Question
The value of $$'a'$$ for which one root of the quadratic equation $$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$ is twice as large as the other is
A.
$$ - \frac{1}{3}$$
B.
$$ \frac{2}{3}$$
C.
$$ - \frac{2}{3}$$
D.
$$ \frac{1}{3}$$
Answer :
$$ \frac{2}{3}$$
Solution :
Let the roots of given equation be $$\alpha $$ and $$2\alpha $$ then
$$\eqalign{
& \alpha + 2\alpha = 3\alpha = \frac{{1 - 3a}}{{{a^2} - 5a + 3}} \cr
& \& \,\,\alpha .2\alpha = 2{\alpha ^2} = \frac{2}{{{a^2} - 5a + 3}} \cr
& \Rightarrow \,\,\alpha = \frac{{1 - 3a}}{{3\left( {{a^2} - 5a + 3} \right)}} \cr
& \therefore \,\,2\left[ {\frac{1}{9}\frac{{{{\left( {1 - 3a} \right)}^2}}}{{{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right] = \frac{2}{{{a^2} - 5a + 3}} \cr
& \frac{{{{\left( {1 - 3a} \right)}^2}}}{{\left( {{a^2} - 5a + 3} \right)}} = 9\,\,{\text{or}}\,\,9{a^2} - 6a + 1 = 9{a^2} - 45a + 27 \cr
& {\text{or}}\,\,39a = 26\,\,\,{\text{or }}a = \frac{2}{3} \cr} $$