Question

IF the equation `(a^2-4a+3)x^2+(a-1)x+(a^2-1)=0` has infinite roots, then the value of `a` is ?

Answer 1

`a=1`

Explanation:

`0 = (a^2-4a+3)x^2+(a-1)x+(a^2-1)`

`color(white)(0) = (a-1)(a-3)x^2+(a-1)x+(a-1)(a+1)`

`color(white)(0) = (a-1)((a-3)x^2+x+(a+1))`

This holds for any value of `x` if `a=1`.

On the other hand, if `a != 1` then we are looking at solutions of:

`(a-3)x^2+x+(a+1) = 0`

If `a=3` then this has exactly one solution `x = -a-1`

If `a != 3` then this is a quadratic equation, so has exactly two solutions counting multiplicity.