If P Q - r x 2- Q r - P x - r P - Q -0 has equal roots then 2- Q - where P - Q - r - R 1-P-1-rB- 1-P-1-rC- P-rD- P r

The correct option is A Product of the roots = 1 ⇒2/Q=1/P+1/r ...

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

Byju's Answer

Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

If P Q - r x ...

Question

$I f P (Q - r) x^{2} + Q (r - P) x + r (P - Q) = 0 h a s e q u a l r o o t s t h e n \frac{2}{Q} = (w h e r e P, Q, r ϵ R)$

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

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

Open in App

Solution

The correct option is A

Product of the roots = 1
$\Rightarrow \frac{2}{Q} = \frac{1}{P} + \frac{1}{r}$

Suggest Corrections

Similar questions

Q. If the roots of the quadratic equation

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

are equal,
Show then

\frac{1}{p} + \frac{1}{r} = \frac{2}{q} = \frac{m}{q}

Find m.

Q. If

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

has equal roots, then

\frac{2}{q} =

If $p > q > 0 a n d p r < - 1 < q r$ , then find the value of

{tan}^{- 1} \frac{p - q}{1 + p q} + {tan}^{- 1} \frac{q - r}{1 + q r} + {tan}^{- 1} \frac{r - p}{1 + r p} .

If every pair of the equations $x^{2} + p x + q r = 0, x^{2} + q x + r p = 0, x^{2} + r x + p q = 0$ have a common root, then the sum of three common roots is

Factorise: $(p^{2} + q^{2} - r^{2})^{2} - 4 p^{2} q^{2}$

Join BYJU'S Learning Program

If P(Q−r)x2+Q(r−P)x+r(P−Q)=0 has equal roots then 2Q=(where P,Q,r ϵ R)

The correct option is A Product of the roots = 1 ⇒2Q=1P+1r

$I f P (Q - r) x^{2} + Q (r - P) x + r (P - Q) = 0 h a s e q u a l r o o t s t h e n \frac{2}{Q} = (w h e r e P, Q, r ϵ R)$

The correct option is A

Product of the roots = 1
$\Rightarrow \frac{2}{Q} = \frac{1}{P} + \frac{1}{r}$