Question

If ordered pair $(\alpha ,\beta )$ where $\alpha ,\beta \in I$ satisfy the equation $2x^2-3xy-2y^2=7$,then value of $\alpha +\beta$ can be

• A. $5$
• B. $4$
• C. $-4$
• D. $3$

Answer 1

The correct option is B $4$

given equation is

$2x^2-3xy-2y^2=7$

$\Rightarrow 2x^2-4xy+xy-2y^2-7$

$\Rightarrow (x-2y)(2x+y)=7$

$\Rightarrow (x-2y)(2x+y)=7\times 1$

$\Rightarrow$ different possible condition by comparing both side can be

$x-2y=7---\left(1\right)$ and

$2x+y=1---\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$

$-2y=7----\left(1\right)\times 12x+y=1---\left(2\right)\times 2\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_x-2y=74x+2y=2+++\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_5x=9x=\frac{9}{5}$

Similarly,

$y=\frac{-13}{5}$

$x-2y=1----\left(3\right)\times 12x+y=7---\left(4\right)\times 2\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_x-2y=14x+2y=14+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_5x=15x=3$

Similarly

$y=1$

So possible value of $(\alpha ,\beta )$, i.e, $(x,y)$ is

$(\frac{9}{5},\frac{-13}{5})$ and $(3,1)$

$\therefore \alpha +\beta =4$ or $\frac{4}{5}$

Answer $\left(B\right)4$