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shelby1290

  • one year ago

Determine the value(s) of k such that each trinomial is a perfect square. a) x^2 + 4x + k b) 4x^2 - 12x + k

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  1. misty1212
    • one year ago
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    two steps only

  2. misty1212
    • one year ago
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    what is half of 4?

  3. shelby1290
    • one year ago
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    @misty1212 half of 4 is 2

  4. shelby1290
    • one year ago
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    what's next?

  5. anonymous
    • one year ago
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    \[b^2-4ac=0\]

  6. anonymous
    • one year ago
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    \[4^2-4(1)(k)=0\]

  7. anonymous
    • one year ago
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    k=?

  8. shelby1290
    • one year ago
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    12?

  9. anonymous
    • one year ago
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    Try again

  10. shelby1290
    • one year ago
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    4^2-4(1)(k) 16-4k=0 divide both sides by -4 k=-4

  11. shelby1290
    • one year ago
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    is that right?

  12. anonymous
    • one year ago
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    \[16-4k=0 \\ \\ 16=4k \\ \\ k=?\]

  13. shelby1290
    • one year ago
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    k = 4 oops

  14. shelby1290
    • one year ago
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    thank you

  15. shelby1290
    • one year ago
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    now how about part b

  16. anonymous
    • one year ago
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    Do the same for the next one \[4x^2 - 12x + k=0 \\ Using \\ ax^2+bx+c=0 \\ \\ b^2-4ac=0\]

  17. anonymous
    • one year ago
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    \[(-12)^2-4(4)(k)=0\]

  18. shelby1290
    • one year ago
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    -144-16k=0 -144=16k divide both sides by 16 k=-9 ?

  19. anonymous
    • one year ago
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    \[(-12)^2= (-12)(-12)=12 \times 12=144\]

  20. anonymous
    • one year ago
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    k=9

  21. shelby1290
    • one year ago
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    oh i see it's because i didn't put (-12)^2 in brackets

  22. shelby1290
    • one year ago
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    thanks for the help though!

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