Dodo1
3-sqrt(x)/9-x. how do you break down?
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UnkleRhaukus
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\[\frac{3-\sqrt x}{9-x}\]
factorize the denominator
use \[(a+b)(a-b)=a^2-ab+ba-b^2=a^2-b^2\]
cancel common factors
Dodo1
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yes i know that but how do you that for sqrt?
UnkleRhaukus
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use \(x=\sqrt x^2\)
Dodo1
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mmm so 3-sqet(x)=(x^2-3^9)?
UnkleRhaukus
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your looking at the numerator (bottom bit of the fraction)
you should be looking at the denominator ( the bottom bit of the fraction)
\[9-x=(\cdots)(\cdots)\]
Dodo1
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9-x...... sqert(3)-(x)??
UnkleRhaukus
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\[9-x=3^2-\sqrt x^2=(\cdots)(\cdots)\]
Dodo1
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(3-sq\[\left( 3-\sqrt{x} \right)\]
Dodo1
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(3-sq\[\left( 3-\sqrt{x} \right)\](3-sq\[\left( 3-\sqrt{x} \right)\]?
Dodo1
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\[\left( 3-\sqrt{x} \right)\]
Dodo1
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\left( 3-sqrt{x} \right) \left( 3-sqrt{x} \right)
Dodo1
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(3-sqrt(x)) (3-sqrt(x)
UnkleRhaukus
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not quite\[a^2-b^2=(a+b)(a-b)\]
so\[9-x=3^2-\sqrt x^2=\]
Dodo1
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3-sqrt(x)?
Dodo1
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hope its right
UnkleRhaukus
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\[9-x=3^2-\sqrt x^2=(3+\sqrt x)(3-\sqrt x)\]
UnkleRhaukus
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so\[\frac{3-\sqrt x}{9-x}=\frac{3-\sqrt x}{(3+\sqrt x)(3-{\sqrt x)}}\]\[\qquad\qquad =\frac{\cancel{3-\sqrt x}}{(3+\sqrt x)\cancel{(3-{\sqrt x)}}}\]
Dodo1
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Thanks!!
Dodo1
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what about x^3+64/x+4
Dodo1
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I got (x+4)(x^2+4x+16)/(x+4)?
UnkleRhaukus
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almost
\[(a^3+b^3)=(a+b)(a^2-ab+b^2)\]
UnkleRhaukus
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once you have fixed the sign error, cancel the common factor
Dodo1
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Thank you it was very helpful!!