Question

If x = 1 is a common root of $a{x}^2+ax+3=0\mathrm{and}{x}^2+x+b=0,\mathrm{then}ab=?$
(a) 3
(b) −3
(c) 4
(d) 6

Answer 1

(a) 3

$$\begin{gathered} \text{It is given that}\mathit{\text{x}}\text{=}1\text{is a common root of (}a{x}^2+ax+3=0)\text{and (}{x}^2+x+b=0). \\ \therefore a{\left(1\right)}^2+a\left(1\right)+3=0\text{and}{\left(1\right)}^2+1+b=0 \\ \Rightarrow a=\frac{-3}{2}\text{and}b=-2 \\ \therefore ab=\frac{-3}{2}\times (-2)=3 \end{gathered}$$