some quadratic problem (1 Viewer)

[wolf7]

sry for the disturbance, but i have some trouble with some question. can someone help me with some questions if it doesnt trouble a

find the values of k for which 5+4x-x^2=k has equal roots

and also

if (2k+3)^2 - 4kx+4 = 0 has equal roots find k

also can ya plz show the working out

 

[Riviet]

-x²+4x+5-k=0
let the equal root be p for both questions
then 2p=-b/a (using sum of roots rule)
2p=4
.: p=2
now p²=c/a (using product of roots rule)
=k-5
p²=k-5
sub in p=2
=>4=k-5
.: k=9

(2k+3)²-4kx+4=0
4k²+12k+9-4kx+4=0
4k²+(12-4x)k+13=0 (1)
2p=(-12+4x)/4 (using sum of roots rule)
2p=x-3
p²=13/4 (using product of roots rule)
.: p=sqrt13/2
.:x=3+sqrt13
sub in (1)
4k²+(12-4(3+sqrt13))k+13=0
expanding and expressing in quadratic form you get:
4k²-(sqrt208)k+13=0
k=[sqrt208+/-(208-208)]/8
k=sqrt13/2
There you go.

 

[Slidey]

Don't worry. If you ever have any other problems feel free to ask them here.

Rearrange it: x^2-4x+k-5=0

equal roots: discriminant =0. Discriminant = b^2-4ac = 16-4(k-5)=36-4k=0, so k=9.

Next:

Can you please rewrite this? This is not a quadratic, as there is no x^2. Unless it is in terms of k?

 

[香港!]

^ Yeap!
This Captain(Slide Rule) is very powerful mathematician!

 

[DeanM]

do you guys find it really hard to follow maths when looking at it on a screen ? i cant follow it.. i need to have pen and paper to do it ...

 

[Slidey]

Interesting you should say that. I sometimes find it hard to follow the working of others when it is on screen, though that is true on paper, too (though to a lesser extent). My own working I can follow flawlessly on screen.

It's an acquired skill - I used to find it hard.