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1. (Original post by musicangel)
For 2b) how did you get 36(4+k) > 0 as I did 144-36k > 0
Also, why is k on the graph instead of x?
With regards to where the function is positive, is that everything to the right of the y axis?

Thank you for all of this!
2(b)

- you're right, that was a silly error on my part!

Because the equation with the x's will have two solutions whenever k^2 - 64 > 0, so you forget x exists for now and k's are the only things that matter to you know. You want to find what values of k makes the graph of y = k^2 - 64 > 0 (positive).

The graph of y against k is a quadratic with a minimum at k= 0, it's negative there and then only becomes positive at k =8 and then it stays positive. So k > 8 is one solution to your inequality.

But the graph is also positive whenever k < some value, can you tell me what value that is?
2. (Original post by musicangel)
Ok - that makes sense.
With qu 4, how could I do this as I get the discriminant to be 4-4q^2 > 0 ==> q^2 >0 Or have I done something wrong
4-4q2 > 0

is correct

but then
1-q2 >0
(1-q)(1+q) >0
diagram
-1<q<1

bedtime now
3. (Original post by TeeEm)
4-4q2 > 0

is correct

but then
1-q2 >0
(1-q)(1+q) >0
diagram
-1<q<1

bedtime now
I was wondering how you got 1-q2 > 0 and the rest of them

I really appreciate all of your help! Thank you!
4. (Original post by Zacken)
2(b)

- you're right, that was a silly error on my part!

Because the equation with the x's will have two solutions whenever k^2 - 64 > 0, so you forget x exists for now and k's are the only things that matter to you know. You want to find what values of k makes the graph of y = k^2 - 64 > 0 (positive).

The graph of y against k is a quadratic with a minimum at k= 0, it's negative there and then only becomes positive at k =8 and then it stays positive. So k > 8 is one solution to your inequality.

But the graph is also positive whenever k < some value, can you tell me what value that is?
Is it when k < -8?
This all makes much more sense now! Thank you so much for this!
5. (Original post by musicangel)
I was wondering how you got 1-q2 > 0 and the rest of them

I really appreciate all of your help! Thank you!
I divided the inequality by 4
then I factorized
6. (Original post by musicangel)
Is it when k < -8?
This all makes much more sense now! Thank you so much for this!
Correct! Good job.

So, now you can say that for the equation y = x^2 + kx + 16, the equation will have two roots if k is any number above 8 or below - 8. So y = x^2 + 9x + 16 has two roots, x^2 - 10x + 16 has two roots but x^2 + 5x + 16 doesn't.

Cool 'innit?

7. (Original post by musicangel)
I was wondering how you got 1-q2 > 0 and the rest of them

I really appreciate all of your help! Thank you!

since 0 divided by anything is still 0.
8. (Original post by Zacken)
Correct! Good job.

So, now you can say that for the equation y = x^2 + kx + 16, the equation will have two roots if k is any number above 8 or below - 8. So y = x^2 + 9x + 16 has two roots, x^2 - 10x + 16 has two roots but x^2 + 5x + 16 doesn't.

Cool 'innit?

Thank you!!
9. (Original post by TeeEm)
I divided the inequality by 4
then I factorized
Oh - ok. Thank you!!
10. (Original post by musicangel)
Oh - ok. Thank you!!
all the best

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