The equation (x^2)-4x+K=0 & (x^2)+Kx-4=0 where k is a real number, have exactly one common root. What is the value of K.

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Solution

x^2 - 4x +k = 0----------------(1)
x^2 + kx - 4 =0----------------(2)
as equations 1 and 2 has common root
Let 'a' be the common roo for the equation, therefore equation 1 and 2 becomes
a^2 - 4a +k = 0-----------------(3)
a^2 +ka - 4 = 0 ----------------(4)
subtracting equation 4 from 3:
a^2 - 4a +k - a^2 - ka +4 =0
-(4 +k)a +k + 4 =0
a = 1
substittuting the value of 'a' in equation 3, we get
1 - 4 +k = 0
k = 3

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