Question

Find the value of a for which the equation has coincident roots

a^2x^2+2(a+1)x+4=0

Answer 1

for any equation to be having coincidentent roots it must follow the condition B^2=4*A*C(symbols have usual meanings)
(2(a+1))^2=4*a^2*4
4(a+1)^2=16a^2
ie,
4(a^2+2a+1)=16a^2
4a^2+8a+4=16a^2
12a^2-8a-4=0
this is another quadratic equ
ation which on solving gives a=-1/3 or a=1
so for a=-1/3 or a=1 the equation has coincidentent roots