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karinewoods17

  • 2 years ago

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

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  1. ilikephysics2
    • 2 years ago
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    quadratic equation, do you know the formula?

  2. karinewoods17
    • 2 years ago
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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.

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

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

  5. ilikephysics2
    • 2 years ago
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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

  6. ilikephysics2
    • 2 years ago
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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

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

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

  9. karinewoods17
    • 2 years ago
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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?

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

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

  12. ilikephysics2
    • 2 years ago
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    why are you both copying and pasting?

  13. karinewoods17
    • 2 years ago
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    I am not allowed to have a decimal answer though.

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

  15. amistre64
    • 2 years ago
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    wolfram does nice graphs :)

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

  17. ilikephysics2
    • 2 years ago
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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

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

  19. ilikephysics2
    • 2 years ago
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    Then make that happen

  20. amistre64
    • 2 years ago
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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 :)

  21. karinewoods17
    • 2 years ago
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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.

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

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

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

  25. karinewoods17
    • 2 years ago
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    Whats all that?! lol

  26. amistre64
    • 2 years ago
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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.

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

  28. karinewoods17
    • 2 years ago
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    True

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

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