MedVision ad

Need help with Roots of Quadratic (1 Viewer)

wllblsstn

New Member
Joined
Aug 11, 2018
Messages
4
Gender
Male
HSC
2019
5. a) Prove that the equation (k^2+l^2)x^2-4m(k-l)x+4m^2, where m does not equal to 0, and k and l are reciprocal, has no real roots.
b) Given that (k^2 -l^2 )x^2+2m(k+l)x+m^2=0 has a double root and that k, l, m does not equal to 0, prove that k=-l
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
5. a) Prove that the equation (k^2+l^2)x^2-4m(k-l)x+4m^2, where m does not equal to 0, and k and l are reciprocal, has no real roots.
b) Given that (k^2 -l^2 )x^2+2m(k+l)x+m^2=0 has a double root and that k, l, m does not equal to 0, prove that k=-l
Have you tried using the discriminant? What do you know about what the discriminant looks like based on whether the quadratic has no real roots, or has a double root?
 

wllblsstn

New Member
Joined
Aug 11, 2018
Messages
4
Gender
Male
HSC
2019
Have you tried using the discriminant? What do you know about what the discriminant looks like based on whether the quadratic has no real roots, or has a double root?
Hmm yes I do realise that b^2-4ac needs to be less than 0 in order for it to have no real roots, and equal to zero in order for it to have double roots, but I believe that step comes after finding 'k' and 'l' as it does mention that they are reciprocal, thus k=1/land yeah.
 

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
Hmm yes I do realise that b^2-4ac needs to be less than 0 in order for it to have no real roots, and equal to zero in order for it to have double roots, but I believe that step comes after finding 'k' and 'l' as it does mention that they are reciprocal, thus k=1/land yeah.
You only need the substitution and it is possible to prove the resulting expression is always negative regardless of the value of .
 

wllblsstn

New Member
Joined
Aug 11, 2018
Messages
4
Gender
Male
HSC
2019
You only need the substitution and it is possible to prove the resulting expression is always negative regardless of the value of .
aah how did I not think of this. Thank you, but are you still able to provide more help, since I still don't see how variables 'm' and 'l' could prove that a discriminant is less than 0. Thanks
 

fan96

617 pages
Joined
May 25, 2017
Messages
543
Location
NSW
Gender
Male
HSC
2018
Uni Grad
2024
aah how did I not think of this. Thank you, but are you still able to provide more help, since I still don't see how variables 'm' and 'l' could prove that a discriminant is less than 0. Thanks
Let .



which is negative for all nonzero real .
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top