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C1 A-level maths: Inequalities problem Watch

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    Show that x^2 + 2kx + 9 ≥ 0 for all real values of x, if k^2 ≤ 9.
    I keep getting/showing that k^2≥ 9, and not the opposite. Can somebody help?
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    Is this the full question?
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    x^2 + 2kx +9 = (x +k)^2 - k^2 +9 (completing the square)

    Therefore, -k^2 + 9 is the minimum value.

    So if k^2 ≤ 9 ,
    0 ≤ 9 - k^2
    0 ≤- k^2 + 9
    0 ≤ the minimum value

    As its minimum value is greater than or equal to zero, all its possible values must be.

    Therefore, 0 ≤ x^2 + 2kx + 9 , when k^2 ≤ 9. Q.E.D.

    Hope this helps.
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    Got it, b^2-4ac≥0 for values of real roots
    a=1
    b=2k
    c=9
    (2k)^2-4x1x9
    ≥0
    4k^2-36
    ≥0
    4k^2
    ≥36 (Then dividing through by 4 changes the direction of the sign from ≥ to ≤)
    therefore making it k^2
    ≤9
    Hope this helps.
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    (Original post by brooky2312)
    Got it, b^2-4ac≥0 for values of real roots
    a=1
    b=2k
    c=9
    (2k)^2-4x1x9≥0
    4k^2-36≥0
    4k^2≥36 (Then dividing through by 4 changes the direction of the sign from ≥ to ≤)
    therefore making it k^2≤9
    Hope this helps.
    Thanks for the worked example, very clear and I understand what you did. But why does dividing by a positive number (here 4) cause the change of sign? From that I thought you only swapped the sign when dividing or multiplying by a negative.
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    x^2 + 2kx + 9 ≥ 0 for all real values of x, therefore discriminant is ≤ 0 and not ≥ 0 like brooky2312 said because there will be either one root or no real roots as the curve needs to be above or touching the x-axis, hence the discriminant should be less than or equal to 0.

    Therefore b^2-4ac ≤ 0
    a=1, b=2k, c=9
    (2k)^2-4(1)(9) ≤ 0
    4k^2-36 ≤ 0
    4k^2 ≤ 36
    k^2 ≤ 9
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    Yes sorry my apologies, Sahil_ is correct. Hope it all helped
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    What year is this question from and on which exam board?
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    I don't know. I got a pack with review questions on all the different C1 topics, and that was one regarding inequalities.
 
 
 
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