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# C1 A-level maths: Inequalities problem watch

1. 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?
2. Is this the full question?
3. 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.
4. 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.
5. (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.
6. 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
7. Yes sorry my apologies, Sahil_ is correct. Hope it all helped
8. What year is this question from and on which exam board?
9. 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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