Help me solve a quadratic inequality!
August 6th, 2016
i) 2 equal roots
ii) 2 real roots, one of which is the reciprocal of the other,
iii) 2 real roots, one of which is twice the other.
b) Find the value of k for which the equation 4(x+1)(x-4) = k has roots which differ by 8
Thanks!
do your homework. if you don’t understand it ask your teacher its that simple
What dont you understand about it? What have you come up with so far?
This might be helpful…i cant help much i cant remember math right now its 4am here. lolz
http://www.purplemath.com/modules/ineqquad.htm
do your homework. if you don't understand it ask your teacher its that simple
He is only asking some questions, you dont have to respond to his thread if you are not willing to help.
This topic has been posted in the wrong section. Rule #3.3: Topics must be submitted to the relevant forums. Please read the forum descriptions before posting.
Moving to proper forum.
1) k=3
2) Not sure, haven’t come across such question with ‘reciprocal of the other’
3) Not really sure
b) Should be k=39
Please tell me whether it’s correct (and also the solution for those that I’m not sure).
Thanks.
dude i really got this hint for you – it will help you i am sure of it:
for the quadratic equation ax^2+bx+c=0,
let roots be alpha and beta.
now “alpha + beta” [i mean sum of roots]
is given by,
-b/a
and product of roots is given by “alpha*beta” = c/a
i think you got my point so now if they ask what is difference of roots,
you find alpha+beta and alpha*beta and then,
do this alpha-beta = {[(alpha + beta)^2-4*(alpha*beta)]}^(1/2)
so you got the three thing alpha + beta
alpha -beta
alpha * beta so you can really find almost everything regarding alhpa and beta.
now if the say the two roots are reciprocal of each other that means roots are alpha and 1/alpha
so product of roots is 1 and you equate c/a in equation to one – thats it you get a solution to what you want.
dont flame me in your mind itself thinking that this guy has not posted the answers instead he just a blah blah blah blah – another math lecture.
a) i.K=3 ii.K=3 iii.K=-27/8
b)k=39
are the answers to your questions.
In
i. A quadratic equations’ (a^2x+bx+c=0) discriminant should be zero to have double roots(equal roots) so (b^2-4ac)=0
ii.Since one root is the others reciprocal their multiplication should equal 1. Which is also equal to c/a.
iii.Inserting 2m and m instead x and solving them simultaneously should give you the answer
b)Again inserting m and m+8 instead of x , and solving the resulting equations simultaneously should do it.
@ :
Thanks for the solutions. I’ve tried no. 2, but my answer is k=-3 by using your way. Since c/a = 1, that means -k/3 = 1, -k = 3 and k = -3. Also, since it said 2 real roots, there’s only 1 by using this way, how about the other?
For no. 3, I tried using alpha and 2 alpha, then solve using sum and product of roots. However, I just can’t get the answer. Perhaps you can help a bit?
For question b, I used sum and product of roots, and the answer is k=39.
Thanks.
@ :
Thanks for the solutions. I've tried no. 2, but my answer is k=-3 by using your way. Since c/a = 1, that means -k/3 = 1, -k = 3 and k = -3. Also, since it said 2 real roots, there's only 1 by using this way, how about the other?
For no. 3, I tried using alpha and 2 alpha, then solve using sum and product of roots. However, I just can't get the answer. Perhaps you can help a bit?
For question b, I used sum and product of roots, and the answer is k=39.
Thanks.
Yes that should be -3 I made a typo. The root is 1 and reciprocal of 1 is also 1. So that should be correct technically.
As for iii, say I’m inserting m and 2m
For x=m; 3m^2+2km-k=0
For x=2m;12m^2+4km-k=0
multiplying the (x=m) part by -4 and summing the equations you get
-4km+3k=0
k is not equal to 0 , so m=0.75
Insert x=0.75in the original equation and solve for k, which is -27/8
for b), expand the given equation, and get it in the general form, ie, ax^2+bx+c=0
Then use difference between roots = sqrt[(-b/a)^2 – 4c/a]
@ :
Thanks for the solutions. I've tried no. 2, but my answer is k=-3 by using your way. Since c/a = 1, that means -k/3 = 1, -k = 3 and k = -3. Also, since it said 2 real roots, there's only 1 by using this way, how about the other?
For no. 3, I tried using alpha and 2 alpha, then solve using sum and product of roots. However, I just can't get the answer. Perhaps you can help a bit?
For question b, I used sum and product of roots, and the answer is k=39.
Thanks.
Yes that should be -3 I made a typo. The root is 1 and reciprocal of 1 is also 1. So that should be correct technically.
As for iii, say I'm inserting m and 2m
For x=m; 3m^2+2km-k=0
For x=2m;12m^2+4km-k=0
multiplying the (x=m) part by -4 and summing the equations you get
-4km+3k=0
k is not equal to 0 , so m=0.75
Insert x=0.75in the original equation and solve for k, which is -27/8
why did you multiply 3m^2+2km-k=0 by -4 to get rid of the 12m^2?
Multiply 3m^2+2km-k=0 by -4, you will get :
-12m^2-8km-4k=0
Solve simultaneously, you will get :
-12m^2-8km-4k=12m^2+4km-k
Rearranging them and you should get :
-4km+3k=0
For the steps after that, I’m not really clear as well :p
why did you multiply 3m^2+2km-k=0 by -4 to get rid of the 12m^2?
To get a value for k. Getting rid of k is a bit harder than that.
For the steps after that, I'm not really clear as well :p
k (-4m+3)=0
To make the equality true either k or -4m+3 should be zero. Since it is said that k is nonzero, -4m+3=0.
So m=0.75 (which is one root of the original equation)
Substituting a root in the original equation should allow you to find k which is -27/8.
wow lmao, ive never seen this sort of thing, whilst doing math at high shcool: and im leaving next wednesday lmao, my school must suck
wow lmao, ive never seen this sort of thing, whilst doing math at high shcool: and im leaving next wednesday lmao, my school must suckHmm…I feel a lot better after seeing this, as I thought that Maths level in my country is very low, and we learn very ‘basic’ things
PS : This is under Additional Maths in my country, but it’s actually basic Maths in most other countries lol