If both the roots of the equation $x^{2} + m x + 1 = 0$ and $(b - c) x^{2} + (c - a) x + (a - b) = 0$ are common then:

m = - 2

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

m = - 1

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

m = 0

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

m = 1

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

Open in App

Solution

The correct option is B $m = - 2$
If both the roots of the equations $x^{2} + m x + 1 = 0$ and $(b - c) x^{2} + (c - a) x + (a - b) = 0$ are common, then

$- m = \frac{- (c - a)}{(b - c)}$ and $1 = \frac{(a - b)}{(b - c)}$

$⟹ m = \frac{c - a}{b - c}$ and $b - c = a - b$

$⟹ m = \frac{c - a}{a - b}$ and $2 b = a + c$

$⟹ m = \frac{c - a}{a - b}$ and $b = \frac{a + c}{2}$

$⟹ m = \frac{c - a}{(a - \frac{a + c}{2})}$

$⟹ m = \frac{2 (c - a)}{(a - c)}$

$∴ m = - 2$ (Option A)

Suggest Corrections

Similar questions

m

x^{2} + m x + 1 = 0

(b - c) x^{2} + (c - a) x + (a - b) = 0

The equation (m being real), $m x^{2} + 2 x + m = 0$ has two distinct roots if -

m x^{2} + 2 x + m = 0

x^{2} - m x - 4 x + 1 = 0

m

(- \infty, - 6)

(- \infty, - 3)

(- 2, \infty)

(2, \infty)

x^{2} - m x + 1 = 0

Join BYJU'S Learning Program