The equation $4 x^{2} + m x y - 3 y^{2} = 0$ represents a pair of real and distinct lines if

m \in R

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

m \in (3, 4)

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

m \in (- 3, 4)

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

m > 4

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

Open in App

Solution

The correct option is A $m \in R$
Using above formula
In given equation $a = 4, h = \frac{m}{2}, g = 0, b = - 3, f = 0, c = 0$
$Δ$ is always zero no matter what value of $m .$
So $m \in R$
Hence, option 'A' is correct.

Suggest Corrections

Similar questions

4 x^{2} + m x y - 3 y^{2} = 0

m \in R .

x^{2} - 12 k x y + y^{2} = 0

2,

k

\frac{1}{30}

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

[1, 5]

m

x^{2} - (m - 3) x + m = 0 (m \in R)

m

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

2 x^{2} + 5 x y + 3 y^{2} + 6 x + 7 y + 4 = 0

{tan}^{- 1}

Join BYJU'S Learning Program