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JacksonJRB

  • one year ago

Solve for x in terms of y. y=x^2-4x+3

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  1. SolomonZelman
    • one year ago
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    ok, I will show a similar problem.

  2. JacksonJRB
    • one year ago
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    ok

  3. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y=2x^2-3x+5 }\) \(\large\color{black}{ \displaystyle 0=2x^2-3x+5-y }\) so a=2 b=-3 c=(5-y) \(\large\color{black}{ \displaystyle x=\frac{-(-3)\pm\sqrt{(-3)^2-4(2)(5-y)}}{2(2)} }\) \(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-8(5-y)}}{4} }\) \(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{9-40+8y}}{4} }\) \(\large\color{black}{ \displaystyle x=\frac{3\pm\sqrt{-31+8y}}{4} }\)

  4. JacksonJRB
    • one year ago
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    So, i just use quadratic formula?

  5. SolomonZelman
    • one year ago
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    Yes

  6. JacksonJRB
    • one year ago
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    Okay, I'll try it. Thanks

  7. SolomonZelman
    • one year ago
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    ok

  8. JacksonJRB
    • one year ago
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    \[x=\frac{ 4+/- \sqrt{4+4y} }{ 2 }\]

  9. SolomonZelman
    • one year ago
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    let me check ...

  10. JacksonJRB
    • one year ago
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    \[x=2+/-\sqrt{1+y}\]

  11. JacksonJRB
    • one year ago
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    Simplifies to that I believe

  12. SolomonZelman
    • one year ago
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    y=x^2-4x+3 0=x^2-4x+3-y \(\large\color{black}{ \displaystyle x=\frac{4\pm\sqrt{(-4)^2-4(1)(3-y)}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{4\pm\sqrt{16-4(3-y)}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{4\pm\sqrt{4(4-(3-y))}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{4\pm2\sqrt{4-(3-y)}}{2} }\) \(\large\color{black}{ \displaystyle x=2\pm\sqrt{4-(3-y)} }\) \(\large\color{black}{ \displaystyle x=2\pm\sqrt{4-3+y} }\) \(\large\color{black}{ \displaystyle x=2\pm\sqrt{1+y} }\) Very Nice!

  13. SolomonZelman
    • one year ago
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    Good job, keep that up:)

  14. JacksonJRB
    • one year ago
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    Thanks for the help

  15. SolomonZelman
    • one year ago
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    you have done it all yourself, i just pointed:)

  16. JacksonJRB
    • one year ago
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    There's another problem that involves fractions though

  17. SolomonZelman
    • one year ago
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    ok, ...

  18. JacksonJRB
    • one year ago
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    Little confused

  19. JacksonJRB
    • one year ago
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    \[y=\frac{ x^2+1 }{ x^2-1 }\]

  20. SolomonZelman
    • one year ago
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    you want x in terms of y?

  21. JacksonJRB
    • one year ago
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    yes

  22. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle y=\frac{x^2+1}{x^2-1} }\) \(\large\color{black}{ \displaystyle y=\frac{x^2-1+2}{x^2-1} }\) \(\large\color{black}{ \displaystyle y=\frac{x^2-1}{x^2-1}+\frac{2}{x^2-1} }\) \(\large\color{black}{ \displaystyle y=1+\frac{2}{x^2-1} }\) can you get it from there?

  23. JacksonJRB
    • one year ago
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    Why did you add 2?

  24. SolomonZelman
    • one year ago
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    I didn't, I rewrote 1 as 2-1

  25. SolomonZelman
    • one year ago
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    -1+2, is same as 1

  26. JacksonJRB
    • one year ago
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    Oh, I see now

  27. JacksonJRB
    • one year ago
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    I think I can finish it

  28. SolomonZelman
    • one year ago
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    ok, go ahead, but take your time...

  29. JacksonJRB
    • one year ago
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    \[x=\sqrt{\frac{ 2 }{ y-1 }+1}\]

  30. SolomonZelman
    • one year ago
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    yes (I think though you need the \(\pm\) there)

  31. JacksonJRB
    • one year ago
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    You're right

  32. JacksonJRB
    • one year ago
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    Thanks

  33. SolomonZelman
    • one year ago
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    But, I mean without the \(\pm\) you have still achieved the task. You solved for x in terms of y. You might not need ± because (-)² or (+)² is all 1.

  34. SolomonZelman
    • one year ago
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    yw in any case.

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