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koli123able

  • 3 years ago

Find 'p' if 3x^2+2x+3p=0 has real & different roots. . . ANSWER = p<1/9

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  1. koli123able
    • 3 years ago
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    I have no idea on solving this question....

  2. Xavier
    • 3 years ago
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    A quadratic has real and different roots if the descriminant is greater than zero: \[\Delta > 0\]

  3. koli123able
    • 3 years ago
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    delta?? u mean discriminant?

  4. ParthKohli
    • 3 years ago
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    Yes.

  5. Xavier
    • 3 years ago
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    Yup

  6. koli123able
    • 3 years ago
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    so how D>0 ??

  7. Xavier
    • 3 years ago
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    What do you mean? Set the discriminant to be greater than zero b^2-4ac > 0

  8. koli123able
    • 3 years ago
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    I mean how is discriminant greater than zero can u explain?

  9. harsimran_hs4
    • 3 years ago
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    if d > 0 then there are real and different roots d = 0 real and equal roots d < 0 imaginary roots

  10. agent0smith
    • 3 years ago
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    Real and different...? I'm guessing you mean real and distinct. But yes you can use the discriminant. When the discriminant is less than zero, it will have complex/imaginary roots, as the quadratic formula takes the square root of the discriminant ( http://www.purplemath.com/modules/quadform.htm ) \[3x^2+2x+3p=0 \]\[ax^2+bx+c = 0\] so here, a=3, b = 2 and c = 3p discriminant is \[b^2-4ac > 0 \]

  11. koli123able
    • 3 years ago
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    If D>0 then sqrt(b^2-4ac) is a real number and roots are real and unequal

  12. koli123able
    • 3 years ago
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    am I right?

  13. harsimran_hs4
    • 3 years ago
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    \[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\] try putting different values of d and check out

  14. harsimran_hs4
    • 3 years ago
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    yes you are!!

  15. koli123able
    • 3 years ago
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    as D>0 and D=b^2-4ac so, (2)^2-4*3*3p>0 4-36p>0 4>36p 1/9>0

  16. koli123able
    • 3 years ago
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    is this ok?

  17. Xavier
    • 3 years ago
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    You lost a p at the end

  18. koli123able
    • 3 years ago
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    oh sorry I typed it in hurry

  19. koli123able
    • 3 years ago
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    p<1/9

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