Question

solve

Answer 1

its ax^2- (a-1) x +3=0

Answer 2

is it (a+1)x ? because (a+1) doesn't make much sense to me

Answer 3

yes
its (a+1) x
i wrote this equation again
plse have a look

Answer 4

okay just give me few minutes to solve

Answer 5

dude you just told me it's (a+1)x and your equation says (a-1)x which one of them ?

Answer 6

equation is
ax^2-(a+1)x+3=0

Answer 7

plse i need it urgent

Answer 8

@amistre64
i need ur help

Answer 9

im in no big rush ... but i would just solve for "x" using the quadratic formula

Answer 10

\[x = \frac{-(a+1)\pm\sqrt{(a+1)^2-4(3a)}}{2a}\]
\[1<\frac{-(a+1)\pm\sqrt{(a+1)^2-4(3a)}}{2a}<2\]

Answer 11

the - in front might be bad since i think you want this at -(a+1)x to start with

Answer 12

let (a+1)/2a = 1.5 to estabish an axis of symmetry

Answer 13

then the sqrt/2a part just has to be less then .5

Answer 14

to get the function has root between 1 and 2 so f(1) <0 and f(2) >0 or vice versa. solve for a.

Answer 15

\[\frac{a+1}{2a}=\frac{1+2}{2}\]\[0\le\frac{\sqrt{(a+1)^2-4(3a)}}{2a}<\frac12\]

Answer 16

so far I just tried to find the values for a that centres x in the stated domain
solve for x=2 you see a=-1/2
I have to study the behaviour of the function to see the other endpoint since f(1) does not equal 0

Answer 17

if the end points can equal 1 and 2, then that other part is <= 1/2

Answer 18

f(1) =4 we don't have a there, so it is always positive. Therefore, f(2) must be negative.

Answer 19

I believe @Loser66 has the most elegant solution to this.

Answer 20

@asnaseer thnks

Answer 21

yw :)

Answer 22

vat is the value of a

Answer 23

my attempts keep coming up as no solution ..

Answer 24

\[(a+1)^2-4(3a)\ge0\]
\[(a^2+2a+1-12a)\ge0\]
\[(a^2-10+1)\ge0\]\[<---(5-2\sqrt6)~~~~~~(5+2\sqrt6)--->\]
in order to have any real roots

Answer 25

but then for those values, i cant get a suitable "a" to stay within 1 and 2

Answer 26

so far I think the maximum value that a can take is -1/2
still I have to evaluate the minimum, I'm trying to find the closest point to the point (1,0) but probalbly I'll use optimization, is that okay with you?

Answer 27

\[1<\frac{a+1}{2a}<2\]
\[2<\frac{a+1}{a}<4\]
\[2<1+\frac{1}{a}<4\]
\[1<\frac{1}{a}<3\]\[\frac13<a<1\]

Answer 28

unless a can be some complex values ... i dont see how a can be less than 1/3 and greater then 1; at the same time being between 1.3 and 1

Answer 29

@Loser66 has already indicated how to obtain the solution.

Answer 30

dude @asnaseer can you give a set of numbers please -_-

Answer 31

I believe the aim of this site is to help teach - not just provide answers. @Loser66 has given sufficient information to @msingh for him to be able to solve it.

Answer 32

there are no real roots for the set: a = \((5-2\sqrt6,5+2\sqrt6)\) but if we have real roots, then they are not between 1 and 2

Answer 33

I dont think it's sufficient Mr @asnaseer ,if you can find an answer I will be so glad to hear that

Answer 34

is the solution set for only one root being between 1 and 2?

Answer 35

to quote from what @Loser66 said above:
f(1) =4 we don't have a there, so it is always positive. Therefore, f(2) must be negative.

therefore you just need to find the value for 'a' that makes f(2) < 0

Answer 36

@Loser66
u told
f(1) <0
4<0
and
f(2) >0
a>-1/2
acc. to u , i got this

Answer 37

you have the inequalities the wrong way around @msingh

Answer 38

i am confused how to start it with can u guide me

Answer 39

@Loser66 showed that f(1)=4
therefore we know that f(1) > 0 for any value of 'a'
agreed?

Answer 40

it means 4>0

Answer 41

yes

Answer 42

now, for there to be a root between 1 and 2, the sign of f(x) must change between x=1 and x=2. we just showed that f(1) > 0, therefore, for there to be a root between x=1 and x=2, f(2) must be negative.

agreed?

Answer 43

yes

Answer 44

therefore, you just need to solve for 'a' such that:\[f(2)\lt0\]if the root is supposed to INCLUDE the end points, then you need to solve for 'a' such that:\[f(2)\le0\]

Answer 45

i got f(2) = -1/2

Answer 46

how?

Answer 47

srry
f(2) = 2a+1

Answer 48

yes - that is better

Answer 49

therefore solve:\[2a+1\lt0\]

Answer 50

a>-1/2

Answer 51

you have the inequality the wrong way around again

Answer 52

a<-1/2

Answer 53

yes :)

Answer 54

so ANY value of 'a' that is less than minus one half will satisfy the conditions set in the problem.

Answer 55

i assume im reading the question wrong ...

\[0=ax^2-(a+1)x+3~:~s.t~~~1<(x_1~and~x_2)<2\]
or
\[0=ax^2-(a+1)x+3~:~s.t~~~1<(x_1~or~x_2)<2\]

Answer 56

I read it as the 2nd form

Answer 57

ahh, the second one is doable; i cant see any solution to the first one

Answer 58

hmm, that's the secret then, he doesn't need both roots inside the interval..
that makes me feel a little rude after knowing that :( , I'm sorry anyways @asnaseer

Answer 59

np - I didn't take any of your responses as being rude! I assumed you were just asking for clarification. :)

Answer 60

that's cool then :)

Answer 61

yw my friend! :)