Very-Short-Answer Questions

It is given that $y=1$ is a root of the equation $a y^{2}+a y+3=0$.

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

$\Rightarrow a+a+3=0$

$\Rightarrow 2 a+3=0$

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

Also, $y=1$ is a root of the equation $y^{2}+y+b=0$.

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

$\Rightarrow 1+1+b=0$

$\Rightarrow b+2=0$

$\Rightarrow b=-2$

$\therefore a b=\left(-\frac{3}{2}\right) \times(-2)=3$

Hence, the value of ab is 3.