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Set of values (C1)

Hi all, just a quick question about a C1 maths past paper question. I'm not entirely sure how to go about answering this question, anyway here it is.
Find the set of values of k k for which the graph of y=x2+2kx+5 y=x^2 + 2kx + 5 does not intersect the x x axis.
Any help on how to answer this question or the method I should use will be much appreciated. :smile:
When a parabola does not intersect the x axis, the discriminant of the quadratic will be negative.
Reply 2
Original post by biologeek
Hi all, just a quick question about a C1 maths past paper question. I'm not entirely sure how to go about answering this question, anyway here it is.
Find the set of values of k k for which the graph of y=x2+2kx+5 y=x^2 + 2kx + 5 does not intersect the x x axis.
Any help on how to answer this question or the method I should use will be much appreciated. :smile:


I believe this means that the curve does not cross the x-axis, so therefore
b24ac<0 b^2-4ac<0
Input the values which give us
4k220<0 4k^2-20<0
Which means
4k2<20[br]k2<5[br]kislessthan5 4k^2<20[br]k^2<5[br]k is less than \sqrt5

Was this the right answer?
Original post by MashB
I believe this means that the curve does not cross the x-axis, so therefore
b24ac<0 b^2-4ac<0
Input the values which give us
4k220<0 4k^2-20<0
Which means
4k2<20[br]k2<5[br]kislessthan5 4k^2<20[br]k^2<5[br]k is less than \sqrt5

Was this the right answer?


No. That isn't how you solve quadratic inequalities.
Reply 4
Original post by Mr M
No. That isn't how you solve quadratic inequalities.


How do you?


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Original post by MashB
How do you?


The easiest way is to graph the inequality.

k2<5k^2 <5 means k25<0k^2-5<0

Sketch y=k25y=k^2-5

Check which values of k produce negative values of y.
Reply 6
Original post by Mr M
The easiest way is to graph the inequality.

k2<5k^2 <5 means k25<0k^2-5<0

Sketch y=k25y=k^2-5

Check which values of k produce negative values of y.


(Just checking) 5<k<5 -\sqrt{5} < k < \sqrt{5} ?
Original post by McLH
(Just checking) 5<k<5 -\sqrt{5} < k < \sqrt{5} ?


Yes.
Reply 8
Original post by Mr M
Yes.


So in a way I was on the right track, just didnt finish it...
k<±5 k<\pm\sqrt5
k=5k=\sqrt5 and k=5 k=-\sqrt5
so plot them on the graph and then that becomes
5<k<5-\sqrt5<k<\sqrt5
Reply 9
Original post by MashB
So in a way I was on the right track, just didnt finish it...
k<±5 k<\pm\sqrt5


That does not lead to this


5<k<5-\sqrt5<k<\sqrt5


The first line is not possible
Reply 10
Original post by TenOfThem
That does not lead to this



The first line is not possible


Could you explain it, I'm confusing myself here now.


This was posted from The Student Room's iPhone/iPad App
Original post by MashB
Could you explain it, I'm confusing myself here now.



You go from

k2<5k^2<5

to

5<k<5-\sqrt{5}<k<\sqrt{5}

there are no intermediate lines
Reply 12
Original post by TenOfThem
You go from

k2<5k^2<5

to

5<k<5-\sqrt{5}<k<\sqrt{5}

there are no intermediate lines


I meant as in explain the answer to op's question.


This was posted from The Student Room's iPhone/iPad App
Original post by MashB
I meant as in explain the answer to op's question.


the answer is in the thread

not sure what else you want
Reply 14
Original post by MashB
Could you explain it, I'm confusing myself here now.


This was posted from The Student Room's iPhone/iPad App

I know I shouldn't post full solutions, but for benefit:

b24ac<0 b^2 -4ac < 0

(2k)2(415)<0 (2k)^2 - (4 * 1 * 5) < 0

4k220<0 4k^2 - 20 < 0

Divide by 4
k25<0 k^2 - 5 < 0

k2<5 k^2 < 5

k=±5 k = \pm\sqrt{5}

Plot :smile:
(edited 11 years ago)
Original post by McLH



k=±5 k = \pm\sqrt{5}


This line is incorrect
Reply 16
Original post by TenOfThem
This line is incorrect


As a solution, but to show where the roots are that I'm plotting before i do a sketch of the graph is it? If so, what should I write?
Original post by McLH
As a solution, but to show where the roots are that I'm plotting before i do a sketch of the graph is it? If so, what should I write?


The answer that I gave in post 12

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