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@Mursalin Habib
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Look at the last. Assuming the mistake that it is $-\sqrt{(5-\frac{9}{2})^{2}}$
we will get $-5+\frac{9}{2}+\frac{9}{2}=4$So, it is proved that $2+2=4$ and not $5$

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

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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## Comments

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TopNewest$\sqrt{(4 - \frac{9}{2}) ^{2}}$ 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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sqrt( x^2 ) is always | 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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exactly..but in the subsequent steps, (a-b)^2 disguises this.

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

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$-\sqrt{(5-\frac{9}{2})^{2}}$ we will get $-5+\frac{9}{2}+\frac{9}{2}=4$So, it is proved that $2+2=4$ and not $5$

Look at the last. Assuming the mistake that it isLog in to reply

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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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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Where is your proof ??

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

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