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some.random.cool.kid

  • one year ago

Solve the system y equals negative x squared, y equals x minus two.

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  1. some.random.cool.kid
    • one year ago
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    @ganeshie8 @Zarkon @zzr0ck3r @Nnesha

  2. some.random.cool.kid
    • one year ago
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    @undeadknight26 @peachpi

  3. some.random.cool.kid
    • one year ago
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    @ParthKohli

  4. Nnesha
    • one year ago
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    \[\huge\rm y=-x^2\]\[\huge\rm \color{ReD}{y}=x-2\] substitute red y for first equation

  5. Nnesha
    • one year ago
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    |dw:1443539180290:dw| and then solve the quadratic equation

  6. some.random.cool.kid
    • one year ago
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    ok

  7. Nnesha
    • one year ago
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    let me know what u get after that

  8. some.random.cool.kid
    • one year ago
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    not sure..

  9. some.random.cool.kid
    • one year ago
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    is it now -x^2 = x -2?

  10. Nnesha
    • one year ago
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    right

  11. some.random.cool.kid
    • one year ago
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    so what do i do next im confused?

  12. Nnesha
    • one year ago
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    set it equal to zero move -x^2 to the right side then solve for x

  13. some.random.cool.kid
    • one year ago
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    ok

  14. some.random.cool.kid
    • one year ago
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    "-x^2" = x -2 so im moving whats in quotes to the right side of the equal sign right? after that do multiply or do some division of any sort to solve for x?

  15. Nnesha
    • one year ago
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    yes right move -x^2 to the right side then u will get the quadratic equation and i'm pretty sure you r familiar with the quadratic equations we solved 3 q yesterday

  16. some.random.cool.kid
    • one year ago
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    yeah ok

  17. some.random.cool.kid
    • one year ago
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    so now it should look like this right = x -2 -x^2 im not sure what happens to the equal thoough and i believe that the quad equation to solve this is- "ax2 + bx + c"

  18. Nnesha
    • one year ago
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    |dw:1443541053008:dw| when u move number/variable from one side to ohter sign would change

  19. Nnesha
    • one year ago
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    how would you cancel out -x^2 from left side ?

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