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1

AIEEE 2002

Product of real roots of equation $${t^2}{x^2} + \left| x \right| + 9 = 0$$
A
is always positive
B
is always negative
C
does not exist
D
none of these

Explanation

Product of real roots $$= {9 \over {{t^2}}} > 0,\forall \,t\, \in R$$

$$\therefore$$ Product of real roots is always positive.
2

AIEEE 2002

If $$\alpha \ne \beta$$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta$$ and $$\beta /\alpha \,\,$$ as its roots is
A
$$3{x^2} - 19x + 3 = 0$$
B
$$3{x^2} + 19x - 3 = 0$$
C
$$3{x^2} - 19x - 3 = 0$$
D
$${x^2} - 5x + 3 = 0$$

Explanation

We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$

$$\Rightarrow \alpha \,\,\& \,\,\beta$$ are roots of

equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$

$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$

Thus, the equation having $${\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }$$ as its roots is

$${x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0$$

$$\Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0$$

or $$3{x^2} - 19x + 3 = 0$$

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