JEE MAIN - Mathematics (2003 - No. 37)
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
is twice as large as the other is
$$ - {1 \over 3}$$
$$ {2 \over 3}$$
$$ - {2 \over 3}$$
$$ {1 \over 3}$$
Explanation
Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then
$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$
and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$
$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$
$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$
$$ = {2 \over {{a^2} - 5a + 3}}$$
$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$
or $$9{a^2} - 6a + 1$$
$$ = 9{a^2} - 45a + 27$$
or $$39a = 26$$ or $$a = {2 \over 3}$$
$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$
and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$
$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$
$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$
$$ = {2 \over {{a^2} - 5a + 3}}$$
$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$
or $$9{a^2} - 6a + 1$$
$$ = 9{a^2} - 45a + 27$$
or $$39a = 26$$ or $$a = {2 \over 3}$$
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