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    Hi all, I was doing some revision last night and i was looking at this inequality quadratic:

    x^2 + 2x + 3 > 0
    I end up with (x+1) ^2 > -2, so what is the answer to this?
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    Well any squared term is going to be greater than or equal to zero, and therefore greater than -2, so any x will satisfy it.
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    no real roots because root of b^2 - 4ac is less than 0 (in other words, the answer to your question is x= + or - infinity). Hope this helps
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    Hm, can i say this then?

    (x+1) ^2 > -2
    (x+1)^2 + 2 > 0
    (x+1) + sqroot 2 > 0
    x > -(1+sqroot2) ?
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    (Original post by slaw)
    Hm, can i say this then?

    (x+1) ^2 > -2
    (x+1)^2 + 2 > 0
    (x+1) + sqroot 2 > 0
    x > -(1+sqroot2) ?
    No.

    As JamesF said, any x will satisfy the inequality.
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    (Original post by slaw)
    Hi all, I was doing some revision last night and i was looking at this inequality quadratic:

    x^2 + 2x + 3 > 0
    I end up with (x+1) ^2 > -2, so what is the answer to this?
    x^2 + 2x + 3 = (x + 1 - i√2)(x + 1 + i√2) I think.
    It has no real roots- the curve never goes below the x-axis.
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    (Original post by slaw)
    Hi all, I was doing some revision last night and i was looking at this inequality quadratic:

    x^2 + 2x + 3 > 0
    I end up with (x+1) ^2 > -2, so what is the answer to this?
    If you really need to prove something, I suppose you could always use

    f'(x) = 2x+2, giving a turning point at (-1,2)
    f''(x) = 2, telling you it's a minimum, so always y=>2

    Aitch
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    Okay thanks guys
 
 
 
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