Find the set of values of a for which each one of the roots of $x^{2} - 4 a x + 2 a^{2} - 3 a + 5 = 0$ is greater than $2$ .

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Solution

Given,
$x^{2} - 4 a x + 2 a^{2} - 3 a + 5 = 0$
Let $α$ and $β$ are the roots of the above equation
Also given that $α, β > 2$
So, $α + β > (2 + 2)$
$⟹ α + β > 4$
Now, $α + β = - (- 4 a / 1) = 4 a$
So, $4 a > 4 a > 1$
Also,
$α β > 4$
Now, $α β = (2 a^{2} - 3 a + 5) / 1 = 4 a 2 a^{2} - 3 a + 5 > 4$
$⟹ 2 a^{2} - 3 a + 1 > 0$
$⟹ (2 a - 1) (a - 1) > 0$
$a ϵ (\infty, 1 / 2) \cup (1, \infty)$
Hence, $a ϵ (1, \infty)$

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