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- #1

[tex]x\leq 1[/tex]

[tex](x - 1)^{2} = y[/tex]

So I know that (x - 1) will always be 0 or a negative, therefore I must take the negative square root of (x - 1):

[tex]-\sqrt{(x - 1)^{2}} = -\sqrt{y}[/tex]

Am I to understand that this is the same as:

[tex]-1 \cdot \sqrt{(x - 1)^{2}} = -1 \cdot \sqrt{y}[/tex]

[tex]=[/tex]

[tex]-1 \cdot (x - 1) = -1 \cdot \sqrt{y}[/tex]

[tex]=[/tex]

[tex]-x + 1 = -1 \cdot \sqrt{y}[/tex]

Or how does this work?