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qudrex

  • 4 years ago

Which expression gives the solutions of –5 + 2x² = –6x?

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  1. myininaya
    • 4 years ago
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    add 6x to both sides

  2. myininaya
    • 4 years ago
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    \[2x^2+6x-5=0\]

  3. myininaya
    • 4 years ago
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    Now use the quadratic formula! :)

  4. myininaya
    • 4 years ago
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    a=2,b=6,c=-5

  5. qudrex
    • 4 years ago
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    whats the quadratic formula?

  6. myininaya
    • 4 years ago
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    \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  7. qudrex
    • 4 years ago
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    i dont know how to use that. D:

  8. myininaya
    • 4 years ago
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    Do you know how to complete the square?

  9. qudrex
    • 4 years ago
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    no. :/ im horrible at math.

  10. myininaya
    • 4 years ago
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    hmmm... so what method do you want to use?

  11. qudrex
    • 4 years ago
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    i dont know any methods to solve that.

  12. ash2326
    • 4 years ago
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    @Qudrex I'll help you.

  13. qudrex
    • 4 years ago
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    okay.

  14. ash2326
    • 4 years ago
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    A standard quadratic equation is given as \[ax^2+bx+c=0\] its roots or its solution is given by the following formula \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  15. ash2326
    • 4 years ago
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    This formula gives us two values of x which satisfies the quadratic equation

  16. ash2326
    • 4 years ago
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    Did you understand till now?

  17. qudrex
    • 4 years ago
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    formulas confuse me.

  18. ash2326
    • 4 years ago
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    what's confusing you?

  19. qudrex
    • 4 years ago
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    the formula i dont know how to use it.

  20. ash2326
    • 4 years ago
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    It's very easy, I'll show you your question is -5+2x^2=6x The first step is to bring all the terms to one side \[-5+2x^2=6x\] Let's subtract both sides by 6x we get \[-5+2x^2-6x=6x-6x\] we'll get \[-5+2x^2-6x=0\] Now let's write this in descending powers of x \[2x^2-6x-5\] Did you understand till now?

  21. qudrex
    • 4 years ago
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    yes. i think i understand that.

  22. ash2326
    • 4 years ago
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    Great now let's compare this with standard equation \[ax^2+bx+c=0\] \[2x^2-6x+5=0\] we have \[a=2,\ b=-6\ and\ c=5 \] Did you get this?

  23. qudrex
    • 4 years ago
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    kinda

  24. ash2326
    • 4 years ago
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    Now its solution is given by \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] Let's substitute a=2, b=-6 and c=-5 \[x=\frac{-(-6)\pm \sqrt{(-6)^2-4\times (2)\times (-5)}}{2\times (2)}\] We get now \[x=\frac{6\pm \sqrt{36-(-40)}}{4}\] \[x=\frac{6\pm \sqrt{36+40}}{4}\] \[x=\frac{6\pm \sqrt{76}}{4}\] Did you understand this?

  25. qudrex
    • 4 years ago
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    yes

  26. ash2326
    • 4 years ago
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    We've almost found the solution, we could simplify it more We have \[x=\frac{6\pm \sqrt{76}}{4}\] \[\sqrt {76}= \sqrt{4\times 19}=2\sqrt {19}\] so \[x=\frac{6\pm 2\sqrt{19}}{4}\] divide the whole equation by 2 we get \[x=\frac{3\pm \sqrt {19}}{2}\] so the two solutions\roots are \[x=\frac{3+ \sqrt {19}}{2}, \frac{3- \sqrt {19}}{2}\]\] Did you understand?

  27. qudrex
    • 4 years ago
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    yeah.

  28. ash2326
    • 4 years ago
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    So whenever you've to find roots of quadratic equation, use this formula:D

  29. qudrex
    • 4 years ago
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    okay!

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