# Need help with Roots of Quadratic (1 Viewer)

#### wllblsstn

##### New Member
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
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
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
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 $l = 1/ k$ and it is possible to prove the resulting expression is always negative regardless of the value of $k$.

#### wllblsstn

##### New Member
You only need the substitution $l = 1/ k$ and it is possible to prove the resulting expression is always negative regardless of the value of $k$.
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
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 $l = 1/k$.

\begin{aligned} \Delta &= (4m)^2\left(k-\frac 1k\right)^2 - 4(4m^2)\left(k^2 + \frac{1}{k^2}\right) \\ &= 16m^2\left(k^2 - 2 + \frac{1}{k^2} - k^2 - \frac{1}{k^2}\right) \\ &= 16m^2(-2) \\ &= -32m^2\end{aligned}

which is negative for all nonzero real $m$.