Question

if the ratio of the roots of the equation .
$lx { }^{2} + nx + n = 0 \: \: is \:$
p:q,. prove that:-

√p/q + √q/p + √n/ l = 0


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Answer 1

Proof :

The given equation is

    lx² + nx + n = 0 .....(1)

The ratio of roots of (1) is p : q

Let, r (≠ 0) be the common multiple

Then, the roots be pr and qr

By the relation between roots and coefficients, from (1), we get

    pr + qr = - n/l .....(2)

    pr * qr = n/l .....(3)

From (2) and (3), we get

    pr * qr = - (pr + qr)

⇒ pqr = - (p + q) , since r ≠ 0

⇒ r = - (p + q)/(pq)

⇒ r = - (1/q + 1/p)

From (2), we get

(p + q) {- (1/q + 1/p)} = - n/l

⇒ - p/q - 1 - 1 - q/p = - n/l

⇒ p/q + 2 + q/p - n/l = 0

⇒ (√p/√q)² + 2 (√p/√q) (√q/√p) +  (√p/√q)² - (√n/√l)² = 0

⇒ {(√p/√q) + (√q/√p)}² - (√n/√l)² = 0

⇒ {(√p/√q) + (√q/√p) + (√n/√l)}

    {(√p/√q) + (√q/√p) - (√n/√l)} = 0

Either {(√p/√q) + (√q/√p) + (√n/√l)} = 0

    or,  {(√p/√q) + (√q/√p) - (√n/√l)} = 0

Taking the first term, we get

    √(p/q) + √(q/p) + √(n/l) = 0

Hence, proved.

Answer 2

Step-by-step explanation:

Given Equation is lx² + nx + n = 0

On comparing with ax² + bx + c = 0, we get

a = l, b = n, c = n

Let, roots of the equation are α and β.

(i) Sum of roots:

⇒ α + β = -b/a

⇒ α + β = -n/l

(ii) Product of roots:

⇒ αβ = c/a

⇒ αβ = n/l

(iii) Ratio of roots:

α/β = p/q

LHS:

$Given:\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{n}{l}}$

$=\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}} + \sqrt{\frac{n}{l}}$

$=(\frac{\sqrt{\alpha^2}+\sqrt{\beta^2}}{\sqrt{\alpha} * \sqrt{\beta}}) + \sqrt{\frac{n}{l} }$

$=(\frac{\alpha+\beta}{\sqrt{\alpha\beta}})+\sqrt{\frac{n}{l}}$

$=-\frac{\frac{n}{l}}{\sqrt{\frac{n}{l}}} + \sqrt{\frac{n}{l}}$

$=-\frac{n}{l\sqrt{\frac{n}{l}}}+\sqrt{\frac{n}{l}}$

$=-\frac{n}{\sqrt{l} \sqrt{n}}+\sqrt{\frac{n}{l}}$

$=-\frac{\sqrt{n}}{\sqrt{l}}+\sqrt{\frac{n}{l}}$

$=-\sqrt{\frac{n}{l}}+\sqrt{\frac{n}{l}}$

$=\boxed{0}$

RHS

Hope it helps