Question

The value of $(x+y{)}^2$ , $x\ne 0$ and $y\ne 0$ would be $0$ when

• A. $x=y=0$
• B. $x=-y$
• C. $x=2y$
• D. $x=-2y$

Answer 1

The correct option is B $x=-y$

$(x+y{)}^2$ would be equal to $0$ in two cases.

$\underset{–}{Case1:}$

When both $x$ and $y$ are equal to $0$.
$(x+y{)}^2=(0+0{)}^2$
$=0^2+0^2+2\times 0\times 0$
$=0+0+0$
$=0$

But this cannot be the case as in the question, it is given that $x\ne 0$ and $y\ne 0$. So, case $1$ is not correct.


$\underset{–}{Case2:}$

$(x+y{)}^2$ would be equal to zero when the term inside the bracket, i.e., $(x+y)$ becomes equal to $0$.

$(x+y{)}^2=0$
$⟹(x+y)=0$

Subtracting $y$ from both the sides of the above equation:
$(x+y)-y=0-y$
$x+(y-y)=-y$
$x+0=-y$
$x=-y$

Case $2$ is accepted as $x\ne 0$ and $y\ne 0$ but $(x+y{)}^2=0$.

$(x+y{)}^2$ would be equal to $0$ when $x=-y$. Therefore, option (b.) is the correct answer.