You visited us 1 times! Enjoying our articles? Unlock Full Access!

Nature of Roots

If 'k' be t...

Question

If $^{'} k^{'}$ be the ratio of the roots of the equation $x^{2} - p x + q = 0$ , the value of $\frac{k}{1 + k^{2}}$ is

\frac{q^{2} - 2 p}{p}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{q}{p^{2} - 2 q}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{p}{q^{2} - p}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{p}{p^{2} - 2 q}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{q}{p^{2} - 2 q}$
We know that equation $a x^{2} + b x + c = 0$
Then sum of roots $= \frac{- b}{a}$
Let the roots of the equation be a and b.
Then, $a + b = p$
$a b = q$
$\frac{a}{b} = k$
Now, $\frac{k}{1 + k^{2}}$
$= \frac{\frac{a}{b}}{1 + (\frac{a}{b})^{2}}$
$= \frac{a b}{a^{2} + b^{2}}$
$= \frac{a b}{(a + b)^{2} - 2 a b}$
$= \frac{q}{p^{2} - 2 q}$

Suggest Corrections

Similar questions

s e c α, c o s e c α

x^{2} - p x + q = 0

p^{2} - 2 q - q^{2} = 0

x^{2} - x s i n α + c o s α = 0

p^{2} = \frac{1 - q^{2}}{1 + q^{2}}

If tan A, tan B are the roots of $x^{2} - P x + Q = 0$ the value of $s i n^{2}$ (A+B)=(where P, Q $ϵ$ R)

t a n A, t a n B

x^{2} - p x + q = 0

s i n^{2} (A + B)

p

q

\frac{p^{2} + p}{q^{2} + q}

p^{t h}

q^{t h}

P = \frac{x}{x + y}, Q = \frac{y}{x + y}

\frac{1}{P - Q} - \frac{2 Q}{P^{2} - Q^{2}}

Join BYJU'S Learning Program