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Find the set of values of x for which C1

The title says it all. This is an inequality. I came up with x is smaller than -2 and bigger than 2.5 and 3 however the answer is x>3. But this excludes the -2 and the 2.5 Please help.
x2−x−6>0x^2-x-6>0 and 10−2x<510-2x<5

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Ahhh I see, I just needed to see them as graphs. x<2.5 I should've seen as a graph. Graphs are so helpful. I got the right answer now.
Btw, does anybody know whether you can, and if so, how to delete threads you've posted?
Original post by MathMeister
The title says it all. This is an inequality. I came up with x is smaller than -2 and bigger than 2.5 and 3 however the answer is x>3. But this excludes the -2 and the 2.5 Please help.
x2−x−6>0x^2-x-6>0 and 10−2x<510-2x<5


From your second equation you have
2x>5 means that you cannot have any values less than 2.5

Then from the first equation you have
x cannot be between -2 and 3
So this now tells you x>3
Original post by MathMeister

how to delete threads you've posted?


there is no need to delete - others may learn from your initial confusion
Original post by MathMeister
...


Remember that both inequalities have to be satisfied.

x2−x−6>0  ⟹  x2−3x+2x−6>0  ⟹  (x+2)(x−3)>0  ⟹  x∈(−∞,−2)∪(3,+∞)\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  ⟹  5<2x  ⟹  2.5<x10 - 2x < 5 \implies 5 < 2x \implies 2.5 < x

These are both satisfied by x>3x > 3

Spoiler

(edited 9 years ago)
I see the importance like many have said of learning/revising hard from the modular books instead of smashing exam papers as a excuse for learning. However, I have recently became even more confused, exactly because they have to satisfy each inequality. And just as I write this I figure it out. Gosh- the textbook is so vital in learning.
Original post by MathMeister
The title says it all. This is an inequality. I came up with x is smaller than -2 and bigger than 2.5 and 3 however the answer is x>3. But this excludes the -2 and the 2.5 Please help.
x2−x−6>0x^2-x-6>0 and 10−2x<510-2x<5

Here, I have assumed that both inequalities need to be satisfied to give a set of values for xx, is that right?

For the quadratic, just factorise as normal - gives crossing points as x=−2x=-2 or x=3 x=3. However, since the inequality says "greater than 0", you want all solutions that are greater than 0. It's easiest to graph a sketch. You'll realise that the bit that is below zero is the region: −2<x<3 -2 < x < 3. Hence the rest of it is the first solution, i.e. x<−2 x < -2 and x>3 x > 3.

This then gets combined with the other inequality, which works out to be x>2.5 x > 2.5. However, when combined, we know that x must be larger than 3. Hence, x can no longer be less than -2, or between 2.5 and 3. This leads to the correct solution, x>3 x > 3.
Original post by TenOfThem
there is no need to delete - others may learn from your initial confusion

That's correct - other people may find this helpful.

Subscribers can delete their own threads, or alternatively, go to Ask A Moderator and one of the mods can delete it for you.
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 x2−kx+(k+3)=0x^2-kx+(k+3)=0 has real roots. I thankfully knew real roots meant that it had the discriminant b squared was greater or equal to 4 ac. So I subbed the parts of the equation into the b squared = 4ac thing and solved for k. I then hoped/ and got a quadratic which meant I could get the answer, however I have a few quetions about the answer; -2<k<6. First of all I don't understand what the question is asking for. I understand the dicriminant, roots , recurring roots knowledge is important for a variety of questions, however I am not sure what the question was 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.
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 x2−kx+(k+3)=0x^2-kx+(k+3)=0 has real roots. I thankfully knew real roots meant that it had the discriminant b squared was greater or equal to 4 ac. So I subbed the parts of the equation into the b squared = 4ac thing and solved for k. I then hoped/ and got a quadratic which meant I could get the answer, however I have a few questions about the answer; -2<k<6. First of all I don't understand what the question is asking for. I understand the dicriminant, roots , recurring roots knowledge is important for a variety of questions, however I am not sure what the question was 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.



Hi,

I'll give it a crack now. It is asking for the set of values for k, that let b2−4ac>0. b^2 - 4ac > 0. In other words, what values of k can be put into the equation, that let the discriminant be more than zero?

Here, we need to set up an inequality; that is: b2−4ac>0 b^2 - 4ac > 0 .

By plugging in the values for a, b and c, and simplifying, we get: k2−4k−12>0 k^2 - 4k - 12 > 0

Solving gives the roots of the quadratic as k=−2 k = -2 or k=6 k = 6 .

Then, graph the curve. Remember, since the discriminant must be greater than zero for the equation to have real roots, the quadratic involving k must be greater than zero.

Hence, you need the part of the curve that is greater than zero, which gives the answer:

k<−2 k < -2 and k>6 k > 6

I've also attached my methodology to this problem. :smile:
Hope I've helped :smile:
(edited 9 years ago)
20140713_142009.jpg155.1KB
Original post by spotify95
Hi


I am very surprised to see a full solution from someone who officially represents TSR
Original post by TenOfThem
I am very surprised to see a full solution from someone who officially represents TSR

Whoopsie, am I not allowed to put a full solution? I'll edit if necessary.

The reason why I posted it was to explain how I got to that answer. It was more of a confirmation, as the FM that asked me the question did get somewhere along the lines of the correct answer (mentioned a quadratic in k). The only thing he got wrong was the inequality sign; he answered -2 < k < 6 when the answer is k < -2 and k > 6.

Graphing this type of question always helps - but don't worry I'll edit if necessary.
(edited 9 years ago)
What I don't understand is why it isn't greater or equal instead of just being greater-Since when b squared = 4ac, there are roots, two of them, which happen to be equal and ''recurring''- as well as b squared being larger than 4ac having real roots. So why isn't b^2 greater or equal to 4ac in the calculation, or is it something to do with the graph?
Also, in the answer book in the modular book and cd, the answer is 2<k<6. ???? Misprint/mistake? This is rather interesting. GeoGebra also agrees with you. I think its a misprint. possibly
Also, what's with the cross in your zero?
(edited 9 years ago)
Original post by MathMeister
What I don't understand is why it isn't greater or equal instead of just being greater-Since when b squared = 4ac, there are roots, two of them, which happen to be equal and ''recurring''- as well as b squared being larger than 4ac having real roots. So why isn't b^2 greater or equal to 4ac in the calculation, or is it something to do with the graph?
Also, in the answer book in the modular book and cd, the answer is 2<k<6. ???? Misprint/mistake? This is rather interesting. GeoGebra also agrees with you. I think its a misprint. possibly
Also, what's with the cross in your zero?

Hey there :smile:

I assumed you wanted real roots, and it is greater than when there are real roots.
You are correct in that it can be equal to zero; however then they are called "equal roots", as the root is the same (it is a touching point rather than a crossing point on the curve)

It probably is a misprint - as 2<k<6 would be right for if it was less than zero. However, less than zero there are no real roots, so that is not possible. I would say a mis print.

The cross in my zero? You mean the line? Probably how I write them, I can make it clearer for you from now on :smile:
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!
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!


No, that is completely wrong.

A real root is just that, it doesn't matter if it's repeated or two distinct roots.

Posted from TSR Mobile
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?



no, this thinking is not correct

a real root is Real rather than Complex

A quadratic can have

•

no Real roots

•

a repeated real root

•

2 real roots

Original post by MathMeister
There are many contradictions. Maybe TenOfThem may hopefully remove all confusion. So many different opinions...
If you are right then spotify95 was wrong in not putting greater than or equal.


Let ∆ represent the discriminant:

Real roots - ∆ => 0
Distinct real roots - ∆ >0
Repeated real roots - ∆=0
Complex roots - ∆ <0
(edited 9 years ago)
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 arithmatic posted (ty)
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 arithmatic posted (ty)


hmmmmm

The words Real Roots implies more than 1 real root

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