Correct option is (C) ± 1
Let \(\alpha\) be common root of equations
\(x^2+4ax+3=0\) and \(2x^2+3ax-9=0\)
\(\therefore\) \(\alpha^2+4a\alpha+3=0\) _______________(1)
\(2\alpha^2+3a\alpha-9=0\) _______________(2)
Multiply equation (1) by 2, we get
\(2\alpha^2+8a\alpha+6=0\) _______________(3)
Subtract equation (2) from (3), we get
\((2\alpha^2+8a\alpha+6)-(2\alpha^2+3a\alpha-9)\) \(=0-0=0\)
\(\Rightarrow5a\alpha+15=0\)
\(\Rightarrow a\alpha=\frac{-15}5=-3\) _______________(4)
Put \(a\alpha=-3\) into equation (1), we get
\(\alpha^2-12+3=0\)
\(\Rightarrow\) \(\alpha^2-9=0\)
\(\Rightarrow\) \(\alpha^2=9\)
\(\Rightarrow\) \(\alpha=\pm3\)
Then from (4), we get
\(a=\frac{-3}\alpha=\frac{-3}{\pm3}=\mp1\)
\(\therefore\) a = \(\pm\,\,1\)
