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karinewoods17

x^2 + x – 4 = 0 How would this be solved by graphing?

  • one year ago
  • one year ago

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  1. ilikephysics2
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    quadratic equation, do you know the formula?

    • one year ago
  2. karinewoods17
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    yes. \[x =\frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}\] But I need the vertex, domain and range, x and y intercepts, and the axis of symmetry.

    • one year ago
  3. ilikephysics2
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    I did that stuff years ago, hang on I will look it up

    • one year ago
  4. karinewoods17
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    Thanks. lol I just have trouble with solving the equation by graphing. :/

    • one year ago
  5. ilikephysics2
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    the x and y intercepts are going to be the equation factored out in the quadratic formula...I'm sure you can do that right

    • one year ago
  6. ilikephysics2
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    the domain will be all x values not equal to the x and y intercepts that you get, and they can not make the denominator 0

    • one year ago
  7. ilikephysics2
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    The axis of symmetry of the parabola y = ax^2 + bx + c is the vertical line x = -b/2a

    • one year ago
  8. amistre64
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    solved by graphing means that we graph the equation and look at where it crosses the x axis

    • one year ago
  9. karinewoods17
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    \[x = \frac{ -1 \pm \sqrt{1^2 - 4 (1) (-4)} }{ 2 (1) }\] I get \[x = \frac{ -1 \pm \sqrt{17} }{ 2 }\] How do I simplify that?

    • one year ago
  10. ilikephysics2
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    you can leave it in those terms if you want, but type it in the calculator

    • one year ago
  11. amistre64
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    http://www.wolframalpha.com/input/?i=x%5E2+%2B+x+%E2%80%93+4

    • one year ago
  12. ilikephysics2
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    why are you both copying and pasting?

    • one year ago
  13. karinewoods17
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    I am not allowed to have a decimal answer though.

    • one year ago
  14. ilikephysics2
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    haha you give her wolfram , nice source and your a moderator...

    • one year ago
  15. amistre64
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    wolfram does nice graphs :)

    • one year ago
  16. amistre64
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    to solve by algebra, you can employ the quadratic equation, but the question disnt ask for that method

    • one year ago
  17. ilikephysics2
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    its basically cheating...people that use that site might not go anywhere in life due to it just giving away answers

    • one year ago
  18. karinewoods17
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    Lmao, thanks to the both of you! @ilikephysics2 I want to be an engineer so your right, lol.

    • one year ago
  19. ilikephysics2
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    Then make that happen

    • one year ago
  20. amistre64
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    "solve by graphing" is prolly the least exact method of finding a solution. But it is a viable in order to see if you ar in the right ballpark. The wolf does go ahead and spit out al the properties such as the exact figures :)

    • one year ago
  21. karinewoods17
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    Yeah its just apart of learning the different methods to solving quadratic equations. I also have to find the answer using factoring, quadratic formula, and completing the square.

    • one year ago
  22. ilikephysics2
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    you will get them and understand them, they are not bad at all

    • one year ago
  23. amistre64
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    completing the square is nice; its the "proof" to the quadratic equation.

    • one year ago
  24. amistre64
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    \[ax^2+bx+c=0\] \[x^2+\frac bax+\frac ca=0\] \[x^2+\frac bax=-\frac ca\] \[x^2+\frac bax+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac ca\] \[(x+\frac b{2a})^2=\frac{b^2-4ac}{4a^2}\] \[x+\frac b{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=-\frac b{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}\] \[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • one year ago
  25. karinewoods17
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    Whats all that?! lol

    • one year ago
  26. amistre64
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    thats the a general quadratic setup, and by running thru the "complete the square" process, we end up with the quadratic formula. completeing the square is the proof to the quadratic formula.

    • one year ago
  27. amistre64
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    since the completeing the square and the quadratic formula are 2 sides of the same coin.

    • one year ago
  28. karinewoods17
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    True

    • one year ago
  29. karinewoods17
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    How do I write x^2 + x - 4 = 0 in standard form?

    • one year ago
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