Question

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:
_____________________________


$let \: \alpha \: and \: \beta \: be \: the \: roots \: of \: the \: given \: quadratic \: equation$
Therefore,

$\alpha + \beta = \frac{ - c}{a} \: and \: \alpha \beta = \frac{c}{a}$

Also,

$\frac{ \alpha }{ \beta } = \frac{p}{q} ..........(given)$
Therefore,

$\sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{ \frac{c}{a} } = \sqrt{ \frac{ \alpha }{ \beta } } + \sqrt{ \frac{ \beta }{ \alpha } } + \sqrt{ \alpha \beta } \\ \\ = \sqrt{ \frac{ \alpha }{ \beta } } + \sqrt{ \frac{ \beta }{ \alpha } } + \sqrt{ \alpha \beta } \\ \\ = \frac{ \alpha + \beta }{ \sqrt{ \alpha \beta } } + \sqrt{ \alpha \beta } \\ \\ = \frac{ \alpha + \beta + \alpha \beta }{ \sqrt{ \alpha \beta } } \\ \\ = \frac{ - c}{a} + \frac{c}{a} \\ - - - - - - - \\ \: \: \: \: \: \: \:\sqrt{ \alpha \beta } \\ \\ = \frac{0}{ \sqrt{ \alpha \beta } } \\ \\ = 0 \\ \\ therefore \\ \sqrt{ \frac{p}{q} } + \sqrt{ \frac{q}{p} } + \sqrt{ \frac{c}{a} } = 0$

Answer 2

▶ Question :-

→ 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 :-

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


▶ 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 .


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

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

==> p + q = - pq .

==> p + q + pq = 0 .

[ Dividing both side by √(pq) . ]

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


[ °•° pq = c/a . ]


✔✔ Hence, it is proved ✅✅.


$\huge \boxed{ \boxed{ \blue{ \mathscr{THANKS}}}}$