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anonymous

  • 5 years ago

x^2-2kx+49 is the square of a binomial. What is the possible value of k?

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  1. anonymous
    • 5 years ago
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    7

  2. anonymous
    • 5 years ago
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    how do you solve that? Can you show your work pleazzz???

  3. anonymous
    • 5 years ago
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    (x+a)^2=x^2+2ax+a^2=x^2-2kx+49 a^2=49 2ax=-2kx a=+/- 7 a=-k k=-a k=+/- 7

  4. anonymous
    • 5 years ago
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    perfect square \[(a-b)^2=a^2-2ab+b^2\] here \[a=x\] \[b^2=49\] so \[b=7\]

  5. anonymous
    • 5 years ago
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    wait rsvitale, where do you get the a from?

  6. anonymous
    • 5 years ago
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    k could be -7 too, (x+7)^2=x^2+14x+49, which would mean k=-7

  7. anonymous
    • 5 years ago
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    a is just a variable for the second term in the binomial

  8. anonymous
    • 5 years ago
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    why is it (a-b)^2 im confused about the beginning of the problem

  9. anonymous
    • 5 years ago
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    thats the general form of a square of a binomial

  10. anonymous
    • 5 years ago
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    so we want to see what the general form looks like expanded and find what k can be

  11. anonymous
    • 5 years ago
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    \[(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2=a^2+2ab+b^2\]

  12. anonymous
    • 5 years ago
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    ok satelite, now explain from there. i get that step now so what do we do next... explain it like you just did.... sorry im not very good at math

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