What should be added to $6 x^{2} - 3 x y + 4 y^{2}$ to get $2 y^{2} + x y - 4 x^{2}$ ?

- x^{2} - y^{2} + 4 y

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

- 10 x^{2} - y^{2} + 4 x y

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

x^{2} - 2 y^{2} + 4 x

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

- 10 x^{2} - 2 y^{2} + 4 x y

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

Open in App

Solution

The correct option is D $- 10 x^{2} - 2 y^{2} + 4 x y$
Let the required expression be $P$
$P = (2 y^{2} + x y - 4 x^{2}) - (6 x^{2} - 3 x y + 4 y^{2})$
$= - 4 x^{2} + 2 y^{2} + x y - 6 x^{2} + 3 x y - 4 y^{2}$
$= - 10 x^{2} - 2 y^{2} + 4 x y$

Suggest Corrections

Similar questions

6 x^{2} - 3 x y + 4 y^{2}

2 y^{2} + x y - 4 x^{2}

3 x^{2} + 2^{2}, - x^{2} - 4 y^{2} + 4 x y

6 x y + 2 x^{2} - 3 y^{2}

- 5 x^{2} + 2 y^{2} - 3 x y

Equation $12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + 2 = 0$ represent a pair of straight lines. Find the pair of angle bisector.

(x + 2 y)^{2} = x^{2} + 4 y^{2} + 4 x y

x^{2} - 4 x y + 4 y^{2} = 0, x^{2} - 2 y^{2} - 2 = 0

Join BYJU'S Learning Program