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anonymous

  • 5 years ago

What are the solutions of x^2 - 3x +9 = 0

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  1. anonymous
    • 5 years ago
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    Imaginary

  2. anonymous
    • 5 years ago
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    Thanks man.

  3. anonymous
    • 5 years ago
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    Sarcastic? :( I didn't technically help you.

  4. anonymous
    • 5 years ago
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    Well not sarcastic i figured you were at least trying to help so i appreciate that. The problem is still confusing me though? :[

  5. anonymous
    • 5 years ago
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    You can either work it out with the quadratic formula or completing the square, but the graph does not cross the x axis, so both solutions are imaginary.

  6. anonymous
    • 5 years ago
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    But i did not get two solutions in the problem? :[ im just totally confused on how to solve it :/

  7. anonymous
    • 5 years ago
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    OK one second:

  8. anonymous
    • 5 years ago
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    \[x^2 - 3x + 9 = 0\] \[\iff \left(x-\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 9 = 0\] \[\iff \left(x-\frac{3}{2}\right)^2 = -\frac{27}{4}\] \[\implies x-\frac{3}{2} = \pm \sqrt{\frac{-27}{4}}\]

  9. anonymous
    • 5 years ago
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    Of course, you could equally just use the formula, but I think this way is more fun. Do you know how to carry on (mainly with the root) to get the imaginary/complex solution?

  10. anonymous
    • 5 years ago
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    I do not.

  11. anonymous
    • 5 years ago
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    Do you see how the working I posted works?

  12. anonymous
    • 5 years ago
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    Yeahh.....

  13. anonymous
    • 5 years ago
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    \[\sqrt{\frac{-27}{4}} = \sqrt{-1}\cdot\sqrt{\frac{27}{4}} = i\cdot\frac{3\sqrt{3}}{2}\]

  14. anonymous
    • 5 years ago
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    Again, it may be more 'standard' and easier to repeat if you use the quadratic formula instead, but it would still need you to take the root of a negative number.

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