Question

If $x$ and $y$ is real and $4y^2+4xy+x+6=0$ , the value of x can be

• A. 0
• B. -2
• C. 1
• D. None of the above

Answer 1

Answer: (B)

-2

$4y^2+4xy+x+6=0,$, x and y are real

$\therefore 16x^2-4\left(4\right)(x+6)\ge 0$ ($\because y$ is real)

$x^2-x-6\ge 0$ ($\therefore △$ must be $\ge 0$)

$(x-3)(x+2)\ge 0$

$\therefore xϵ(-\infty ,-2]\bigcup [3,\infty )$