First slide

Theory of expressions

Question

Suppose $a \in R .$ If $3 x^{2} + 2 (a^{2} + 1) x + (a^{2}$ $- 3 a + 2) = 0$ possesses roots of opposite signs, then a lies in the interval:

Easy

A

$(- \infty, - 1)$
B

(–1, 1)
C

(1, 2)
D

(2, 3)

Solution

As the roots are of opposite signs, the product of roots must be negative, that is,
$\begin{aligned} \frac{a^{2} - 3 a + 2}{3} < 0 \Rightarrow (a - 1) (a - 2) < 0 \end{aligned}$
$\Rightarrow a \in (1, 2)$

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