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 ?

1 Answer
George C.
Jul 31, 2018

#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.

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