Question

HOLA friends ☺☺

→ If the roots of the quadratic equation:
$a {x}^{2} + cx + c = 0$ are in the ratio p : q, then show that

$\sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{ \frac{c}{a} } = 0.$

Answer 1

▶ Answer :-

→$\sf \sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{ \frac{c}{a} } = 0 .$

▶ Step-by-step explanation :-

➡ Given :-

→ Given quadratic equation :- $\sf a {x}^{2} + cx + c = 0 .$

→ Roots of the given are in the ratio p : q .

➡ To prove :-

→ $\sf \sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{ \frac{c}{a} } = 0 .$

$\huge \pink{ \mid \underline{ \overline{ \sf Solution :- }} \mid}$

We have,

Roots of the given equation are p and q .

$\begin{lgathered}\sf \because Sum \: of \: roots = \frac{ - B}{A} . \\ \\ \sf \implies p + q = \frac{ - c}{a} ......(1). \\ \\ And \\ \\ \sf \because Product \: of \: roots = \frac{C}{A} . \\ \\ \sf \implies pq = \frac{c}{a} . \\ \\ \big[ \sf both \: side \: multiply \: by \: - . \big] \\ \\ \sf \implies - pq = \frac{ - c}{a} ......(2).\end{lgathered}$

▶ From equation (1) and (2) , we get .

$\implies$ p + q = - pq .

$\implies$ p + q + pq = 0 .

[ Dividing both side by √(pq) . ]

$\begin{lgathered}\sf \implies \frac{p}{ \sqrt{pq} } + \frac{q}{ \sqrt{pq} } + \frac{pq}{ \sqrt{pq} } = \frac{0}{ \sqrt{pq} } . \\ \\ \sf \implies \frac{ \cancel{ \sqrt{p}} \times \sqrt{p} }{ \cancel{\sqrt{p}} \times \sqrt{q} } + \frac{ \cancel{\sqrt{q}} \times \sqrt{q} }{ \sqrt{p} \times \cancel{\sqrt{q} }} + \frac{ \cancel{ \sqrt{pq}} \times \sqrt{pq} }{ \cancel{\sqrt{pq} } } = 0. \\ \\ \sf \implies \sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{pq} = 0. \\ \\ \boxed{ \boxed{\green{\sf \therefore\sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{ \frac{c}{a} } = 0.}}}\end{lgathered}$

[ $\because$ pq = c/a . ]

Hence, it is proved .

Answer 2

$\huge \green{ \boxed{ \orange{ \boxed{ \red{ \boxed{ \blue{\boxed{\boxed{ \mathbb{ \pink{ \underline{ \underline{ \mid \overline{ \star HELLO \star \mid}}}}}}}}}}}}}}$