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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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TopNewestThere's a subtle mistake in there.

Here's a cryptic hint that might help:

Is \(\sqrt{x^2}\) always equal to \(x\)?

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exactly..but the usage of (a-b)^2 somehow disguises this.

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You can't write (4-9/2) as square root of (4-9/2)^2. That is the mistake.

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Why not?

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Ram Prakash is right. \(4-\frac{9}{2}\) is not equal to \(\sqrt{(4-\frac{9}{2})^2}\). This is what I hinted at in my initial comment.

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Since \(4-9/2\) is negative, while \(\sqrt{(4-9/2)^2}\) is positive.

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\( \sqrt{(4 - \frac{9}{2}) ^{2}} \) is not equal to \( 4 - \frac{9}{2}\)

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thanks

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\( \sqrt x^2 = |x| \) not \( x\).

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You cant make a negative number positive by just squaring it and taking root. I'm talking about 2nd step, (4-(9/2))

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thanks

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\((4-\frac{9}{2})=-\frac{1}{2}\)

but,

\((5-\frac{9}{2})=\frac{1}{2}\)

in this proof we wrote--

\(2+2=4-\frac{9}{2}+\frac{9}{2}...........step (i)\)

\(=\sqrt{(5-\frac{9}{2})^2}+\frac{9}{2}..........step(viii)\)

\(=5-\frac{9}{2}+\frac{9}{2}..........step(ix)\)

so,what we actually doing is

\(-\frac{1}{2}+\frac{9}{2}=\frac{1}{2}+\frac{9}{2}\)..............consider step (i) and (ix)

this is where we made mistake.

So in this case at the time of removing square root of \( \sqrt{(5-\frac{9}{2})^2}\)

we need to consider \(\sqrt{(5-\frac{9}{2})^2}=-(5-\frac{9}{2})\)

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\( \sqrt {(5-\frac{9}{2})^{2}} \) is not equal to \( -(5-\frac{9}{2})\)

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ok..i understood thanks.

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Careful! Gypsy's last statement is incorrect.

\(\sqrt{(5-\frac{9}{2})^2}\) is

notequal to \(-(5-\frac{9}{2})\).However, \(\sqrt{(4-\frac{9}{2})^2}=-(4-\frac{9}{2})\).

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I think so that the square roots can't be split...

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It is wrong because \(\sqrt{x^2}\) is not always equal to \(x\). It has two values, that are \(x\) and \(-x\)

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You should be careful about what you write. \(\sqrt{x^2}\) [the principal square-root of \(x^2\)] does not have two values.

Take this for example: what is \(\sqrt{(1-\sqrt{3})^2}\). Is it \((1-\sqrt{3})\)? Or is it \(-(1-\sqrt{3})\)? Or is it both?

According to you, both of these should be correct. Are they?

Also see the solutions for this problem where almost all of them make this same mistake.

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Mmm, that got me.

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sqrt( x^2 ) is always | x |

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exactly..but in the subsequent steps, (a-b)^2 disguises this.

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Where is your proof ??

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it's an external link. Click on the heading of this discussion.

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Did you see the term: 2 x 4 x 9/2? if we cancel 2 then the answer will be 39 but when we multiply the numerators, it is equal to 29. So maybe it is the mistake

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No,the calculation is\( 2 \times 4 \times \frac{9}{2}\).If we cancel the 2's,then it becomes \( 4 \times 9 =36\) and if we multiply the numerators it is \( \frac{72}{2} = 36\). So this is not the mistake.

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No sir, i don't think so.

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