Find the value of $k$ for which the expression $12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k$ is product of two linear factors.

3

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

2

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

5

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

6

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

Open in App

Solution

The correct option is A $2$
Let $f (x, y) = 12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k$
$= 12 x^{2} - (10 y - 11) x + (2 y^{2} - 5 y + k)$
$△$ must be $\geq 0$ and must be perfect square
$△ = {(10 y - 11)}^{2} - 4 (12) (2 y^{2} - 5 y + k)$
$= 100 y^{2} - 220 y + 121 - 96 y^{2} + 240 y - 48 k$
$= 4 y^{2} + 20 y + (121 - 48 k)$
$= {(2 y)}^{2} + 2 (2 y) (5) + (121 - 48 k)$
For $△$ to be perfect square,
$121 - 48 k = 25$
$48 k = 96$
$k = 2$

Suggest Corrections

Similar questions

For what value of k does the equation $12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k = 0$ represents a pair of straight lines?

12 x^{2} - 10 x y + {2 y}^{2} + 11 x - 5 y + k = 0

12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k = 0.

λ

12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + λ = 0

k

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

Join BYJU'S Learning Program