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anonymous

  • 5 years ago

What would be the most logical first step to solve this quadratic equation? x2+2x+13=8

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  1. anonymous
    • 5 years ago
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    what level is this? (that equation only has imaginary roots and, without stereotyping, most people who know about imaginary numbers can solve quadratics)

  2. anonymous
    • 5 years ago
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    just use ur quadtratic equatin and substitute.

  3. anonymous
    • 5 years ago
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    so would it be the square root of both sides .

  4. anonymous
    • 5 years ago
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    is it x^2?

  5. anonymous
    • 5 years ago
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    Did you make up this question yourself?

  6. anonymous
    • 5 years ago
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    no its algebra twilight math class .

  7. anonymous
    • 5 years ago
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    its not complicated, just put in the quadtratic formula, the values of Ax +Bx +C=0, where in ur case, its x^2 +2x +5 =0..... or am i missing something...

  8. anonymous
    • 5 years ago
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    Hmm. Well the first step would be to either set it to 0, or complete the square. It won't factorise, so I'd complete the square. But you'll still need to square root a negative number.

  9. anonymous
    • 5 years ago
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    i dont know how to do that sorry im not qood at math worse subject evr .

  10. anonymous
    • 5 years ago
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    Note \[\sqrt{-1} = i \]

  11. anonymous
    • 5 years ago
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    where do u see a negative value?

  12. anonymous
    • 5 years ago
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    nvm is see it now...

  13. anonymous
    • 5 years ago
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    { -2 ± Sq Root ( 4 - 20 ) } / 2

  14. anonymous
    • 5 years ago
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    unless u use imaginary numbers, thats as simplified as it gets...

  15. amistre64
    • 5 years ago
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    subtract the right hand side from both sides of the equation would be the first step

  16. amistre64
    • 5 years ago
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    x2+2x+13 -8=8 -8 x^2 +2x +5 = 0

  17. amistre64
    • 5 years ago
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    the next step would be to "complete the square".... if you dont know the quadratic formula that is :) x^2 +2x +5 = 0 ; subtract 5 from both sides x^2 +2x + ___ = -5 + ____ we need to find a suitable number to fill in these blanks with; we need to create a "perfect" square trinomial that will factor into: (x+?)^2 we know that the product of a binomial and itself is: (a+b)(a+b) = a^2 +2ab + b^2 what we have is: a^2 +a(2__ ) + (___)^2

  18. anonymous
    • 5 years ago
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    I know the quadratic formula, but would still complete the square.

  19. amistre64
    • 5 years ago
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    if I recall correctly, completing the sqaure is how we get the quadratic formula :)

  20. anonymous
    • 5 years ago
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    Affirmative.

  21. anonymous
    • 5 years ago
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    lol.

  22. anonymous
    • 5 years ago
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    man im in university calc, and i dont even remember how to complete the square...havent done that since 9th grade haha

  23. amistre64
    • 5 years ago
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    (a+b)(a+b) = a^2 +2ab + b^2 what we have is: a^2 +a(2__ ) + (___)^2 we see that if we divide the midde term by 2a we will get a suitable "b" right? x^2 +2x +5 ....2x/2x = 1...b=1 b^2 = 1

  24. anonymous
    • 5 years ago
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    pellet, university calc. That's harcore.

  25. anonymous
    • 5 years ago
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    hardcore* :@

  26. amistre64
    • 5 years ago
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    x^2 +2x +1 = -5+1 (x+1)^2 = -4 already see a problem lol x+1 = +-sqrt(-4) x = -1 +- sqrt(-4)

  27. anonymous
    • 5 years ago
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    its also really hard, and i dont know why i need it if im in software engineering, but ok...

  28. anonymous
    • 5 years ago
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    Because Maths is everything.

  29. amistre64
    • 5 years ago
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    x = -1 +- 2i :) but thats a complex number and not a "real" number

  30. anonymous
    • 5 years ago
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    \[x^2 + 2x + 13 = 8 \] \[\iff (x+1)^2 = -4 \] \[\implies (x+1) = \pm \sqrt{-4} = \pm \sqrt{-1}\sqrt{4} = \pm 2i \] \[\implies x = -1 \pm 2i \]

  31. anonymous
    • 5 years ago
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    sorry had that typed out but realised you were explaining it so waited - but didn;t want all that typing to go to waste.

  32. amistre64
    • 5 years ago
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    lol ..... you did good ;)

  33. amistre64
    • 5 years ago
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    does that make any sense miranda?

  34. anonymous
    • 5 years ago
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    ... :(

  35. anonymous
    • 5 years ago
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    im sure thats a yes ;)

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