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    Prove that the complete solution of the inequality

    x^2 + 4x + 5 > 0

    is the set of real numbers?

    What the hell? I don't really understand the last line at all, any ideas?

    Thanks
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    Try to sketch it, finding the stationary points and if they are minimas or maximas. Then find the value of x^2+4x+5 at the local minima. You should find only one stationary point, which is a local minima and so the lowest value of x^2+4x+5, which will be greater than 0, hence all other real numbers will be bigger than 0, as all other numbers have to be bigger than the local minima if there are no other turning points.

    You can just interpret 'in the set of real numbers' as 'exists'. For example, the square root of -1 'doesn't exist' so is not a real number, it is an imaginary number.
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    An even easier way is to rewrite the inital equation as (x+2)^2 + 1 which is > 0 for all x.
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    do the discriminant?
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    (Original post by theone)
    An even easier way is to rewrite the inital equation as (x+2)^2 + 1 which is > 0 for all x.
    Exactly, the brackets always come out positive, and if they come out as zero, the +1 sorts it out. QED
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    (Original post by _Devour_You)
    do the discriminant?
    Nope, not in this case, it won't tell you anything...

    (Original post by theone)
    Nope, not in this case, it won't tell you anything...
    It will tell you how many times the parabola crosses the x-axis (0). If you then just work out the equation for any value of x and find that it is positive, then it must be positive for all values of x as the graph does not cross the x-axis.
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    Thanks for the replies. I tried completing the square, just confused with the language.
 
 
 
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Updated: December 7, 2003
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