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DLS

  • 2 years ago

PLEASE HELP!! If the line y=root 3x interests the curve x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0 At 3 points A,B,C .Then find OA*OB*OC...

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  1. DLS
    • 2 years ago
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    |dw:1353211730667:dw| SMT lyk this?

  2. DLS
    • 2 years ago
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    @sauravshakya

  3. sauravshakya
    • 2 years ago
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    Is O the origin?

  4. DLS
    • 2 years ago
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    yes

  5. sauravshakya
    • 2 years ago
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    Is it |dw:1353224657415:dw|

  6. DLS
    • 2 years ago
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    yes

  7. sauravshakya
    • 2 years ago
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    Is it |dw:1353225679747:dw|

  8. sauravshakya
    • 2 years ago
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    @DLS

  9. DLS
    • 2 years ago
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    whatts that:o

  10. sauravshakya
    • 2 years ago
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    y=root 3x ..............i x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0 ............ii Now, From i and ii, (1+3root3)x^3 + (14+3root3)x^2 + (4+5root3)x-1=0 x^3 + (14+3root3)/(1+3root3)x^2 + (4+5root3)/(1+3root3)x -1/(1 + 3root3) =0 Now product of all the roots of this cubic equation is -1/(1+3root3) Now, OA*OB*OC = 8*3root3*product of all the roots of the above cubic equation =-24root3 /(1+3root3)

  11. DLS
    • 2 years ago
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    its 8/3root3 +1

  12. DLS
    • 2 years ago
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    \[x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0\] is the equation :(

  13. DLS
    • 2 years ago
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    \[y=\sqrt{3}x\] is the line

  14. sauravshakya
    • 2 years ago
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    oh yes it is 8/(3root3 +1) y=root 3x ..............i x^3+y^3+3xy+5x^2+3y^2+4x+5y-1=0 ............ii Now, From i and ii, (1+3root3)x^3 + (14+3root3)x^2 + (4+5root3)x-1=0 x^3 + (14+3root3)/(1+3root3)x^2 + (4+5root3)/(1+3root3)x -1/(1 + 3root3) =0 Now product of all the roots of this cubic equation is -{-1/(1+3root3)} NOTE: it is -ve since it is a cubic equation which has odd number of roots Now, OA*OB*OC = 8*product of all the roots of the above cubic equation =8/(1+3root3)

  15. DLS
    • 2 years ago
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    explain properly? ://

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