If $2(a+b+c)=t^2+u^2+v^2$ and the roots of $x^2+tx-a=0$ are $u,v$ and the roots of equation $x^2+ux-b=0$ are $v,t$ then I need to show that the equation whose roots are $t,u$ is $x^2+vx-c=0$ I am not able to guess any approach so sorry. Please do not solve the problem . I want an idea for the question. I have tried expressing $t,u$ in terms of $a,b,c$ but not of much help. Please do upvote the question if you find it helpful.
Asked
Active
Viewed 24 times
1
-
Please give an idea regarding the question – Akash Roy Jul 03 '18 at 09:44
-
Please help me Seems no one paying attention to this question – Akash Roy Jul 03 '18 at 10:33
-
$x^2 + t x - a = (x-u)(x-v)$ leads to interesting relations between $t$, $u$, $v$ and $a$. – random Jul 03 '18 at 11:02
-
I got the answer please dont answer the question now – Akash Roy Jul 03 '18 at 11:21
-
Thanks Random for help – Akash Roy Jul 03 '18 at 11:21
1 Answers
0
Here is a hint:
The quadratic expression whose roots are $t$ and $u$ is $(x-t)(x-u)=x^2-(t+u)x+tu,$ so what you want to show is that $-(t+u)=-t-u=v$ and $-c=ut.$
Since you know the roots of the two previous quadratics, it means that $x^2+ux-b=(x-v)(x-t)$ and $x^2+tx-a=(x-u)(x-v).$ After expanding and comparing coefficients, you will get four relations, two of which are equivalent and prove the first relationship.
These three relations, when combined with $2(a+b+c)=t^2+u^2+v^2$ then yield the second equation to be proved.
Allawonder
- 13,327