Question

Q2. If the ratio of roots of the equation lx² + nx + n = 0 is p:q, then prove that √p/q + √q/p + √n/l = 0.​

Answer 1

Kindly refer the attachment for answer!

Answer 2

Topic :-

Quadratic Equation

Given :-

The ratio of roots of the equation lx² + nx + n = 0 is p : q.

To Prove :-

$\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}=0$

Concept Used :-

$If\:\alpha\:and\:\beta\:are\:roots\:of\:equation,$

$ax^2+bx+c=0\;then$

$\alpha+\beta=\dfrac{-b}{a}$

$\alpha\beta=\dfrac{c}{a}$

Proof :-

Given equation : lx² + nx + n = 0

Let as assume that,

$\alpha = \lambda p\;and$

$\beta=\lambda q\:are\:the\:roots\:of\:given\:equation.$

It satisfies the condition that ratio of roots of the equation should be p : q.

$\dfrac{\alpha}{\beta}=\dfrac{\lambda p}{\lambda q}=\dfrac{p}{q}$

Now,

Compare the given equation with ax² + bx + c = 0.

On comparing we get,

a = l

b = n

c = n

Calculating Sum and Product of roots,

$\alpha+\beta=\lambda p+\lambda q=\lambda(p+q)=\dfrac{-b}{a}=\dfrac{-n}{l}$

$\therefore\:p+q=\dfrac{-n}{\lambda l}$

$\alpha\beta=\lambda p\cdot \lambda q=\lambda^2pq=\dfrac{c}{a}=\dfrac{n}{l}$

$\therefore\:pq=\dfrac{n}{\lambda^2l}$

Simplifying LHS,

$\sqrt{\dfrac{p}{q}}+\sqrt{\dfrac{q}{p}}+\sqrt{\dfrac{n}{l}}$

Taking LCM,

$\dfrac{\sqrt{p}\cdot\sqrt{p}+\sqrt{q}\cdot\sqrt{q}}{\sqrt{pq}}+\sqrt{\dfrac{n}{l}}$

$\dfrac{p+q}{\sqrt{pq}}+\sqrt{\dfrac{n}{l}}$

Substitute the values of (p + q) and pq,

$\dfrac{\dfrac{-n}{\lambda l}}{\sqrt{\dfrac{n}{\lambda^2l}}}+\sqrt{\dfrac{n}{l}}$

$\dfrac{-n}{\lambda l}\sqrt{\dfrac{\lambda^2l}{n}}+\sqrt{\dfrac{n}{l}}$

$\dfrac{-n\lambda}{\lambda l}\sqrt{\dfrac{l}{n}}+\sqrt{\dfrac{n}{l}}$

$\dfrac{-n\not{\lambda}}{\not{\lambda}l}\sqrt{\dfrac{l}{n}}+\sqrt{\dfrac{n}{l}}$

$\dfrac{-\sqrt{n}\cdot\sqrt{n}\cdot\sqrt{l}}{\sqrt{l}\cdot\sqrt{l}\cdot\sqrt{n}}+\sqrt{\dfrac{n}{l}}$

Cancelling few terms, we get,

$-\sqrt{\dfrac{n}{l}}+\sqrt{\dfrac{n}{l}}$

0

RHS,

0

We observe that LHS = RHS.

Hence, Proved !!