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Why does square root of (x^2-6x+9) = 3-X???

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Why does square root of (x^2-6x+9) = 3-X??? [#permalink]

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10 Nov 2013, 10:06
Just as the title, stated above, why does the square root of (x^2-6x+9) = 3-x???? Shouldn't it equal x-3??? Thank you.
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Re: Why does square root of (x^2-6x+9) = 3-X??? [#permalink]

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10 Nov 2013, 20:57
The outcome of sqrt(x^2-6x+9) depends on the value of x. If x >= 3, then sqrt(x^2-6x+9)=x-3, however if x<3, then sqrt(x^2-6x+9)=3-x. For example, if x=4, then sqrt(x^2-6x+9)=sqrt(16-24+9)=sqrt(1)=1, which is indeed equal to x-3=4-3=1. However, if x=0, then sqrt(x^2-6x+9)=sqrt(0-0+9)=sqrt(9)=3, which is indeed equal to 3-x=3-0=3.

The reason for this simplification is based on the identity sqrt(x^2)=|x|. In this particular case sqrt(x^2-6x+9) = sqrt[(x-3)^2]=|x-3| and the absolute value expression can be rewritten based on the value of x, if x>=3, then |x-3|=x-3, and if x<3, then |x-3|=3-x.

In summary, sqrt(x^2-6x+9) = x-3 if x>=3 and sqrt(x^2-6x+9) = 3-x if x<3. In the question you encountered, it was probably a data sufficiency question where they added the constraint x<3.
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Re: Why does square root of (x^2-6x+9) = 3-X??? [#permalink]

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12 Nov 2013, 23:38
Expert's post
carpediem437 wrote:
Just as the title, stated above, why does the square root of (x^2-6x+9) = 3-x???? Shouldn't it equal x-3??? Thank you.

$$(x^2-6x+9) = (x - 3)^2$$

Note that:
$$\sqrt{y^2} = |y|$$ (it is not y, it is mod y)
$$\sqrt{(x - 3)^2} = |x - 3|$$

Now |x - 3| can be (x - 3) or (3 - x). It depends on the value of x.

Now note how you handle mods:
|y| = y if y > 0
|y| = -y is y < 0

So, |x - 3| = x - 3
if (x - 3) >= 0 i.e. x >= 3

|x - 3| = -(x - 3) = 3 - x
If (x - 3) < 0 i.e. x < 3

So, the value of $$\sqrt{(x^2-6x+9)}$$ depends in the value of x. It is (x - 3) if x >= 3 and it is (3 - x) if x < 3.
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Re: Why does square root of (x^2-6x+9) = 3-X???   [#permalink] 12 Nov 2013, 23:38
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