $$\sqrt{x^2}=|x|$$.

So "is $$\sqrt{(x-5)^2}=5-x$$?" becomes: is $$|x-5|=5-x$$?

$$|x-5|=5-x$$ is true only for $$x\leq{5}$$, as in this case $$\{LHS=|x-5|=5-x\}=\{RHS=5-x\}$$. So we have that if $$x\leq{5}$$, then $$|x-5|=5-x$$ is true.

Basically question asks is $$x\leq{5}$$?

(1) $$-x|x| > 0$$ --> $$|x|$$ is never negative (positive or zero), so in order to have $$-x|x| > 0$$, $$-x$$ must be positive $$-x>0$$ --> $$x<0$$, so $$x$$ is less than 5 too. Sufficient.

(2) $$5-x>0$$ --> $$x<5$$. Sufficient.

Hope it helps.

Hi Bunuel,

$$-x|x| > 0$$ --> $$|x|$$ is never negative (positive or zero), so in order to have $$-x|x| > 0$$, $$-x$$ must be positive $$-x>0$$ --> $$x<0$$, so $$x$$ is less than 5 too

Can't we solve it as follows:
As $$|x|$$ is never negative -> $$-x|x| > 0$$ = -x*x = -x^2 >0 = x^2<0 (multiplying by -ve sign and flipping sign)
x^2<0 =>
$$sqrt{x^2}$$ <0
=> lxl <0 (as $$sqrt{x^2}$$ =lxl )

Since lxl cannot be negative and lxl <0 that implies X<0

I have reached to same conclusion as yours but wanted to confirm if my approach is right. Please explain.Thanks
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Re: Inequalities with absolute value [#permalink]  07 Sep 2010, 04:01
Expert's post
oldstudent wrote:
Bunuel wrote:
metallicafan wrote:
Is $$sqrt(x-5)^2 = 5-x?$$

(1) -xlxl > 0
(2) 5-x > 0

PS. Is always this true?: $$sqrt{x^2}$$ = lxl ?

Is $$\sqrt{(x-5)^2}=5-x$$?

Remember: $$\sqrt{x^2}=|x|$$.

So "is $$\sqrt{(x-5)^2}=5-x$$?" becomes: is $$|x-5|=5-x$$?

$$|x-5|=5-x$$ is true only for $$x\leq{5}$$, as in this case $$\{LHS=|x-5|=5-x\}=\{RHS=5-x\}$$. So we have that if $$x\leq{5}$$, then $$|x-5|=5-x$$ is true.

Basically question asks is $$x\leq{5}$$?

(1) $$-x|x| > 0$$ --> $$|x|$$ is never negative (positive or zero), so in order to have $$-x|x| > 0$$, $$-x$$ must be positive $$-x>0$$ --> $$x<0$$, so $$x$$ is less than 5 too. Sufficient.

(2) $$5-x>0$$ --> $$x<5$$. Sufficient.

Hope it helps.

Hi Bunuel,

$$-x|x| > 0$$ --> $$|x|$$ is never negative (positive or zero), so in order to have $$-x|x| > 0$$, $$-x$$ must be positive $$-x>0$$ --> $$x<0$$, so $$x$$ is less than 5 too

Can't we solve it as follows:
As $$|x|$$ is never negative -> $$-x|x| > 0$$= -x*x = -x^2 >0 = x^2<0 (multiplying by -ve sign and flipping sign)
x^2<0 =>
$$sqrt{x^2}$$ <0

=> lxl <0 (as $$sqrt{x^2}$$ =lxl )

Since lxl cannot be negative and lxl <0 that implies X<0

I have reached to same conclusion as yours but wanted to confirm if my approach is right. Please explain.Thanks

This approach is not right.

The red parts are not correct.

$$-x|x| > 0$$ can not be written as $$-x*x>0$$, as $$|x|\geq{0}$$ does not mean $$x$$ itself can not be negative --> $$|x|=x$$ if $$x\geq{0}$$ and $$|x|=-x$$ if $$x\leq{0}$$, so when you are writing $$x$$ instead of $$|x|$$ you are basically assuming that $$x\geq{0}$$ and then in the end get the opposite result $$x<0$$.

Next, $$x^2<0$$ has no solution, square of a number can not be negative, so no $$x$$ can make this inequality hold true.
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Re: Inequalities with absolute value [#permalink]  07 Sep 2010, 22:42
I think for this question we dont need any statement... without statement itself it is possible to say if the equality is correct or wrong. Can someone comment on this... Bunuel please?
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Re: Inequalities with absolute value [#permalink]  08 Sep 2010, 05:59
Expert's post
amitjash wrote:
I think for this question we dont need any statement... without statement itself it is possible to say if the equality is correct or wrong. Can someone comment on this... Bunuel please?

This not correct. $$x$$ must be less than or equal to 5 inequality $$\sqrt{(x-5)^2}=5-x$$ to hold true. For example if $$x=10>5$$ then $$\sqrt{(x-5)^2}=5\neq{-5}=5-x$$
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Re: Inequalities with absolute value [#permalink]  16 Oct 2010, 08:01
+1 for D...nice explanation Bunuel..
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Is sqrt(x-5)^2 = 5-x? [#permalink]  22 Jan 2013, 00:01
Analyzing the question first
$$\sqrt{{(x-5)}^2} = 5-x$$
$$|x-5| = 5-x$$ The question is asking whether 5-x >= 0? or x <= 5?

1. -x|x| > 0 means x is negative which is always less than 5.. SUFFICIENT
2. 5-x < 0 means x < 5... SUFFICIENT

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Is sqrt(x-5)^2 = 5-x?   [#permalink] 22 Jan 2013, 00:01
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