Theory of quadratic equations

[Vishalrox]

theory of quadratic equations...

theres a quadratic equation lx^2 + nx + n where its roots are in the ratio p:q .we need to prove that
√(p/q) + √(q/p) + √(n/l) = 0

what i did was..i introduced a proportionality constant k... so pk + qk = -(n/l)
while pq(k^2) = n/l ...solved these two equations and got the value of k as -(p+q)/pq...
and i substituted..i got the roots of the equation in terms of p and q...i got the value of n/l = ((p+q)^2)/pq...coming to the equation which we need to prove...i squared the whole Left Hand Side...expanded it...and substituting whatever i got...i got..4((p+q)^2)/pq...i.e.,i got 4(n/l)...so now we got to prove that 4(n\l) is 0...but if 4(n\l) = 0 then n/l will be 0...which in turn accounts for n = 0...which lands us into a trivial case...all roots and all other coefficients other than leading coefficient (l) to be 0...but i don't think that would be the right way to solve this problem...can anyone help me on this on a different method...?

 

[Millennial]

Did you ever try actually solving the equation for its roots p and q?

 

[Infinitum]

There is something wrong in what you are trying to prove. It should be $-\sqrt{n/l}$, because if all positive, it means all the square root terms should be individually zero, which isn't possible.

Simply reduce the expression you are trying to prove,

$$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} - \sqrt{\frac{n}{l}}$$

Is equal to,

$$\frac{p+q}{\sqrt{pq}} - \sqrt{\frac{n}{l}}$$

Use n/l from the Vieta formula and that should give you your result.

 

[Vishalrox]

The question was from Hall & Knight...i know the problem is wrong...so we have to prove the trivial case...so i did it...everything i did...i did using Vieta's theorem...

 

[Vishalrox]

Did you ever try actually solving the equation for its roots p and q?

I found the values of p and q...as i found the value k...and by the way p and q are not the roots...only pk and qk are the roots which i found...