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Dodo1

  • 3 years ago

3-sqrt(x)/9-x. how do you break down?

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  1. UnkleRhaukus
    • 3 years ago
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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

  2. Dodo1
    • 3 years ago
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    yes i know that but how do you that for sqrt?

  3. UnkleRhaukus
    • 3 years ago
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    use \(x=\sqrt x^2\)

  4. Dodo1
    • 3 years ago
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    mmm so 3-sqet(x)=(x^2-3^9)?

  5. UnkleRhaukus
    • 3 years ago
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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) \[9-x=(\cdots)(\cdots)\]

  6. Dodo1
    • 3 years ago
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    9-x...... sqert(3)-(x)??

  7. UnkleRhaukus
    • 3 years ago
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    \[9-x=3^2-\sqrt x^2=(\cdots)(\cdots)\]

  8. Dodo1
    • 3 years ago
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    (3-sq\[\left( 3-\sqrt{x} \right)\]

  9. Dodo1
    • 3 years ago
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    (3-sq\[\left( 3-\sqrt{x} \right)\](3-sq\[\left( 3-\sqrt{x} \right)\]?

  10. Dodo1
    • 3 years ago
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    \[\left( 3-\sqrt{x} \right)\]

  11. Dodo1
    • 3 years ago
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    \left( 3-sqrt{x} \right) \left( 3-sqrt{x} \right)

  12. Dodo1
    • 3 years ago
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    (3-sqrt(x)) (3-sqrt(x)

  13. UnkleRhaukus
    • 3 years ago
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    not quite\[a^2-b^2=(a+b)(a-b)\] so\[9-x=3^2-\sqrt x^2=\]

  14. Dodo1
    • 3 years ago
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    3-sqrt(x)?

  15. Dodo1
    • 3 years ago
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    hope its right

  16. UnkleRhaukus
    • 3 years ago
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    \[9-x=3^2-\sqrt x^2=(3+\sqrt x)(3-\sqrt x)\]

  17. UnkleRhaukus
    • 3 years ago
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    so\[\frac{3-\sqrt x}{9-x}=\frac{3-\sqrt x}{(3+\sqrt x)(3-{\sqrt x)}}\]\[\qquad\qquad =\frac{\cancel{3-\sqrt x}}{(3+\sqrt x)\cancel{(3-{\sqrt x)}}}\]

  18. Dodo1
    • 3 years ago
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    Thanks!!

  19. Dodo1
    • 3 years ago
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    what about x^3+64/x+4

  20. Dodo1
    • 3 years ago
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    I got (x+4)(x^2+4x+16)/(x+4)?

  21. UnkleRhaukus
    • 3 years ago
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    almost \[(a^3+b^3)=(a+b)(a^2-ab+b^2)\]

  22. UnkleRhaukus
    • 3 years ago
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    once you have fixed the sign error, cancel the common factor

  23. Dodo1
    • 3 years ago
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    Thank you it was very helpful!!

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