#### Vishalrox

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 dont 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......

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