Question
If one root of the equation $$\left( {l - m} \right){x^2} + lx + 1 = 0$$ is double the other and $$l$$ is real, then what is the greatest value of $$m ?$$
A.
$$ - \frac{9}{8}$$
B.
$$ \frac{9}{8}$$
C.
$$ - \frac{8}{9}$$
D.
$$ \frac{8}{9}$$
Answer :
$$ \frac{9}{8}$$
Solution :
Given equation is
$$\left( {l - m} \right){x^2} + lx + 1 = 0$$
Roots are $$\alpha ,\beta .$$
∵ One root is double the other.
$$\beta = 2\alpha $$
Sum of roots $$ = \alpha + \beta .$$
$$\eqalign{
& 3\alpha = \frac{{ - l}}{{l - m}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \left( {2\alpha } \right) = \frac{1}{{\left( {l - m} \right)}} \cr
& \Rightarrow {\alpha ^2} = \frac{{{l^2}}}{{9{{\left( {l - m} \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{\alpha ^2} = \frac{1}{{l - m}} \cr
& \Rightarrow 2\frac{{{l^2}}}{{9{{\left( {l - m} \right)}^2}}} = \frac{1}{{\left( {l - m} \right)}} \cr
& \Rightarrow \frac{{2{l^2}}}{{9\left( {l - m} \right)}} = 1 \cr
& \Rightarrow 2{l^2} = 9\left( {l - m} \right) \cr
& \Rightarrow 2{l^2} - 9l + 9m = 0 \cr} $$
For $$l$$ to be real discriminant should be $${b^2} - 4ac \geqslant 0$$
$$\eqalign{
& 81 - 4 \times 2 \times 9m \geqslant 0 \cr
& m \leqslant \frac{9}{8}. \cr} $$