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anonymous

  • one year ago

Evaluate x^2-6x+13=0 and tell whether or not it is extraneous

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  1. anonymous
    • one year ago
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    We can solve this equation by completing the square since it doesn't factor. 1) x^2-6x =13 2) x^2-6x+ 9 =13+9 3) (x-3)^2=22 4) x-3=+/-sq rt( 22) 5) x=3 +/- sq rt(22)

  2. anonymous
    • one year ago
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    @mackenzie_willa

  3. wolf1728
    • one year ago
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    To determine if it has real or imaginary roots, you take a part of the quadratic formula: Square Root (b^2 -4*a*c) In this quadratic a=1 b=-6 and c=13 Square Root (6^2 -4*1*13) Square Root (36 -52) Square Root (-16) You must take a negative square root and therefore the roots are complex (or imaginary).

  4. wolf1728
    • one year ago
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    geny55's solution is wrong To complete the square the equation should be set up as 1) x^2-6x =MINUS 13 2) x^2-6x+ 9 = -13+9 and then continue from there

  5. anonymous
    • one year ago
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    hey @wolf1728 can you help me with like 12 math questions please? :)

  6. wolf1728
    • one year ago
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    TWELVE math questions? I could help you with a couple

  7. anonymous
    • one year ago
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    but they are easy math questions :) please help with 12 :)

  8. wolf1728
    • one year ago
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    Well at least a few. By the way, why not start a new topic?

  9. anonymous
    • one year ago
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    or how about science questions? are you good in science?

  10. wolf1728
    • one year ago
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    I know science pretty well too

  11. anonymous
    • one year ago
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    ok awesome.. ill tag you :)

  12. wolf1728
    • one year ago
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    Okay

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