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haterofmath

  • 2 years ago

√((18)/(36))=(√(2))/(2)

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  1. haterofmath
    • 2 years ago
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    \[\frac{ \sqrt{18} }{ \sqrt{36} }=\frac{ \sqrt{2} }{ 2 }\]

  2. phi
    • 2 years ago
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    yes. you can see it is true by factoring each number into its prime factors 18 is 2*9 = 2*3*3 36 is 2*18= 2*2*9= 2*2*3*3 when these are under a square root \[ \frac{\sqrt{2\cdot 3\cdot 3}}{\sqrt{2\cdot 2\cdot 3 \cdot 3} }\] you can "pull out" pairs. in the top, 3*3 out of the square root becomes 3 in the bottom both 2 and 3 come out \[ \frac{3 \sqrt{2}}{2\cdot 3} \] we can divide top and bottom by 3 to simplify to \[ \frac{\sqrt{2}}{2} \]

  3. phi
    • 2 years ago
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    of course, if you know 6*6=36 you can just replace the bottom sqrt(36) with 6 right away.

  4. haterofmath
    • 2 years ago
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    so the question ask to simplify

  5. phi
    • 2 years ago
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    sqrt(2)/2 is as simple as it gets. you could write it as 1/sqrt(2) but people do not sqrt's in the denominator.

  6. haterofmath
    • 2 years ago
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    ok. so i have another problem...it says to multiply and simplify \[(3\sqrt{10}-2\sqrt{2})^2=98-24\sqrt{5}\]

  7. phi
    • 2 years ago
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    it looks like they gave you the answer. But to show how to do it, use FOIL on ( 3 sqrt(10) - 2 sqrt(2) )* ( 3 sqrt(10) - 2 sqrt(2) )

  8. phi
    • 2 years ago
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    here is an example of how to do this http://www.khanacademy.org/math/algebra/polynomials/v/multiplying-binomials-with-radicals

  9. haterofmath
    • 2 years ago
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    so it will just be \[(3\sqrt{10}-2\sqrt{2})(3\sqrt{10}-2\sqrt{2})=98-24\sqrt{5}\]

  10. phi
    • 2 years ago
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    yes, but I think they want you to do the multiplication and get the answer they show you.

  11. haterofmath
    • 2 years ago
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    how would you do that?

  12. phi
    • 2 years ago
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    This video http://www.khanacademy.org/math/algebra/polynomials/v/multiplying-binomials and the other one posted go into the details.

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