Question

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

(a) 3
(b) −7/2
(c) 6
(d) −3

Answer 1

Since, y = 1 is a root of the equations $a{y}^2+ay+3=0$.
So, it satisfies the given equation.

$$\begin{gathered} \therefore a{\left(1\right)}^2+a\left(1\right)+3=0 \\ \Rightarrow 2a+3=0 \\ \Rightarrow a=-\frac{3}{2}...\left(1\right) \end{gathered}$$

Since, y = 1 is a root of the equations $y^2+y+b=0$.
So, it satisfies the given equation.

$$\begin{gathered} \therefore {\left(1\right)}^2+\left(1\right)+b=0 \\ \Rightarrow 2+b=0 \\ \Rightarrow b=-2...\left(2\right) \end{gathered}$$

From (1) and (2),
$$\begin{gathered} ab=\left(-\frac{3}{2}\right)\left(-2\right) \\ =3 \end{gathered}$$

Thus, ab is equal to 3.

Hence, the correct option is (a).