Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

qudrex

  • 2 years ago

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

  • This Question is Closed
  1. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    add 6x to both sides

  2. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[2x^2+6x-5=0\]

  3. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Now use the quadratic formula! :)

  4. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    a=2,b=6,c=-5

  5. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whats the quadratic formula?

  6. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  7. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont know how to use that. D:

  8. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Do you know how to complete the square?

  9. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no. :/ im horrible at math.

  10. myininaya
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    hmmm... so what method do you want to use?

  11. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i dont know any methods to solve that.

  12. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @Qudrex I'll help you.

  13. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay.

  14. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    This formula gives us two values of x which satisfies the quadratic equation

  16. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Did you understand till now?

  17. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    formulas confuse me.

  18. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    what's confusing you?

  19. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the formula i dont know how to use it.

  20. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes. i think i understand that.

  22. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    kinda

  24. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  26. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah.

  28. ash2326
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    So whenever you've to find roots of quadratic equation, use this formula:D

  29. qudrex
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay!

  30. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.