Find the set of values of x for which C1

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So the answer is in fact

k\le2, k\ge6

GeoGebra agrees with me.

(edited 9 years ago)

Reply 21

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Original post by MathMeister

So the answer is in fact

x\le2, x\ge6

You mean k, x doesn't come into it.

Reply 22

spotify95

Original post by MathMeister

So, to conclude this question, a real root is when there are 2 x-intercepts. A real root cannot be an equal root (vice versa) as an equal root does not cross the x-axis. Am I correct in thinking this?
Also, (sorry) another question I have came up with, is why the graphical representation of the expression tells you what values of k for which the equation has real roots. I see the inequality, but fail to see how this relates to k. I don't see.... ahhh wait... I figured it out. This has been the best maths problem so far. Wopee! I completely and utterly understand it all now. I understand the question and answer entirely. Please may somebody answer the underlined question however, still.
Thank you!

Sorry about the confusion here. A real root can be an equal root; when this happens, the graph touches the x-axis, at one point, rather than crossing it at two different points.

The reason why I didn't tell you to put the greater than or equal to sign in, is that when

b^2 - 4ac

equals zero, there are equal roots, which with the inequalities, rarely happens.

Reply 23

MathMeister

Original post by Khallil

Remember that both inequalities have to be satisfied.

\begin{aligned} x^2 - x - 6 > 0 & \implies x^2 - 3x + 2x - 6 > 0 \\ & \implies (x+2)(x-3) > 0 \\ & \implies x \in \left( -\infty, -2 \right) \cup \left( 3, +\infty \right) \end{aligned}

10 - 2x < 5 \implies 5 < 2x \implies 2.5 < x

These are both satisfied by

x > 3

Spoiler

I'm sorry Khalil but this maths is gibberish to me. I don't get the infinity thing. :/

Reply 24

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Original post by MathMeister

I'm sorry Khalil but this maths is gibberish to me. I don't get the infinity thing. :/

That means a set of numbers

The closed interval (-infty, 2] means that x can take any value between minus infinity and two (counting two because it's closed on the right side).

It's basically just another way of saying k <= 2

Reply 25

MathMeister

Original post by spotify95

b^2 - 4ac

equals zero, there are equal roots, which with the inequalities, rarely happens.

What GeoGebra gives is this... And yes... arithmeticae, I meant k, not x. :smile:

I suppose it does happen in this case?

Graph.png26.2KB

Reply 26

spotify95

Original post by MathMeister

Prsom*

... to all-In this case then surely spotify95 was wrong in not putting greater than or equal sign in his working. If that's the case then maybe he did not know this. Somebody help?
Also, I will try to understand the delta thing arithmetic posted (ty)

Apologies for this, MathMeister. :frown:

What I meant, was that the answer was going to have distinct, real roots, and not equal, real roots. It is possible to have equal, real roots - however, in the inequality we set up in that example (with the discriminant) I assumed distinct, real roots. Think of it this way.
If

b^2 - 4ac \ge 0

, then the equation for k could have equal roots as well as distinct roots. The equal roots thing very rarely happens with inequalities.

However, now I do agree that the answer should have been greater than or equal to, rather than just greater than etc. So apologies if I confused anyone!

Reply 27

spotify95

Original post by MathMeister

What GeoGebra gives is this... And yes... arithmeticae, I meant k, not x. :smile:

I suppose it does happen in this case?

Yes, I think that's correct. :smile:

In this case it was greater than or equal to, as it didn't matter whether the roots were distinct or equal.

(edited 9 years ago)

Reply 28

user2020user

Original post by Arithmeticae

...

My hero, qtsie tootsie :holmes:

Reply 29

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Original post by Khallil

My hero, qtsie tootsie :holmes:

Has Zen hacked into your account or something? :lol:

Reply 30

MathsNerd1

Original post by spotify95

Spoiler

Hi there, just so you know for next time, we're not allowed to post full solutions as this doesn't help the user out, instead we provide hints and guide the user through the question to make sure they know how to do it on their own.

I understand this time you've helped out the user by continuing to provide assistance, but please don't post full solutions again otherwise I might have to give you a warning, I hope you understand why. :smile:

Reply 31

Zen-Ali

Original post by Arithmeticae

...

Nah, we all speak like this now (Z, K and I). :lol:

Reply 32

spotify95

Original post by MathsNerd1

Okay thanks - I was wondering whether posting full solutions was allowed on here or not (I thought it was ok, but got a message from another member saying it might not have been) and you've now confirmed it isn't. Thanks for letting me know :smile:

Reply 33

spotify95

Original post by MathMeister

I actually done most of the thinking for myself. I just needed help really with the roots thing. I don't see how anybody can figure out that real roots means equal and different. Everybody learns that. It's not something you can figure out. I also done the full working, up to and including the inequalities and graph. Just backin' y'all up.

Ok thanks - I saw that, in your question, you went along with your method of finding the quadratic etc. :smile:

My working out was basically just showing you that you had indeed gone along the correct pathway, just that we'd got it a little bit wrong with the inequalities :smile:

Glad I could help you :smile:

If you have any other questions on which you are totally stuck on, I might drop in a few hints, but I'm not risking getting a nice collection of cards :nah:

Reply 34

MathsNerd1

Original post by spotify95

Yeah, like TenOfThem said, we don't allow any full solutions, thanks for understanding. :smile:

Reply 35

spotify95

Original post by MathsNerd1

Yeah, like TenOfThem said, we don't allow any full solutions, thanks for understanding. :smile:

No problem, thanks for letting me know. I'd rather you tell me that they aren't allowed, than get a card of some sort for it :yep:

I'll stick to just posting hints then, to get the user to work out the right answer. :smile:

Reply 36

Mr Gradgrind

Original post by MathMeister

Alas, I have another question in dire need of answering. Please may somebody help. 4a) Find the set of values for k for which the equation

x^2-kx+(k+3)=0

has real roots [...] First of all I don't understand what the question is asking for [...] It's asking for what part of the equation has real roots, but the whole equation- the equation in itself has real roots, which is why I am confused with what they are asking me for. Please may somebody help?! Ahh wait, it's asking for the range of values for k in which the equation would satisfy the b squared = 4ac. But I do not understand. Brb. Going out. Please help explain the question and answer.

I'll have a go at explaining the question.

Think of it like this:

x^2-kx+(k+3)=0

can become different equations depending on what k equals.

e.g. if k=0, it becomes

x^2+3=0

This has no real roots (i.e. no real solutions – you can think of the word "root" as meaning "solution" in this context), which can easily be seen by rearranging it to

x^2=-3

Imagine setting k equal to different numbers, seeing what the equation becomes, and deciding if it has any real roots.

If k=1 the equation becomes

x^2-x+4=0

(no real roots)
If k=2 the equation becomes

x^2-2x+5=0

(no real roots)
If k=6 the equation becomes

x^2-6x+9=0

(aha, a real root i.e. 3)
If k=7 the equation becomes

x^2-7x+10=0

(real roots i.e. 2 or 5)

So some values of k turn

x^2-kx+(k+3)=0

into an equation which has real roots and other values of k turn it into an equation which doesn't have real roots.

The question is asking you to find the complete set of values of k which turn it into an equation with real roots (which, as posters have pointed out, is under -2 or over 6 inclusive).