If $(1 + m^{2}) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0$ . Then numbers of value(s) of m for which this equation has no solution.

3

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

\infty

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

2

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

1

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

Open in App

Solution

The correct option is B $\infty$
We know that $D = b^{2} - 4 a c$

$D = 4 (1 + 3 m)^{2} - 4 (1 + m^{2}) (1 + 8 m)$

$= 4 (1 + 9 m^{2} + 6 m - (1 + 8 m + m^{2} + 8 m^{3}))$

$= 4 (8 m^{2} - 2 m - 8 m^{3})$

$= - 8 (4 m^{3} - 4 m^{2} + m)$

$= - 8 m (4 m^{2} - 4 m + 1)$

$= - 8 m (2 m - 1)^{2} < 0$ ......... $∵ (2 m - 1)^{2} i s a l w a y s + v e$

Hence infinitely many values.

Suggest Corrections

Similar questions

m

(1 + m^{2}) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0

(1 + m) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0

m

(1 + m) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0

(m \in R)

(1 + m) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0

m \in R - {- 1})

m

2 : 3

(1 + m) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0

m \in R - {- 1})

^{'} m^{'}

Join BYJU'S Learning Program