Question

find value of n for which (n-3)x2-(5-n)x+1=0 has coincident roots

Answer 1

Answer:

Ihave not understand your question

Answer 2

Answer:

roots are coincident when discriminant = 0

dis = b^2 - 4ac = 0

b = -(5-n), a = (n-3) , c = 1

(-(5-n))^2 = 4(n-3)×1

25-10n+n^2 = 4n -12

n^2 - 14n +37 = 0 n= (-(-14) +- √(14^2-4×1×37))/2

n= (14+2√3)

n = 14-2√3 Ans