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qudrex

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

  • 2 years ago
  • 2 years ago

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  1. myininaya
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    add 6x to both sides

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

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

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

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

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

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

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

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

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

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

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

    • 2 years ago
  13. qudrex
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    okay.

    • 2 years ago
  14. ash2326
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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 \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

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

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

    • 2 years ago
  17. qudrex
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    formulas confuse me.

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

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

    • 2 years ago
  20. ash2326
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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?

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

    • 2 years ago
  22. ash2326
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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?

    • 2 years ago
  23. qudrex
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    kinda

    • 2 years ago
  24. ash2326
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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?

    • 2 years ago
  25. qudrex
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    yes

    • 2 years ago
  26. ash2326
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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?

    • 2 years ago
  27. qudrex
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    yeah.

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

    • 2 years ago
  29. qudrex
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    okay!

    • 2 years ago
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