Find the set of values of 'a' for which zeroes of the quadratic polynomial
$(a^{2} + a + 1) x^{2} + (a - 1) x + a^{2}$ are located on either side of 3.

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Solution

Given quadratic equation is $(a^{2} + a + 1) x^{2} + (a - 1) x + a^{2} = 0$
$a^{2} + a + 1 = a^{2} + 2 (a) (\frac{1}{2}) + \frac{1}{4} + \frac{3}{4} = (a + \frac{1}{2})^{2} + \frac{3}{4} > o ⟹$ it is a upward parabola
also given that zeroes of the quadratic polynomial lies on either sides of $3 ⟹ f (3) < 0$
$(a^{2} + a + 1) (3)^{2} + (a - 1) (3) + a^{2} < 0$
$10 a^{2} + 12 a + 6 < 0$
$5 a^{2} + 6 a + 3 < 0$
above expression always above X-axis that is $> 0$ because its discriminant is less than zero so this never becomes less than zero
$∴ a$ has no value satisfying the above condition
$⟹ a \in ϕ$

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