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

Perpendicular Distance of a Point from a Plane

lf α, β are...

Question

lf $α, β$ are the roots of the equation $x^{2} + p x + q = 0$ , then the equation whose roots $\frac{α}{β}$ and $\frac{β}{α}$ is

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

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

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

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

q x^{2} - (p^{2} - 2 q) x - q = 0

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

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

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

Open in App

Solution

The correct option is D $q x^{2} - (p^{2} - 2 q) x + q = 0$
$x^{2} - (\frac{α}{β} + \frac{β}{α}) x + 1 = 0$
$x^{2} - (\frac{{(α + β)}^{2} - 2 α β}{α β}) x + 1 = 0$
$α + β = - p$
$α β = q$
Now,
$x^{2} - \frac{(p^{2} - 2 q)}{q} x + 1$
$q x^{2} - (p^{2} - 2 q) x + q = 0$

Suggest Corrections

Similar questions

p \neq 0

p^{3} \neq 2 q

p^{3} \neq - q

α

β

α + β = - p

α^{3} + β^{3} = q,

\frac{α}{β}

\frac{β}{α}

(α + \sqrt{β})

(α - \sqrt{β})

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

α, β, p

q

(p^{2} - 4 q) (p^{2} x^{2} + 4 p x) - 16 q = 0

(α, β), (β, γ), (γ, α)

x^{2} - 2 p x + 2 = 0; x^{2} - 2 q x + 3 = 0; x^{2} - 2 r x + 6 = 0

α, β, γ

p + q + r

If $α, β$ be the roots of $x^{2}$ +px+q=0 and $α + h, β + h$ are the roots of $x^{2}$ +rx+s=0 , then

\frac{x + 1}{x^{2} - p x + q} = \frac{A}{x - α} + \frac{B}{x - β}

α

β

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

\frac{A - B}{A + B} = \frac{p + 2}{\sqrt{p^{2} - 4 q}}

Join BYJU'S Learning Program