Question

If $1$ is a root of the equations $a{y}^2+ay+3=0$ and $y^2+y+b=0$, then $ab$ equals:

• A. $3$
• B. $-\frac{7}{2}$
• C. $6$
• D. $-3$

Answer 1

The correct option is A $3$

$a{y}^2+ay+3=0$

as $1$ is root then

$\Rightarrow a(1{)}^2+a(1)+3=0$

$\Rightarrow a+a=-3$

$\Rightarrow 2a=-3$

$\Rightarrow a=-\frac{3}{2}$

$y^2+y+b=0$

$\Rightarrow (1{)}^2+1+b=0$

$\Rightarrow 2+b=0$

$\Rightarrow b=-2$

$\therefore ab=-\frac{3}{2}\times -2=3$