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ParthKohli

  • 2 years ago

What are the roots of \(y = x^2\)?

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  1. goformit100
    • 2 years ago
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    |dw:1364477055858:dw||dw:1364477070806:dw|

  2. ParthKohli
    • 2 years ago
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    A friend of mine suggests that people don't think that zero is a real solution.

  3. Mertsj
    • 2 years ago
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    The roots are the x intercepts. Set y to 0 and you will see that x=0 is a double root which indicates a point of tangency to the x axis at x = 0

  4. Mertsj
    • 2 years ago
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    People can think whatever they want. Doesn't make it true.

  5. ParthKohli
    • 2 years ago
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    \[x^2 = 0 \implies x = 0\]

  6. goformit100
    • 2 years ago
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    You friend is a crazy guy.

  7. ParthKohli
    • 2 years ago
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    OK, thank you!

  8. Mertsj
    • 2 years ago
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    |dw:1364476304306:dw|

  9. Mertsj
    • 2 years ago
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    yw

  10. ParthKohli
    • 2 years ago
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    Thanks, enough to explain.

  11. UnkleRhaukus
    • 2 years ago
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    \[y=x^2\]\[x^2-y=0\]\[x_{1,2}=\frac{0\pm\sqrt{0^2-4(1)(-y)}}{2(1)}\\=\frac{\pm\sqrt{4y}}{2}\\ =\pm \sqrt y\]

  12. ParthKohli
    • 2 years ago
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    Nice solution, but by "roots", I meant zeroes :-)

  13. UnkleRhaukus
    • 2 years ago
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    \[0=\pm\sqrt0\]

  14. ParthKohli
    • 2 years ago
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    Ah, lol!

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