If 1 is a root of the equations $ay^{2}+ay+3=0$ and $y^{2}+y+b=0$, then ab is equal to:
$( A) \ 3$
$( B) -\frac{7}{2}$
$( C) \ 6$
$( D) -3$
Given: Two equations$\ ay^{2} +ay+3=0\ $and $y^{2} +y+b=0$, and their root is 1.
To do: To find the value of ab.
Solution: here given equations $ay^{2} +ay+3=0$ and $y^{2} +y+b=0$,
If 1 is the root of the given equations
Then this root will satisfy both the equations
On substituting $y=1$ in both the equations
$a( 1)^{2} +a( 1) +3=0\ and\ ( 1)^{2} +1+b=0$
$\Rightarrow a+a+3=0\ and\ 1+1+b=0\ $
$\Rightarrow 2a+3=0\ and\ 2+b=0$
$\Rightarrow a=( \frac{-3}{2})\ \ and\ b=-2\ $
$\Rightarrow ab=\left( -\frac{3}{2}\right) \times ( -2) =3$
$\therefore$ Option $( A)$ is correct.
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