Room for Leaky Nun and Abcd - 2017-09-04 (page 1 of 2)

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6:01 AM

Hello @LeakyNun

0

Q: Finding the range of values of a parameter to satisfy the given condition.

Find the range of values of a for which the one of the roots of the equation: $(2a+1)x^2-ax+a-2=0$ is greater than and the other smaller than unity (i.e. $1$). Now, $\Delta>0$ because there are two distinct roots. Thus, $a \in \left(\dfrac{6-2\sqrt23}{7},\dfrac{6+2\sqrt23}{7}\righ...

algebra-precalculus quadratics

Please see @LeakyNun ^

@Abcd seen

@LeakyNun Any idea how to proceed?

Are you thinking @LeakyNun?

0

A: Finding the range of values of a parameter to satisfy the given condition.

Let $2a+1>0$, i.e. $a>-\dfrac12$. Then, $f(1)<0$: $(2a+1)-a+a-2=2a-3<0$, i.e. $a<\dfrac32$. We obtain $-\dfrac12<a<\dfrac32$. Let $2a+1<0$, i.e. $a<-\dfrac12$. Then, $f(1)>0$: $(2a+1)-a+a-2=2a-3>0$, i.e. $a>\dfrac32$. Contradiction. In conclusion, $-\dfrac12<a<\dfrac32$.

@LeakyNun Answer is $(-1/2, 1/2 not (3/2))$

right, miscalculations

edited.

6:10 AM

@LeakyNun thank you :)

do you know why I answered so quickly?

12 hours ago, by Abcd

`prove that the set of values of k for which $18x^2 -6(2k+1) +k(k+1)= 0$ may have one root less than k and othe root greater than k are given by $(0,\frac{5}{7})$

@LeakyNun I don't.

because we've already done it here.

@LeakyNun that's a different question.1 Is that the question I gave up on?

@Abcd the concept is the same.

no it isn't

6:12 AM

OH.

I have to be more alert and attentive and try to remember things @LeakyNun. I will try.

alright

What can we interpret from our current $\Delta$ @LeakyNun? We can't say it's positive or negative.

@Abcd ?

???

@LeakyNun Please tell about the graph of this. We used f(k) < 0 in earlier question.

What did we use here and why?

of what?

where?

????

6:16 AM

Just a minute.

12 hours ago, by Leaky Nun

it hits the x-axis once before $k$ and once after $k$

@LeakyNun I liked this reasoning^

Please give a similar reason for our current question.

...

it hits the x-axis once before 1 and once after 1?

12 hours ago, by Leaky Nun

since $f(x)$ is an upward parabola

@LeakyNun Is it upward parabola or downward? How?

or did you check for both conditions?

I did

@LeakyNun Okay, this was the part I was confused about. It was obvious in the previous question that f(x) will be upward parabola. But here it wasn't that's why I was stuck though I had tried to use the same concept.

@LeakyNun Somebody down-voted your answer :(

@Abcd I don't care

6:22 AM

Was my set of $a$((6+2root23)/7) useless or did you intersect with that too @LeakyNun ?

@Abcd do you see that in my answer?

@LeakyNun No. Okay.

$\Delta = -ve $ and coefficent of $x^2= -ve$ $\implies$ downward parabola i.e f(x)<0 always

$\Delta = -ve$ and coefficient of $x^2 =+ve$ $\implies$ upward parabola$ i.e f(x)>0 always

right

ok

$\Delta = +ve $ and coefficent of $x^2= -ve$ $\implies$ downward parabola

$\Delta = +ve $ and coefficent of $x^2= +ve$ $\implies$ upward parabola

Correct @LeakyNun? Need to note down all fur point

four*

no

the direction of the parabola only depends on the coefficient of x^2

6:34 AM

@LeakyNun then?

positive = upward

What's wrong?

@LeakyNun That's what I have written

2 mins ago, by Abcd

$\Delta = +ve $ and coefficent of $x^2= +ve$ $\implies$ upward parabola

3 mins ago, by Abcd

$\Delta = +ve $ and coefficent of $x^2= -ve$ $\implies$ downward parabola

oops

@LeakyNun We got two sets, $a \epsilon$$\left(-1/2,1/2\right)$ and $a\epsilon$\phi$. Do we do intersection or union? Why?

I know that you did union.

because it is or

6:40 AM

ok

Find the value of $a$ for which the inequation $x^2+ax+a^2+6a<0$ is satisfied for $x \epsilon (1,2)$ Please give hint @LeakyNun

parabola is upward

draw a graph

@LeakyNun Yes I interpreted that.

Therefore a belongs to $\phi$ right @LeakyNun?

what?

why

@LeakyNun Because our function can't be negative.

And inequation says it should be negative..

why not?

6:53 AM

@LeakyNun Because it's upward.

Oh. I didn't check for $D$. Yes, it can have some portion in negative region too.

@LeakyNun how?

get a paper and a pen

draw the axis

done that

@LeakyNun What next? After marking points 1 and 2

draw an upward parabola that is negative at (1,2)

@LeakyNun does it have to intersect y axis?

When does quadratic parabola intersect y axis @LeakyNun

chat.stackexchange.com/transcript/message/39802733#39802733 Never right @LeakyNun?

7:16 AM

when x=0...

yeah, depends on the intercept

please tell the approximate graph @LeakyNun. I am somewhat confused.

show me what you have drawn

@LeakyNun molbile's battery dead. I don't have camera.

mobile*

draw it online

@LeakyNun Wherew?

where*

7:21 AM

google it

learn to solve your own problem

user image

I have drawn two possibilities @LeakyNun

Please consider them proper curves @LeakyNun

I couldn't make proper curves.

I am doubtful about the triangular-looking one. Is it incorrect @LeakyNun?

Are you dissatisfied with my graphs @LeakyNun?

just because I didn't reply doesn't mean I'm dissatisfied.

I'm just busy.

your graphs are good.

but you've mistaken 1,2 for 2,3

@LeakyNun Oh yes.

now what can you say about f(1) and f(2)?

@LeakyNun is the triangular looking graph (which in reality is a curve ) correct?

7:35 AM

yes

@LeakyNun Both are negative

go on

?

continue

@LeakyNun I know the meaning of go on. I don't know what to say next about f(1 and 2)

7:43 AM

you just said they are negative

so substitute

7:59 AM

$a^2+7a+1\le 0$ and $a^2+8a+4\le 0$ @LeakyNun

What to do next @LeakyNun?

solve them?

@LeakyNun How?

quadratic inequalities...

find the two roots

@LeakyNun Found

2 hours later…

9:39 AM

@LeakyNun Aren't the factors of second inequation $(a-(-4\pm 2\sqrt3))$?

should be

@LeakyNun Solution doesn't agree. It says: $(a-(4-2\sqrt3))$ and $(a+(-4+2\sqrt3))$

Is solution mistaken @LeakyNun?

yes

$a^2+8a+4 = a^2+8a+16-12 = (a+4)^2-(2\sqrt3)^2 = (a+4+2\sqrt3)(a+4-2\sqrt3)$

@LeakyNun Yes, I used factor theorem.

How to get for sign (= resembling inverted A) using mathjax @LeakyNun?

$\forall$ \forall

9:51 AM

Also, @LeakyNun Is there anyone else too in the room? I am wondering how the message got 2 stars. Creepy.

@LeakyNun ok

@Abcd lol

10:31 AM

$x^2+x+k=0$

How can we say $f(k)$ > 0 @LeakyNun?

we can't.

@LeakyNun Condition: roots of the equation exceed k.

Still we can't @LeakyNun?

Give me the whole damn question.

@LeakyNun If the roots of the equation $x^2+x+k=0$ exceed k, then find k.

I found two conditions @LeakyNun

D>=0

and

-b/2a (=vertex of parabola's x coordinate)> k

you also found that f(k)>0.

10:39 AM

@LeakyNun I didn't /

you did

draw a graph.

@LeakyNun Oh yes :)

Is that enough @LeakyNun?

three conditions are enough.

@LeakyNun Why? How?

@Abcd What? Where?

10:44 AM

-9<4, -9<3, -9<2 aren't enough for instance. The best ones are ones with equality like -9<=-9 @LeakyNun

??

@LeakyNun k<=1/4, k<-1/2, k>-2, k<0

???

@LeakyNun Nothing, I got the right answer.

11:05 AM

FInd the value of a for which $(6-a)x-ax^2-2>0$ for at least one positive real x.

@LeakyNun

$-ax^2+(6-a)x-2>0$

divide into two cases and draw the graphs

@LeakyNun not possible because I don't know whether $\Delta$ is > or < or = zero

are you thinking @LeakyNun ?

you don't need to know

@LeakyNun I will obtain around 8 graphs then :/

@LeakyNun Solution does for $f(x)= ax^2+(a-6)x+2<0$

@Abcd you don't need to draw every possible case

11:22 AM

wrong question I feel

It's a vague question @LeakyNun, isn't it?

mathematics is abstract.

11:39 AM

Given that $ax^2+bx+c= 0$ has complex roots , how is it that $(4a-c)^2>=0$? @LeakyNun

Are you thinking/busy @LeakyNun?

who told you that?

@LeakyNun Solution says this.

The question is:

If the roots of the quadratic equation $ax^2+bx+c=0$ are complex, then prove that $16a^2+c^2>2b^2$

I swear if it is another detail you don't say

Solution says, "We know that (4a-c)^2>=0" @LeakyNun

of course

it's the square of a real number

of course it can't be negative

11:47 AM

@LeakyNun Okay.

12:15 PM

@LeakyNun can you give permalink to the y = numerator/denominator (then we multiplied both sides by denominator) question we did yesterday?

I can't find it.

We had introduced y @LeakyNun

$y = (ax^2+3x+4)/(a+3x-4x^2)$

You said it's simple algebra.

16 hours ago, by Leaky Nun

Let $y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$ (introducing yet another variable :D)

@LeakyNun It's not given that for all y there'a root for x/

@Abcd then it wouldn't take all values

@LeakyNun Please given an example. I feel that it can still take all values.

@Abcd of what?

tell me what it means by "take all value"

12:24 PM

@LeakyNun How?

@LeakyNun It can be any number in the range $(-\infty, +\infty)$. Am I right?

@Abcd does $x^2$ take all value?

@LeakyNun yes It does.

@Abcd no it doesn't

@LeakyNun please elaborate on the two points/.

1 min ago, by Abcd

@LeakyNun It can be any number in the range $(-\infty, +\infty)$. Am I right?

please verify this too.

$x^2$ doesn't take all values.

I can't verify a vague statement.

For example, $x^2$ doesn't take the value $-1$

12:27 PM

Oh yes

@LeakyNun Range of $x^2 = [0,+\infty)$

so what does it mean for a function to take all values?

@LeakyNun It can be any real number in the range +infty to -infty

what does it mean?

symbolically?

@LeakyNun $y \epsilon Z$ where Z is set of complex numbers

...

$x^2$ also $\in \Bbb R$.

????

12:31 PM

@LeakyNun You tell please

$x^2$ also $\in \Bbb C$

@Abcd then explain in words

what does it mean?

@LeakyNun y can be ANY number in the world

why doesn't $x^2$ take all values?

@LeakyNun x^2 is always greater than zero

@Abcd for an arbitrary function $f(x)$ how can we tell if it takes all values?

12:35 PM

y can take any value only when $f(a+3x-4x^2) \ne 0$

@LeakyNun You tell please.

I said arbitrary function.

@Abcd You think please.

I don't know.

@LeakyNun Please, I haven't studied functions yet.

it means for any $y$ there is an $x$ such that $f(x)=y$.

@LeakyNun Okay, then?

17 mins ago, by Abcd

@LeakyNun It's not given that for all y there'a root for x/

then it is.

12:42 PM

In simpler words, you mean $y=0$ , there's an $x$ @LeakyNun

for y=0*

4 mins ago, by Leaky Nun

it means for any $y$ there is an $x$ such that $f(x)=y$.

This is exactly what I mean. No more no less.

Can I conclude from that for y = 0 , there's an x which implies that there's a root! @LeakyNun

?????????????

@LeakyNun Roots are places where y = 0

therefore for y = 0, there's a possibility of root(s)

I said $y$ is a constant.

12:48 PM

@LeakyNun then how does it have roots?

how does $x^2=3$ have roots?

1:00 PM

Please explain graphically if possible @LeakyNun

Not possible @LeakyNun?

Draw a graph yourself.

Draw one for a function that takes all values

and one for a function that doesn't

@LeakyNun a graph of what?

@LeakyNun How do I give you a link to desmos

save it

@LeakyNun desmos.com/calculator/bmglu27tf0

which function takes all values?

1:07 PM

@LeakyNun first, second doesn't take value for x=0

that isn't what it means

it isn't about x.

it is about y.

@LeakyNun y is not defined at x=0 so it doesn't take all values

...

draw another graph that doesn't take all values

desmos.com/calculator/nytwhxlzpy @LeakyNun

it does take all values.

1:09 PM

tan x doesn't take value for pi/2

It isn't about what x can be.

It's about what y can be.

@LeakyNun desmos.com/calculator/lz0lav4bjt

Takes only one value, not all values.

bingo.

@LeakyNun then?

so why doesn't it take all values?

1:13 PM

@LeakyNun Because it is defined that way

eh

which value does it not take, for instance?

@LeakyNun 3,

why not?

@LeakyNun It can only be 2

does $x^2$ take all values?

1:17 PM

@LeakyNun No, it doesn't take values like -1, -2 etc

so how do we check if $f(x)$ take all values?

@LeakyNun I think you mean "give" instead of "take"

attain.

@LeakyNun based on the definition of the function.

take doesn't mean accept here.

1:19 PM

@LeakyNun yes, better.

17 hours ago, by Abcd

Find the value of a for which $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ takes all values for all real values of x

change "takes" to "attains"

@LeakyNun okay then as I said before f(x) can be anything between -infinity and plus infinity, can't it?

@Abcd it can always be.

the thing is, f(x) needs to be able to actually attain all values there

@LeakyNun then/

?*

Can you justify your statement regarding roots?

does $x^2=3$ have a root?

so $f(x)=y$ has a root for every $y$.

1:25 PM

@LeakyNun what does that mean graphically?

it means the line $y=y_0$ will hit the graph for every $y_0$

@LeakyNun WHAT is $y_0$ now?'

an arbitrary constant.

I am frustrated now.

so am I. so what?

1:33 PM

@LeakyNun Can I ask someone else for explanation?

can't you come up with anything by yourself?

If $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ attains all values, then $\dfrac{ax^2+3x-4}{a+3x-4x^2}=b$ has a solution for every $b \in \Bbb R$, i.e. the line $y=b$ hits the graph $y=\dfrac{ax^2+3x-4}{a+3x-4x^2}$ for every $b \in \Bbb R$.

2:00 PM

@LeakyNun What is wrong in stating that the range of f(x)= y = $\dfrac{ax^2+3x-4}{a+3x-4x^2}$ is $(-\infty,+\infty)$. It means exactly the same as what you have typed.

@Abcd who said it's wrong?

@LeakyNun Does it mean exactly the same?

I need your verification/

you swapped the two ends of the interval

but it's the same.

@LeakyNun You should have said it this way :"( :"( :"(

2 hours ago, by Leaky Nun

I can't verify a vague statement.

2 hours ago, by Abcd

1 min ago, by Abcd

@LeakyNun It can be any number in the range $(-\infty, +\infty)$. Am I right?

I said the same thing earlier @LeakyNun

it isn't the same

the value of $x^2$ can also be any number in the range $(-\infty,+\infty)$

whatever

2:05 PM

:'(.

Wasted 2 precious hours. :/

I am such a freaking idiotic fool!

@LeakyNun Star that^ because of it's truth/

I star what I want.

whatever

3:05 PM

if $\dfrac{x^2+ax+3}{x^2+x+a}$ takes all real values for possible real values of x then prove that $4a^3+39<0$
Using the same steps as before I reached
$4a^3\le 36a-48$
(don't worry, I am sure of my calculations since I calculated twice :)). How do I continue from here @LeakyNun?

Are you thinking @LeakyNun?

Are you busy @LeakyNun?

3:27 PM

I was taking a shower.

ok

@LeakyNun What to do next?

I also got $4a^3+39\le9(4a-1)$

@LeakyNun Stuck.

$x^2y+xy+ay=x^2+ax+3$

$(y-1)x^2+(y-a)x+(ay-3)=0$

$\Delta_x = (y-a)^2 - 4(y-1)(ay-3) \ge 0$

$y^2-2ay+a^2-4ay^2+(4a+12)y+12 \ge 0$

@LeakyNun Trust my calculation please

7 mins ago, by Abcd

I also got $4a^3+39\le9(4a-1)$

$(1-4a)y^2+(2a+12)y+(a^2+12)\ge0$

$1-4a>0$ and $\Delta_y = (2a+12)^2-4(1-4a)(a^2+12) \le 0$

@LeakyNun How is $1-4a>0$. Just tell me that then I will get my answer.

3:38 PM

@Abcd it needs to curve upward

$a<\dfrac14$ and $16(a^3+15a+16)\le0$

@LeakyNun Why?

@Abcd because it needs to be above the x-axis

@LeakyNun $\Delta_y \le 0$??

3 mins ago, by Leaky Nun

$(1-4a)y^2+(2a+12)y+(a^2+12)\ge0$

This needs to be true for all $y$

which means the parabola is always above the x-axis

if it curved downward, it would eventually touch the x-axis and then become negative

so it needs to curve upward

Understood @LeakyNun. thanks.

3:41 PM

and the discriminant cannot be positive either, or else it would cut the x-axis at two distinct points and the region between would be negative

4:09 PM

Find the value of a so that two of the roots of the equation:
$(a-1)(x^2+x+1)^2= (a+1)(x^4+x^2+1)$
are real and distinct @LeakyNun.

no idea

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