Sum of the real values of 'a' for which the equation $(a^{2} - 3 a + 2) x^{2} + (a^{2} - 4 a + 3) x + (a^{2} - 6 a + 5) = 0$ has three distinct roots

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Solution

The correct option is C
1

The quadratic equation has three distinct roots implies it is an identity.

Then $a^{2} - 3 a + 2 = 0, a^{2} - 4 a + 3 = 0$

$a^{2}$ - 6a + 5 = 0. $∴$ a = 1.

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