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anonymous

  • 5 years ago

Can someone please double check my answer? Solve the system by method of substitution. y=x y=x^3+4x^2+6x Answer choices: -(0,0) -No real solution I think it is (0,0) but would like to double check

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  1. shadowfiend
    • 5 years ago
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    You don't even *really* need the substitution method, but if you do use it you end up with: \[y = y^3 + 4y^2 + 6y\] \[0 = y^3 + 4y^2 + 5y\] \[0 = y(y^2 + 4y + 5y)\] This means that either y must be 0 or \(y^2 + 4y + 5y\) must be 0.

  2. shadowfiend
    • 5 years ago
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    Sorry, that was \(y^2 + 4y + 5\) must be 0, and the equation should be \(0 = y(y^2 + 4y + 5)\). If y = 0 and y = x, then x = 0. So (0, 0) is *a* solution. If you try to solve \(y^2 + 4y + 5\) and find where it is 0, you'll find it doesn't have a real solution (the discriminant \(b^2 - 4ac\) will be <0), so the only real solution is (0, 0).

  3. shadowfiend
    • 5 years ago
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    Hope that helps!

  4. anonymous
    • 5 years ago
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    Thanks now what if the equation was: y=-x y=x^3+x^2+5x would it be (0,0) or no real solution? I was thinking no real solution....

  5. shadowfiend
    • 5 years ago
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    Well, what do you get if you plug into the bottom equation?

  6. anonymous
    • 5 years ago
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    well i thought that because you can't have x=-0... that's just not possible...Right?

  7. shadowfiend
    • 5 years ago
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    Oh, sure it is. -0 is the same as 0.

  8. anonymous
    • 5 years ago
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    oh ok so it's (0,0) again cool thanks

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