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samnatha

  • 3 years ago

prove that x²+2xy+3y²≥0 for all x, y E R

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  1. asnaseer
    • 3 years ago
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    hint: Expand \((x+y)^2\)

  2. samnatha
    • 3 years ago
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    i'm sorry i'm still lost

  3. asnaseer
    • 3 years ago
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    what can you say about the number \((x+y)^2\)? can it ever be negative?

  4. samnatha
    • 3 years ago
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    yes

  5. asnaseer
    • 3 years ago
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    how?

  6. asnaseer
    • 3 years ago
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    remember x, y E R

  7. samnatha
    • 3 years ago
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    well i thought anything squared would be true bit i can't get the two brackets the same as i can't get 3y squared

  8. asnaseer
    • 3 years ago
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    try and first expand \((x+y)^2\) and then see what else you need to add to this in order to get your original expression of \(x^2+2xy+3y^2\)

  9. samnatha
    • 3 years ago
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    so the answere would be (x+y)^2 + y^2

  10. asnaseer
    • 3 years ago
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    not quite - what do you get when you expand \((x+y)^2\) ?

  11. malevolence19
    • 3 years ago
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    (x+y)^2+2y^2

  12. samnatha
    • 3 years ago
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    x^2 + 2xy + y ^2

  13. asnaseer
    • 3 years ago
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    correct - now what do you need to add to this in order to get to your original expression?

  14. samnatha
    • 3 years ago
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    2y to be squared

  15. asnaseer
    • 3 years ago
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    perfect!

  16. asnaseer
    • 3 years ago
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    so we end up with:\[x^2+2xy+3y^2=(x+y)^2+2y^2\]now - can you see that this can never be negative?

  17. samnatha
    • 3 years ago
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    ok thanks while i have you i have to do this using the same axsis and scales graph the functions f(x)=|x| and g(x)= |2x - 3|

  18. asnaseer
    • 3 years ago
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    yw :) could you please post that as a separate question. thanks.

  19. samnatha
    • 3 years ago
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    using the same axsis and scales graph the functions f(x)=|x| and g(x)= |2x - 3|

  20. asnaseer
    • 3 years ago
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    <--- I meant post in the list to the left

  21. samnatha
    • 3 years ago
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    oh ok sorry

  22. asnaseer
    • 3 years ago
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    np :) it just helps posting each question separately as others can then see each question more easily and also learn from them. :)

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