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KJ4UTS

  • one year ago

Use the discriminant to determine the number of real roots of... x^2-5x-7=0 x^2-2x+1=0 Choices are: 0 1 2

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  1. Nnesha
    • one year ago
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    what is discriminant ? do you know ?

  2. KJ4UTS
    • one year ago
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    Well my work is (I just want to check): x^2-5x-7=0 -5^2-4(1)(-7)=3>0 2 roots x^2-2x+1=0 -2^2-4(1)(1)=-8<0 1 root

  3. KJ4UTS
    • one year ago
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    b^2-4ac

  4. KJ4UTS
    • one year ago
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    x^2-5x-7=0 2 roots x^2-2x+1=0 1 root

  5. Nnesha
    • one year ago
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    \(\huge\color{reD}{\rm b^2-4ac}\) `Discriminant` you can use this to find if the equation is factorable or not if ` b^2-4ac > 0` then there are 2 real zeros if ` b^2-4ac = 0` then there is one real root if ` b^2-4ac < 0` then you will get two complex roots (no -x-intercept)

  6. Nnesha
    • one year ago
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    you should use discriminant

  7. Nnesha
    • one year ago
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    ohh wait nmv

  8. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @KJ4UTS Well my work is (I just want to check): x^2-5x-7=0 -5^2-4(1)(-7)=3>0 2 roots x^2-2x+1=0 -2^2-4(1)(1)=-8<0 1 root \(\color{blue}{\text{End of Quote}}\) b^2 so it wohuld be (-2)^2 -4(1)(1)

  9. Nnesha
    • one year ago
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    so (-2)^2 -4 = ?

  10. KJ4UTS
    • one year ago
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    -8

  11. Nnesha
    • one year ago
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    no (-2)^2 is same as -2 times -2

  12. Nnesha
    • one year ago
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    or in other words when you take `even` power of negative exponent you will get positive answer always !

  13. KJ4UTS
    • one year ago
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    0

  14. Nnesha
    • one year ago
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    no \[\color{ReD}{(-2)^2}-4(1)(1)\] i'm just talkign about (-2)^2

  15. Nnesha
    • one year ago
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    talking *

  16. Nnesha
    • one year ago
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    yes right

  17. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Nnesha \(\huge\color{reD}{\rm b^2-4ac}\) `Discriminant` you can use this to find if the equation is factorable or not if ` b^2-4ac > 0` then there are 2 real zeros if ` b^2-4ac = 0` then there is one real root if ` b^2-4ac < 0` then you will get two complex roots (no -x-intercept) \(\color{blue}{\text{End of Quote}}\) now read this when b^2-4ac =0 how many roots will u get ?

  18. KJ4UTS
    • one year ago
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    x^2-2x+1=0 -2^2-4(1)(1)=0=0 Which would still be 1 root?

  19. Nnesha
    • one year ago
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    yes right

  20. Nnesha
    • one year ago
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    now what about first one ? b^2-4ac = ??

  21. KJ4UTS
    • one year ago
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    Was my work right -5^2-4(1)(-7)=3>0 2 roots

  22. Nnesha
    • one year ago
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    no that's not correct ....

  23. KJ4UTS
    • one year ago
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    53

  24. KJ4UTS
    • one year ago
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    I think what I was doing wrong for both was not putting parenthesis (-5)^2 for both problems into my calculator but 53>0 therefore I think it is still 2 roots?

  25. Nnesha
    • one year ago
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    yes right

  26. KJ4UTS
    • one year ago
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    Ok thank you @Nnesha for your time and help, and for also pointing out where I went wrong :)

  27. Nnesha
    • one year ago
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    np and yes you should put the parentheses (-2)^2 like this :=) good job!

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