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    The quadratic equation x^2+ 2kx + 2(k + 4) = 0 has distinct real roots. Show that k^2 – 2k – 8 > 0.
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    (Original post by kitkat132000)
    The quadratic equation x^2+ 2kx + 2(k + 4) = 0 has distinct real roots. Show that k^2 – 2k – 8 > 0.
    The discriminant is greater than 0 for 2 distinct roots.
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    • By applying the discriminant, b^2-4ac...
    • (2k)^2-4(1)(2k+8)>0
    • Can you do the rest?
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    (Original post by AlexOD)
    Full Solution
    Yeah easy-peasy but full solutions are not allowed on this forum.
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    Oh sorry I didn't know that...
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    Ok thanks! So is the answer 34>0 which proves that the quadratic is greater than 0.
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    (Original post by kitkat132000)
    Ok thanks! So is the answer 34>0 which proves that the quadratic is greater than 0.
    If you're given a particular value of k to test and it turns out to be 34 on the LHS, then yes. If not, then you simply need to arrive at the required answer without doing anything else. Just saying this as I do not know where you pulled the 34 from.
 
 
 
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